Multiple Integrals And Applications

Vector Integration Application Exercise -5

 

1. State and prove Gauss’s divergence theorem.

Solution:

Gauss’s divergence theorem: If F is a differentiable vector point function and S is a closed surface enclosing a region V, then \(\int_S\)F. N dS \(\int_V\)  div F dV, where N is the outward drawn unit normal vector to S.

Proof: Let S be a closed surface. Let us choose the coordinate axes so that any line parallel to the axes meets the surface in almost two points. Let R be the projection of S on xy-plane.  Let S₁ and S₂ be the lower and upper parts of S.

 

Let z=f(x,y) and z = g(x, y) be the equations of S₁ and S₂ which can be put in the form f(x,y)≤z≤ g(x, y)
Let F = F₁i+F₂ j+F₃k where F_1, F₂, and F₃ are scalar point functions.

∴ \(\int_v \text{div} \mathbf{F} d V=\int_V \nabla \cdot \mathbf{F} d V=\iint_V\left(\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}\right) d V=\int_V \frac{\partial F_1}{\partial x} d V+\int_V \frac{\partial F_2}{\partial y}+\int_V \frac{\partial F_3}{\partial z} d V .\)

Now \(\int_V \frac{\partial F_3}{\partial z} d V=\iiint_V \frac{\partial F_3}{\partial z} d x d y d z=\iiint_R\left[F_3(x, y, z)\right]_f^g d x d y\)

⇒ \(\iint_R\left[F_3(x, y, g)-F_3(x, y, f)\right] d x d y .\)

For the upper part S₂, dx dy = dS cos γ = N . k dS, since the normal to S₂ makes an acute angle y with k.

∴ \(\iint_R F_3(x, y, g) d x d y=\int_{s_2} F_3 \mathbf{N} \cdot \mathbf{k} d S\)

For the lower part S₁ dx dy =- cos γ dS =− N . k dS, since the normal to S₁ makes an obtuse angle y with k.

∴ \(\iint_R F_3(x, y, f) d x d y=-\int_{s_1} F_3 \mathbf{N} . \mathbf{k} d S\)

⇒ \(\int_V \frac{\partial F_3}{\partial z} d V=\int_{s_2} F_3 \mathbf{N} . \mathbf{k} d S+\int_{s_1} F_3 \mathbf{N} \cdot \mathbf{k} d S=\int_S F_3 \mathbf{k} . \mathbf{N} d S\)

Similarly \(\int_V \frac{\partial F_2}{\partial y} d V=\int_S F_2 \mathbf{j} \cdot \mathbf{N} d S \text { and } \int_V \frac{\partial F_1}{\partial x} d V=\int_S F_1 \mathbf{i} \cdot \mathbf{N} d S\)

∴ \(\int_V\left(\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}\right) d V=\int_S F_1 \mathbf{i} \cdot \mathbf{N} d S+\int_S F_2 \mathbf{j} \cdot \mathbf{N} d S+\int_S F_3 \mathbf{k} \cdot \mathbf{N} d S\)

⇒ \(\int_V \nabla \cdot \mathbf{F} d V=\int_S\left(F_1 \mathbf{i}+F_2 \mathbf{j}+F_3 \mathbf{k}\right) \cdot \mathbf{N} d S \Rightarrow \int_V \nabla \cdot \mathbf{F} d V=\int_S \mathbf{F} \cdot \mathbf{N} d S\)

2. If F is a continuously differentiable vector point function and S is a closed surface enclosing a region V then prove that \(\int_S\)N×F dS=\(\int_V\) ∇×F dV.

Solution: Let f=a×F where a is any constant vector.

By Gauss’s divergence theorem \(\int_S\)f.N dS= \(\int_V\)∇.f dV.

⇒ \(\int_S(\mathbf{a} \times \mathbf{F}) \cdot \mathbf{N} d S=\int_V \nabla \cdot(\mathbf{a} \times \mathbf{F}) d V \Rightarrow \int_S \mathbf{a} \cdot(\mathbf{F} \times \mathbf{N}) d S=-\int_V \nabla \cdot(\mathbf{F} \times \mathbf{a}) d v\)

⇒ \(-\int_S \mathbf{a} \cdot(\mathbf{N} \times \mathbf{F}) d S=-\int_V(\nabla \times \mathbf{F}) \cdot \mathbf{a} d V \Rightarrow \mathbf{a} \cdot \int_S(\mathbf{N} \times \mathbf{F}) d S=\mathbf{a} \cdot \int_V \nabla \times \mathbf{F} d V\)

⇒ \(\int_S(\mathbf{N} \times \mathbf{F}) d S=\int_V \nabla \times \mathbf{F} d V\) [∵ a is any constant vector]

3. If φ  is a continuously differentiable scalar point function and S is a closed surface enclosing a region V then prove that \(\int_S\)N φ dS= \(\int_V\) ∇φ dV.

Solution: Let f=a φ where a is any constant vector.

By Gauss’s divergence theorem, \(\int_S \mathbf{f} \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{f} d V\)

⇒ \(\int_S a \varphi \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{a} \varphi d V \Rightarrow \int_S \mathbf{a} \cdot \varphi \mathbf{N} d S=\int_V \nabla \varphi \cdot \mathbf{a} d V\)

⇒ \(\mathbf{a} \cdot \int_S \varphi \mathbf{N} d S=\mathbf{a} \cdot \int_V \nabla \varphi d V \Rightarrow \int_S \varphi \mathbf{N} d S=\int_V \nabla \varphi d V\) [∵ a is any constant vector]

4. Apply Gauss’s theorem to prove that  \(\int_S\)r. N dS = 3 V.

Solution: \(\int_S\)r.NdS=\(\int_V\)div r dV=\(\int_V\)∇.r dV=\(\int_V\)3 dV=3V, where V is the volume of the region bounded by the closed surface S.

5. Prove that for any closed surface S, \(\iint_S\)N dS = 0.

Solution:\(\iint_S\)N dS=\(\iint_S\)N 1 dS=\(\int_V\)(∇ 1)dV=\(\int_V\)0 dV=0

6. For any closed surface S, prove that \(\iint_S\)Curl F . N dS = 0.

Solution: By Gauss’s  divergence theorem,

\(\iint_S\)F.N dS=\(\iint_S\)(∇×F).N dS=\(\int_V\)div(∇×F) dV=\(\int_V\)0 dV=0.

7. If S is any closed surface enclosing a volume V and F = xi + 2yj+ 3zk, prove that \(\iint_S\)F.N dS=6v.

Solution:  By Gauss’s divergence theorem,

∴ \(\iint_S \mathbf{F} \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{F} d V=\int_V\left[\frac{\partial}{\partial x}\{x\}+\frac{\partial}{\partial y}\{2 y\}+\frac{\partial}{\partial z}\{3 z\}\right] d V=\int_V(1+2+3) d V=6 V\)

8. If F = xi- 2yi + 3zk and S is a closed surface enclosing a volume V, show that \(\int_S\)F.N dS=2v.

Solution: 

By Gauss’s divergence theorem,

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\int_V \text{div} \mathbf{F} d V=\int_V(\nabla \cdot \mathbf{F}) d V=\int_V\left[\frac{\partial}{\partial x}\{x\}+\frac{\partial}{\partial y}\{-2 y\}+\frac{\partial}{\partial z}\{3 z\}\right] d V\)

⇒ \(\int_V(1-2+3) d V=2 \int_V d V=2 V\)

9. Computed \(\oint_S\)(ax²+by²+cz²) dS over the sphere x² +y² + z² = 1.

Solution: Let φ =x²+y²+z²-1

Normal vector to the surface, ∇φ =\(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)

=2xi+2yj+2zk

Unit normal vector, \(\mathbf{N}=\frac{2 x \mathbf{i}+2 y \mathbf{j}+2 z \mathbf{k}}{\sqrt{4 x^2+4 y^2+4 z^2}}=\frac{2 x \mathbf{i}+2 y \mathbf{j}+2 z \mathbf{k}}{2 \sqrt{x^2+y^2+z^2}}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

⇒ \(\mathbf{F} \cdot \mathbf{N}=a x^2+b y^2+c z^2 \Rightarrow \mathbf{F} \cdot(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})=a x^2+b y^2+c z^2 \Rightarrow \mathbf{F}=a x \mathbf{i}+b y \mathbf{j}+c z \mathbf{k}\)

⇒ \(\nabla \cdot \mathbf{F}=\frac{\partial}{\partial x}(a x)+\frac{\partial}{\partial y}(b y)+\frac{\partial}{\partial z}(c z)=a+b+c\)

Volume of the sphere \(x^2+y^2+z^2=1 \text { is } 4 \pi / 3\)

By Gauss’s divergence theorem,

⇒ \(\oint_S\left(a x^2+b y^2+c z^2\right) d S=\int_S \mathbf{F} \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{F} d V=\int_V(a+b+c) d V=(a+b+c) V\)

⇒ \((a+b+c) \frac{4}{3} \pi=\frac{4 \pi}{3}(a+b+c)\)

10. If F = axi + byj+ czk and a, b, c are constants, show that ∫F.N dS =\(\frac{4}{3} \pi\) (a + b + c) where S is the surface of the  unit sphere.

Solution:

∴ \(\nabla \cdot \mathbf{F}=\frac{\partial}{\partial x}\{a x\}+\frac{\partial}{\partial y}\{b y\}+\frac{\partial}{\partial z}\{c z\}=a+b+c\)

Volume of the sphere, \(V=\frac{4}{3} \pi(1)^3=\frac{4}{3} \pi\)

By Gauss’s divergence theorem, \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{F} d V=\int_V(a+b+c) d V\)

⇒ \((a+b+c) \int_V d V=(a+b+c) V=\frac{4}{3} \pi(a+b+c)\)

11. Show that \(\iint_S\)(ax dy dz + by dz dx + cz dx dy) = \(\frac{4}{3} \pi\)(a + b + c), where S is the surface of the sphere x² +y² + z²=1. where S is the surface of the sphere.

Solution: 

By Gauss’s divergence theorem,

⇒ \(\iint_S(a x d y d z+b y d z d x+c z d x d y)=\int_s[a x \mathbf{N} \cdot \mathbf{i} d S+\text { by N.j } d S+c z \mathbf{N} \cdot \mathbf{k} d S]\)

⇒ \(\int_s(a x \mathbf{i}+b y \mathbf{j}+c z \mathbf{k}) \cdot \mathbf{N} d S=\int_V\left[\frac{\partial}{\partial x}(a x)+\frac{\partial}{\partial y}(b y)+\frac{\partial}{\partial z}(c z)\right] d V=(a+b+c) \int_V d V\)

⇒ \((a+b+c) V \text { where } V \text { is the volume of } x^2+y^2+z^2 = 1\)

⇒ \((a+b+c) \frac{4}{3} \pi=\frac{4 \pi}{3}(a+b+c) \text {. }\)

12. Apply divergence theorem to evaluate \(\iint_S\) x dy dz+y dz dx + z dx dy where S is the surface x² +y² + z² = 1

Solution: By Gauss’s divergence Theorem,

\(\iint_S\)(x dy dz+y dz dy+z dx dy)= \(\int_S\)[x N.idS+y N.jdS+z N.k dS]

⇒ \(\int_S(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) \cdot \mathbf{N} d S=\int_V\left[\frac{\partial}{\partial x}(x)+\frac{\partial}{\partial y}(y)+\frac{\partial}{\partial z}(z)\right] d V=(1+1+1) \int_V d V\)

⇒ \(3 V \text { where } V \text { is the volume of } x^2+y^2+z^2=1=3 \times \frac{4}{3} \pi=4 \pi \text {. }\)

13. Apply Gauss’s divergence theorem to compute the double integral\(\iint_S\) (x+z) dy dz+(y+z) dz dx+(x+y) dxdy where S is the surface of the sphere x² +y² + z² = 4.

Solution: 

By Gauss’s divergence theorem, \(\iint_S(x+z) d y d z+(y+z) d z d x+(x+y) d x d y\)

⇒ \(\int_S(x+z) \mathbf{N} \cdot \mathbf{i} d S+(y+z) \mathbf{N} \cdot \mathbf{j} d S+(x+y) \mathbf{N} \cdot \mathbf{k} d S\)

⇒ \(\int_S[(x+z) \mathbf{i}+(y+z) \mathbf{j}+(x+y) \mathbf{k}] \cdot \mathbf{N} d S\)

⇒ \(\int_v\left[\frac{\partial}{\partial x}(x+z)+\frac{\partial}{\partial y}(y+z)+\frac{\partial}{\partial z}(x+y)\right] d V=\int_V(1+1+0) d V=2 V=2 \frac{4 \pi}{3}(8)=\frac{64 \pi}{3}\)

14. If F = xi−yj+ (z²−1)k find the value of \(\int_S\) F . N dS where S is the closed surface bounded by the planes z = 0, z = 1 and the cylinder x² +y² = 4

Solution: 

∴ \(\mathbf{F}=x \mathbf{i}-y \mathbf{j}+\left(z^2-1\right) \mathbf{k}\)

⇒ \(\text{div} \mathbf{F}=\nabla \cdot \mathbf{F}=\frac{\partial}{\partial x}\{x\}+\frac{\partial}{\partial y}\{-y\}+\frac{\partial}{\partial z}\left\{z^2-1\right\}=1-1+2 z=2 z\)

The limits of the region bounded by the given surface are z = 0 to z = 1, \(y=-\sqrt{4-x^2} \text { to } y=\sqrt{4-x^2}\) and x = -2 to x = 2.

By Gauss’s divergence theorem

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\int_V d i v \mathbf{F} d V=\int_V 2 z d V=\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} \int_{z=0}^{z=1} 2 z d x d y d z\)

⇒ \(\left.=\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} z^2\right]_{z=0}^{z=1} d x d y=\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} 1 d x d y\)

⇒ \(\left.\int_{x=-2}^{x=2} y\right] _{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} d x =\int_{x=-2}^{x=2} 2 \sqrt{4-x^2} d x=4 \int_0^2 \sqrt{4-x^2} d x\)

⇒ \(4\left[\frac{x \sqrt{4-x^2}}{2}+\frac{4}{2} \text{Sin}^{-1} \frac{x}{2}\right]_0^2=4\left[2 \frac{\pi}{2}-0\right]=4 \pi\)

15. If F = x i− y j + (z²−1) k, V is the volume of the cylinder bounded by z = 0, z = 1 and x² +y² = a² and S is the surface of the cylinder, show that \(\int_S\) F . N dS = π a²

Solution: Given=x i− y j + (z²−1)k, Now F₁=x,F₂=−y,F₃=z−1

∴ \(\frac{\partial F_1}{\partial x}=1, \frac{\partial F_2}{\partial y}=-1, \frac{\partial F_3}{\partial z}=2 z . \text{div} \mathbf{F}=\nabla \cdot \mathbf{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=1-1+2 z=2 z\)

By the Gauss divergence theorem,

⇒ \(\int_S \mathbf{F} \mathbf{N} d S=\int_v \text{div} \mathbf{F} d V=\int_V 2 z d V=\int_{x=-a}^{x=a} \int_{y=\sqrt{a^2-x^2}}^{y=-\sqrt{a^2-x^2}} \int_{z=0}^{z=1} 2 z d x d y d z\)

⇒ \(\left.\int_{x=-a}^{x=a} \int_{y=-\sqrt{a^2-x^2}}^{y=\sqrt{a^2-x^2}} z^2\right]_{0}^{1} d x d y=\int_{x=-a}^{x=a}\int_{-y=\sqrt{a^2-x^2}}^{y=\sqrt{a^2-x^2}} d x d y \left.= \int_{x=-a}^{x=a} y\right]_{y=-\sqrt{a^2-x^2}}^{y=\sqrt{a^2-x^2}} dx
\)

⇒ \(\left.\int_{x=-a}^{x=a} 2 \sqrt{a^2-x^2} d x=x \sqrt{a^2-x^2}+a^2 \text{Sin}^{-1}(x / a)\right]_{-a}^a=a^2[\pi / 2-(-\pi / 2)]=\pi a^2\)

16. By transforming into triple integral, evaluate \(\int_S\) ( x³dy dz + x²y dz dx + x²z dx dy) where S is the closed surface consisting of the cylinder x²+y² = a² and the circular disc z = 0 and z = b.

Solution:

Let \(F_1=x^3, F_2=x^2 y, F_3=x^2 z. \quad \frac{\partial F_1}{\partial x}=3 x^2, \frac{\partial F_2}{\partial y}=x^2, \frac{\partial F_3}{\partial z}=x^2\)

⇒ \(\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=3 x^2+x^2+x^2=5 x^2\)

By Gauss’s divergence theorem \(\iint_S\left(x^3 d y d z+x^2 y d z d x+x^2 z d x d y\right)\)

⇒ \(\iiint_V 5 x^2 d x d y d z=4 \int_{x=0}^{x=a} \int_{y=0}^{y=\sqrt{a^2-x^2}} \int_{z=0}^{z=b} 5 x^2 d x d y d z\)

⇒ \(20 \int_{x=0}^{x=a} \int_{y=0}^{y=\sqrt{a^2-x^2}}\left[x^2 z\right]_{z=0}^{z=b} d x d y=20 b \int_{x=0}^a \int_{y=0}^{\sqrt{a^2-x^2}} x^2 d x d y\)

⇒ \(\left.20 b \int_0^a x^2 y\right]_0^{\sqrt{a^2-x^2}}=20 b \int_0^a x^2 \sqrt{a^2-x^2} d x\)

Put x = a sin θ

∴ dx = a cos θ dθ

x = 0 ⇒ θ = 0

x = a ⇒ θ = π/2

⇒ \(20 b \int_0^{\pi / 2} a^2 \sin ^2 \theta \sqrt{a^2-a^2 \sin ^2 \theta} a \cos \theta d \theta=20 a^4 b \int_0^{\pi / 2} \sin ^2 \theta \cos ^2 \theta d \theta\)

⇒ \(20 a^4 b \int_0^{\pi / 2} \sin ^2 \theta\left(1-\sin ^2 \theta\right) d \theta=20 a^4 b\left[\int_0^{\pi / 2} \sin ^2 \theta-\int_0^{\pi / 2} \sin ^4 \theta d \theta\right]\)

⇒ \(20 a^4 b\left[\frac{1}{2} \cdot \frac{\pi}{2}-\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right]=20 a^4 b \cdot \frac{\pi}{16}=5 a^4 b \frac{\pi}{4}\)

17. If F = 2xyi +yz²j + xzk and S is a rectangular parallelopiped bounded by x = 0, y = 0,z = 0,x = 2,y= 1 and z = 3 verily Gauss’s divergence theorem.

Solution:  Consider the six faces of the rectangular parallelopiped bounded by x=o,y=0,z=0,x=2,y=1, and z=3.

Case (1): For the face OADB, the outward normal

N = -k, z = 0, dS = dx dy.

∴ \(\int_{S_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{x=0}^{x=2} \int_{y=0}^{y=1}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(-\mathbf{k}) d x d y=\int_{x=0}^{x=2} \int_{y=0}^{y=1}-x z d x d y=0 .\)

Case (2): For the face OBEC, the outward normal, N = -i, x = 0, dS = dx dz.

∴ \(\int_{S_2} \mathbf{F} \cdot \mathbf{N} d S=\int_{y=0}^{y=1} \int_{z=0}^{z=3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(-\mathbf{i}) d y d z=\int_{y=0}^{y=1} \int_{z=0}^{z=3} 2 x y d y d z=0\)

Case (3): For the face OCFA, the outward normal, N = -j, y = 0, dS = dz dx

∴ \(\int_{S_3} \mathbf{F} \cdot \mathbf{N} d S=\int_{x=0}^{x=2} \int_{z=0}^{z=3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+z x \mathbf{k}\right) \cdot(-\mathbf{j}) d x d z=\int_{x=0}^{x=2} \int_{z=0}^{z=3}-y z^2 d x d z=0\)

Case (4): For the face ADGF, the outward normal, N = I, x = 2, dS = dy dz

⇒ \(\int_{S_4} \mathbf{F} \cdot \mathbf{N} d S=\int_{y=0}^{y=1} \int_{z=0}^{z=3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot \mathbf{i} d y d z=\int_{y=0}^{y=1} \int_{z=0}^{z=3} 2 x y d y d z\)

⇒ \(\left.\left.=\int_{y=0}^{y=1} \int_{z=0}^{z=3} 4 y d y d z=\int_{y=0}^{y=1} 4 y z\right]_{z=0}^{z=3} d y=\int_0^1 12 y d y=6 y^2\right]_0^1=6\)

Case (5): For the faces BDGE, the outward normal, N = j, y = 1, dS = dz dx

⇒ \(\int_{S_5} \mathbf{F} \cdot \mathbf{N} d S=\int_{x=0}^{x=2} \int_{z=0}^{z=3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(\mathbf{j}) d z d x\)

⇒ \(\left.\left.=\int_{x=0}^{x=2} \int_{z=0}^{z=3} y z^2 d z d x=\int_{x=0}^{x=2} \int_{z=0}^{z=3} z^2 d z d x=\int_{x=0}^{x=2} z^3 / 3\right]_0^3 d x=\int_{x=0}^{x=2} 9 d x=9 x\right]_0^2=18\)

Case (6):  For the faces CEGF, the outward normal, N = k, z = 3, dS = dx dy

⇒ \(\int_{S_6} \mathbf{F} \cdot \mathbf{N} d S=\int_{x=0}^{x=2} \int_{y=0}^{y=1}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+z x \mathbf{k}\right) \cdot \mathbf{k} d x d y\)

⇒ \(\left.\left.\int_{x=0}^{x=2} \int_{y=0}^{y=1} x z d x d y=\int_{x=0}^{x=2} \int_{y=0}^{y=1} 3 x d x d y=\int_{x=0}^{x=2} 3 x y\right]_{y=0}^{y=1} d x=\int_{x=0}^{x=2} 3 x d x=3 x^2 / 2\right]_0^2=6 .\)

∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d S\) sum of the six integrals over the six faces

⇒ 0 + 0 + 0 + 6 + 18 + 6 = 30.

⇒ \(\int_V \nabla \cdot \mathbf{F} d V=\int_V\left[\frac{\partial}{\partial x}(2 x y)+\frac{\partial}{\partial y}\left(y z^2\right)+\frac{\partial}{\partial z}(x z)\right] d V=\int_V\left(2 y+z^2+x\right) d V\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=1} \int_{z=0}^{z=3}\left(x+2 y+z^2\right) d x d y d z=\int_{x=0}^{x=2} \int_{y=0}^{y=1}\left[x z+2 y z+\frac{z^3}{3}\right]_{z=1}^{z=3} d x d y\)

⇒ \(\left.\int_{x=0}^{x=2} \int_{y=0}^{y=1}(3 x+6 y+9) d x d y=\int_{x=0}^{x=2}\left[3 x y+3 y^2+9 y\right]\right]_{y=0}^{y=1} d x=\int_0^2(3 x+12) d x\)

∴ \(\left[\frac{3 x^2}{2}+12 x\right]_0^2=6+24=30\) ∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{F} d V\)

∴ Gauss’s divergence theorem is verified.

18. Evaluate \(\int_S\)F.N dS where F = 2xy i +yz² +xzk and S is the surface of the parallelopiped formed by x=0 ,  y = 0, z = 0  x = 2,y = 1, z = 3

Solution: Consider the parallelopiped O A B C P Q R S surrounded by x=0,y=0, z=0, x=2,y=1, and z=3.

1. For the face PQAR, I is the outward normal.

N = i, x = 2, dS = dy dz.

∴ \(\int_{R_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{y=0}^{y=1} \int_{z=0}^{z=3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot \mathbf{i} d y d z\)

⇒ \(=\int_{y=0}^{y=1} \int_{z=0}^{z=3} 2 x y d y d z=\int_{y=0}^{y=1}[4 y z]_{z=0}^{x=3} d y=\int_{y=0}^{y=1} 12 y d y=\left[6 y^2\right]_0^1=6\)

2. For the faces OBSC, -i is the outward normal. N = -i, x = 0, dS = dy dz

∴ \(\int_{R_2} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_2}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(-\mathbf{i}) d y d z=-\iint_R 2 x y d y d z=0 .\)

3. For the face BQPS, j is the outward normal. ∴ N = j, y = 1 and dS = dx dz.

∴ \(\int_{R_3} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot \mathbf{j} d x d z=\iint_{R_3} y z^2 d x d z\)

⇒ \(=\int_{x=0}^{x=2} \int_{z=0}^{z=3} z^2 d x d z=\int_{x=0}^{x=2}\left[\frac{z^3}{3}\right]_{z=0}^{z=3} d x=\int_{x=0}^{x=2} 9 d x=[9 x]_0^2=18\)

4. For the face OARC, -j is the outward normal. ∴ N = -j, y = 0 and dS = dx dz

∴ \(\int_{R_4} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_4}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(-\mathbf{j}) d x d z=-\iint_{R_4} y z^2 d x d z=0\)

5. For the face PRCS, k is the outward normal. ∴ N = k, z = 3 and dS = dx dy.

∴ \(\int_{R_5} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_5}\left(2 x y \mathbf{i}+y z^2 \mathbf{J}+x z \mathbf{k}\right) \cdot \mathbf{k} d x d y=\iint_{R_5} x z d x d y\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=1} 3 x d x d y=\int_{x=0}^{x=2}[3 x y]_{y=0}^{y=1} d x=\int_{x=0}^{x=2} 3 x d x=\left[\frac{3 x^2}{2}\right]_0^2=6\)

6. For the face OAQB, -k is the outward normal. N = -k, z = 0 and dS = dx dy

∴ \(\int_{R_6} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_6}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(-\mathbf{k}) d x d y=-\iint_{R_6} x z d x d y=0\)

∴\(\int_s \mathbf{F} \cdot \mathbf{N} d S=6+0+18+0+6+0=30\)

19. Evaluate\(\iint_S\) (x dydz +y dzdx + z dxdy) taken over the outer surface of the cube[0,a;0,a;0,a].

Solution:  By Gauss’s Divergence theorem,

⇒ \(\iint_S(x d y d z+y d z d x+z d x d y)=\iiint_v\left[\frac{\partial}{\partial x}\{x\}+\frac{\partial}{\partial y}\{y\}+\frac{\partial}{\partial z}\{z\}\right] d x d y d z\)

⇒ \(\iiint_V(1+1+1) d x d y d z=3 \int_{x=0}^{x=a} \int_{y=0}^{y=a} \int_{z=0}^{z=a} d x d y d z=3 \int_{x=0}^{x=a} \int_{y=0}^{y=a}[z]_{z=0}^{z=a} d x d y\)

∴ \(3 \int_{x=0}^{x=a} \int_{y=0}^{y=a} a d x d y=3 \int_{x=0}^{x=a}[a y]_{y=0}^{y=a} d x=3 \int_{x=0}^{x=a} a^2 d x\left[3 a^2 x\right]_{x=0}^{x=a}=3 a^3\)

20. Evaluate \(\int_S\)F . N dS where F = 2x²y i -y²j + 4xz²k taken over the region in the first octant bounded by y² + z²= 9 and x = 2.

Solution:  \(\int_S\)F . N dS=\(\int_V\)∇.F dV=\(\int_V\)∇.(2x²yi-y²j+4xz² k)dv

⇒ \(\int_v\left[\frac{\partial}{\partial x}\left(2 x^2 y\right)+\frac{\partial}{\partial y}\left(-y^2\right)+\frac{\partial}{\partial z}\left(4 x z^2\right)\right] d V=\int_V(4 x y-2 y+8 x z) d V\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=3} \int_{z=0}^{z=\sqrt{9-y^2}}(4 x y-2 y+8 x z) d x d y d z\)

= \(\int_{x=0}^{x=2} \int_{y=0}^{y=3}\left[4 x y z-2 y z+4 x z^2\right]_{=0}^{z=\sqrt{9-y^2}} d x d y\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=3}(2 x-1) 2 y \sqrt{9-y^2}+4 x\left(9-y^2\right) d x d y\)

⇒ \(\int_{x=0}^{x=2}\left[(2 x-1) \frac{\left(9-y^2\right)^{3 / 2}}{-3 / 2}+4 x\left(9 y-\frac{y^3}{3}\right)\right]_{y=0}^{y=3} d x=\int_{x=0}^{x=2}\left[4 x(27-9)+\frac{2(2 x-1)}{3} \times 27\right] d x\)

∴ \(\int_0^2[72 x+36 x-18] d x=\int_0^2(108 x-18) d x=\left[54 x^2-18 x\right]_0^2=216-36=180\)

21. Evaluate by Gauss divergence theorem for \(\iint_S\) 4xz dy dz -y² dz dx+yz dx dy where S is the surface of the cube bounded by the planes  x=0,x=1,y=0,y=1,z=0,z=1

Solution:  

Let \(F_1=4 x z, F_2=-y^2, F_3=y z\)

⇒ \(\frac{\partial F_1}{\partial x}=4 z, \frac{\partial F_2}{\partial y}=-2 y, \frac{\partial F_3}{\partial z}=y \cdot \frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=4 z-2 y+y=4 z-y\)

By Gauss’s divergence theorem, \(\iint_S 4 x z d y d z-y^2 d z d x+y z d x d y\)

⇒ \(\left.\iiint_V(4 z-y) d x d y d z=\int_{x=0}^{x=1} \int_{y=0}^{y=1} \int_{z=0}^{z=1}(4 z-y) d x d y d z=\int_{x=0}^{x=1} \int_{y=0}^{y=1}\left[2 z^2-y z\right]\right]_{z=0}^{z=1} d x d y\)

∴ \(\int_{x=0}^{x=1} \int_{y=0}^{y=1}(2-y) d x d y=\int_{x=0}^{x=1}\left[2 y-\frac{y^2}{2}\right]_{y=0}^{y=1} d x=\int_{x=0}^{x=1} \frac{3}{2} d x=\left[\frac{3 x}{2}\right]_{x=0}^{x=1}=\frac{3}{2} .\)

22. Find the value of \(\int_S\)(F x ∇φ) .N dS, where F = x² i +y²j + z²k,  φ= xy+yz + zx, S is the surface bounded by x = ± 1 ,y = ± 1, z = ± 1.

Solution:

Given \(\mathbf{F}=x^2 \mathbf{i}+y^2 \mathbf{j}+z^2 \mathbf{k}, \varphi=x y+y z+z x . \quad \nabla \varphi=(y+z) \mathbf{i}+(z+x) \mathbf{j}+(x+y) \mathbf{k}\)

⇒ \(\mathbf{F} \times \nabla \varphi=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
x^2 & y^2 & z^2 \\
y+z & z+x & x+y
\end{array}\right|\)

⇒ \(\mathbf{i}\left(x y^2+y^3-z^3-x z^2\right)-\mathbf{j}\left(x^3+x^2 y-y z^2-z^3\right)+\mathbf{k}\left(x^2 z+x^3-y^3-y^2 z\right)\)

∴ \(\nabla \cdot(F \times \nabla \varphi)=\frac{\partial}{\partial x}\left(x y^2+y^3-z^3-x z^2\right)-\frac{\partial}{\partial y}\left(x^3+x^2 y-y z^2-z^3\right)[latex]

+[latex]\frac{\partial}{\partial z}\left(x^2 z+x^3-y^3-y^2 z\right)\)

= \(y^2-z^2-x^2+z^2+x^2-y^2=0\)

By Gauss’s divergence theorem;\(\int_S\)(F×∇φ).N dS=\(\int_V\)∇.(F×∇φ)dV=0.

23. Verify Gauss divergence theorem for F = 4xzi -y²j  +yzk taken over the curve bounded by x = 0, x- 1,y = 0,y= 1,z = 0,z= 1.

Solution:  Consider the cube OABCPQRS surrounded by the following faces.

1. For the face PQAR, I is the outward normal.

N = i, x = 1, ds = dy dz.

∴ \(\int_{R_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{y=0}^{y=1} \int_{z=0}^{z=1}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot \mathbf{i} d y d z\)

⇒ \(=\int_{y=0}^{y=1} \int_{z=0}^{z=1} 4 x z d y d z=\int_{y=0}^{y=1} \int_{z=0}^{z=1} 4 z d y d z=\int_{y=0}^{y=1}\left[2 z^2\right]_{z=0}^{z=1} d y=\int_{y=0}^{y=1} 2 d y=[2 y]_0^1=2\)

2. For the face OBSC, -i is the outward normal. ∴ N = -i, x = 0, and dS = dy dz

∴ \(\int_{R_2} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_2}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot(-\mathbf{i}) d y d z=-\iint_{R_2} 4 x z d y d z=0\)

3. For the face BQPS, j is the outward normal. ∴ N = j, y = 1, and dS = dx dz

∴ \(\int_{R_3} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_3}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot \mathbf{j} d x d z\)

= \(-\iint_{R_3} y^2 d x d z=-\int_{x=0}^{x=1} \int_{z=0}^{z=1} d x d z=-[x]_0^1[z]_0^1=-1\)

4. For the face OARC, -j is the outward normal. ∴ N = -j, y = 0 and dS = dx dz

∴ \(\int_{R_4} \mathbf{F} \cdot \mathbf{N} d S=\int_{R_4}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot(-\mathbf{j}) d S=\iint_{R_4} y^2 d x d z=0\)

5. For the face PRCS, k is the outward normal. ∴ N = k, z = 1, and dS = dx dy

∴ \(\int_{R_5} \mathbf{F} \cdot \mathbf{N} d S=\int_{R_5}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot \mathbf{k} d S\)

= \(\iint_{R_5} y z \cdot d x d y=\int_{x=0}^{x=1} \int_{y=0}^{y=1} y d x d y=[x]_0^1 \cdot\left[\frac{y^2}{2}\right]_0^1=\frac{1}{2} .\)

⇒ \(\int_s \mathbf{F} \cdot \mathbf{N} d S=2+0-1+0+\frac{1}{2}+0=\frac{3}{2}\)

⇒ \(\int_s \mathbf{F} \cdot \mathbf{N} d S=2+0-1+0+\frac{1}{2}+0=\frac{3}{2}\)

⇒ \(\int_V \nabla \cdot \mathbf{F} d V=\int_V\left[\frac{\partial}{\partial x}(4 x z)+\frac{\partial}{\partial y}\left(-y^2\right)+\frac{\partial}{\partial z}(y z)\right] d V=\int_V(4 z-2 y+y) d V\)

⇒ \(=\int_{x=0}^{x=1} \int_{y=0}^{y=1} \int_{z=0}^{z=1}(4 z-y) d x d y d z=\int_{x=0}^{x=1} \int_{y=0}^{y=1}\left[2 z^2-y z\right]_{z=0}^{z=1} d x d y=\int_{x=0}^{x=1} \int_{y=0}^{y=1}(2-y) d x d y\)

⇒ \(=\int_{x=0}^{x=1}\left[2 y-\frac{y^2}{2}\right]_{y=0}^{y=1} d x=\int_{x=0}^{x=1}\left(2-\frac{1}{2}\right) d x=\left[3 \frac{x}{2}\right]_{x=0}^{x=1}=\frac{3}{2}\) ∴ \(\int_S F \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{F} d V\)

∴ Gauss’s divergence theorem is verified.

24. Verify Gauss’s divergence theorem to evaluate \(\int_S\){(x³ −yz) i- 2x²yi + zk }. N dS over the surface of a cube bounded by the coordinate planes x=y=z=a.

Solution:

∴ \(\mathbf{F}=\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k} . \quad \text{div} \mathbf{F}=\nabla \cdot \mathbf{F}=3 x^2-2 x^2+1=x^2+1\)

⇒ \(\int_V \text{div} \mathbf{F} d V=\int_V\left(x^2+1\right) d V=\int_{x=0}^{x=a} \int_{y=0}^{y=a} \int_{z=0}^{z=a}\left(x^2+1\right) d x d y d z\)

⇒ \(=\int_{x=0}^{x=a} \int_{y=0}^{y=a}\left[\left(x^2+1\right) z\right]{ }_{z=0}^{z=a} d x d y=\int_{x=0}^{x=a} \int_{y=0}^{y=a} a\left(x^2+1\right) d x d y=\int_{x=0}^{x=a}\left[a\left(x^2+1\right) y\right]_{y=0}^{y=a} d x\)

⇒ \(\left.=a^2 \int_{x=0}^{x=a}\left(x^2+1\right) d x=a^2\left(\frac{x^3}{3}+x\right)\right]_0^a=a^2\left(\frac{a^3}{3}+a\right)=\frac{a^5}{3}+a^3\)

Case (1): For the face OAQB, the outward normal, N = -k, dS = dx dy, z=0

∴ \(\int_{s_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{s_1}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right] \cdot(-\mathbf{k}) d S=-\int_{s_1} z d S=0\)

Case (2): For the face CSPR, the outward normal, N =k, dS = dx dy, z = a

∴ \(\int_{s_2} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_2}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right] \cdot \mathbf{k} d S\)

⇒ \(=\int_{S_2} z d S=a \int_{x=0}^{x=a} \int_{y=0}^{y=a} d x d y=a \int_{x=0}^{x=a} [y]_{y=0}^{y=a} d x=a^2 \int_0^a d x=\left[a^2 x\right]_{x=0}^{x=a}=a^3\)

Case (3): For the face OASC, the outward normal, N = -j, dS = dx dy, y = 0

∴ \(\int_{s_3} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_3}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right) \cdot(-\mathbf{j}) d S=\int_{s_3}\left(2 x^2 y\right) d S=0\)

Case (4): For the face BQPR, the outward normal, N = j, dS = dx dz, y = a

∴ \(\int_{s_4} \mathbf{F} \cdot \mathbf{N} d S=\int_{s_4}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right) \cdot \mathbf{j} d S=\int_{s_4}-2 x^2 y d S\)

⇒ \(\left.-2 a \int_{x=0}^{x=a} \int_{z=0}^{z=a} x^2 d x d z=-2 a \int_{x=0}^{x=a}\left[x^2 z\right]_{z=0}^{z=a} d x=-2 a^2 \int_0^a x^2 d x=-2 a^2 \frac{x^{3}}{3}\right]_{0}^{a}=-\frac{2 a^5}{3}\)

Case (5): For the face OBRC, the outward normal, N = -i, dS = dy dz, x = 0

∴ \(\int_{s_5} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_5}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right) \cdot(-\mathbf{i}) d S=-\int_{S_5}\left(x^3-y z\right) d S\)

⇒ \(\left.\int_{y=0}^{y=a} \int_{z=0}^{z=a} y z d y d z=\int_{y=0}^{y=a}\left[y z^2 / 2\right]\right]_{z=0}^{z=a} d y=\frac{a^2}{2} \int_{y=0}^{y=a} y d y=\left[\frac{a^2}{2} \cdot \frac{y^2}{2}\right]_{y=0}^{y=a}=\frac{a^4}{4}\)

Case (6): For the face AQPS, the outward normal, N = i, dS = dy dz, x = a

∴ \(\int_{s_5} \mathbf{F} \cdot \mathbf{N} d S=\int_{s_5}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right] \cdot \mathbf{i} d S\)

= \(\int_{s_5}\left(x^3-y z\right) d S=\int_{y=0}^{y=0} \int_{z=0}^{y=a}\left(a^3-y z\right) d y d z\)

⇒ \(\int_{y=0}^{y=a}\left[a^3 z-y z^2 / 2\right]_{z=0}^{z=a} d y=\int_{y=0}^{y=a}\left[a^4-\frac{a^2 y}{2}\right] d y=\left[a^4 y-\frac{a^2 y^2}{4}\right]_{y=0}^{y=a}=a^5-\frac{a^4}{4}\)

∴ \(\int_s \mathbf{F} \cdot \mathbf{N} d S=\int_{s_1} \mathbf{F} \cdot \mathbf{N} d S+\int_{s_2} \mathbf{F} \cdot \mathbf{N} d S+\int_{s_3} \mathbf{F} \cdot \mathbf{N} d S+\int_{s_4} \mathbf{F} \cdot \mathbf{N} d S\)

+\(\int_{s_5} \mathbf{F} \cdot \mathbf{N} d S+\int_{s_6} \mathbf{F} \cdot \mathbf{N} d S\)

= \(0+a^3+0-\frac{2 a^5}{3}+\frac{a^4}{4}+a^5-\frac{a^4}{4}=a^3+\frac{a^5}{3}\)

∴ \(\int_V \text{div} \mathbf{F} d V=\int_s \mathbf{F} \cdot \mathbf{N} d S\)

∴ Gauss’s divergence theorem is verified.

25. Verify Gauss’s divergence theorem for F = (x² -yz) i- 2x² y J + 2k taken over the cube bounded by the planes x = 0, x= a, y = 0, y =a, z = 0, z = a.

Solution: 

∴ \(\int_V \text{div} \mathbf{F} d V=\int_V \nabla \cdot \mathbf{F} d V=\int_V\left[\frac{\partial}{\partial x}\left(x^2-y z\right)+\frac{\partial}{\partial y}\left(-2 x^2 y\right)+\frac{\partial}{\partial z}(2)\right] d V\)

⇒ \(\int_V\left(2 x-2 x^2\right) d V=\int_{x=0}^{x=a} \int_{y=0}^{y=a} \int_{z=0}^{z=a}\left(2 x-2 x^2\right) d x d y d z\)

⇒ \(\int_{x=0}^{x=a} \int_{y=0}^{y=a}\left(2 x-2 x^2\right) d x d y [z]_{z=0}^{z=a}=\int_{x=0}^{x=a} \int_{y=0}^{y=a} a\left(2 x-2 x^2\right) d x d y\)

⇒ \(\left.\int_{x=0}^{x=a} a\left(2 x-2 x^2\right) d x \quad y\right]_{y=0}^{y=a}\quad\)

= \(\int_{x=0}^{x=a} a^2\left(2 x-2 x^2\right) d x=a^2\left[x^2-2 x^3 / 3\right]_0^a=a^2\left[a^2-2 a^3 / 3\right]\)

⇒ \(a^4-2 a^5 / 3\)

Case (1): For the face ADGF, N = i, dS = dy dz and x = a

∴ \(\int_{S_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_1}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right] \cdot \mathbf{i} d S=\int_{S_1}\left(x^2-y z\right) d S\)

⇒ \(\int_{y=0}^{y=a} \int_{z=0}^{z=a}\left(a^2-y z\right) d y d z=\int_{y=0}^{y=a}\left[a^2 z-y z^2 / 2\right]_{z=0}^{z=a}dy\)

⇒ \(\left.\int_0^a\left(a^3-a^2 y / 2\right) d y=a^3 y-a^2 y^2 / 4\right]_0^a=a^4-a^4 / 4=3 a^4 / 4\)

Case (2): For the face OBEC, N = -i, dS = dy dz and x = 0.

∴ \(\int_{S_2} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_2}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right] \cdot(-\mathbf{i}) d S=\int_{y=0}^{y=a} \int_{z=0}^{z=a}(0-y z)(-1) d y d z\)

⇒ \(\left.\left.\int_{y=0}^{y=a} \int_{z=0}^{z=a} y z d y d z=\int_{y=0}^{y=a} y z^2 / 2\right]_{z=0}^{z=a} d y=\int_{y=0}^{y=a}\left[a^2 y / 2\right] d y=\frac{a^2 y^2}{4}\right]_0^a=\frac{a^4}{4}\)

Case (3): For the face BEGD, N = I, dS = dx dz and y = a

∴ \(\int_{S_3} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_3}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right] \cdot(\mathbf{j}) d S=\int_{x=0}^{x=a} \int_{z=0}^{z=a}-2 a x^2 d x d z\)

⇒ \(\left.\int_{x=0}^{x=a}\left[-2 a x^2 z\right]_{z=0}^{z=a} d x=\int_0^a-2 a^2 x^2 d x=-2 a^2 x^3 / 3\right]_0^a=-2 a^5 / 3 .\)

Case (4): For the face OCFA, N = -j, dS = dx dz, y = 0

∴ \(\int_{S_4} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_4}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right] \cdot(-\mathbf{j}) d S=0\)

Case (5): For the face CFGE, N = k, dS = dx dy, z = a

∴\(\int_{S_5} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_5}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right) \cdot \mathbf{k} d S\)

⇒ \(\left.\left.\int_{x=0}^{x=a} \int_{y=0}^{y=a} 2 d x d y=\int_{x=0}^{x=a} 2 y\right]_{y=0}^{y=a} d x=\int_{x=0}^{x=a} 2 a d x=2 a x\right]{ }_0^a=2 a^2\)

Case (6): for this face OADB, N = -k, dS = dx dy, z = 0.

∴ \(\int_{S_6} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_6}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right) \cdot(-\mathbf{k}) d S\)

⇒ \(\left.\left.\int_{x=0}^{x=a} \int_{y=0}^{y=a}(-2) d x d y=\int_{x=0}^{x=a}(-2 y)\right]_{y=0}^{y=a} d x=\int_{x=0}^{x=a}-2 a d x=-2 a x\right]_0^a=-2 a^2\)

∴ \(\mathbf{F} \cdot \mathbf{N} d S=\int_{S_1} \mathbf{F} \cdot \mathbf{N} d S+\int_{S_2} \mathbf{F} \cdot \mathbf{N} d S\)

+ \(\int_{S_3} \mathbf{F} \cdot \mathbf{N} d S+\int_{S_4} \mathbf{F} \cdot \mathbf{N} d S+\int_{S_5} \mathbf{F} \cdot \mathbf{N} d S+\int_{S_6} \mathbf{F} \cdot \mathbf{N} d S\)

= \(3 a^4 / 4+a^4 / 4-2 a^5 / 3+0+2 a^2-2 a^2=a^4-2 a^5 / 3\)

∴ \(\int_V \text{div} \mathbf{F} d V=\int_S \mathbf{F} \cdot \mathbf{N} d S .\)

∴ Gauss’s divergence theorem was verified.

26. State and prove Green’s theorem in a plane.

Solution:   Let S be a closed region in the plane enclosed by a curve C if P and Q are continuous and differentiable scalar functions of x and y in S, then\(\int_C\) P dx + Q dy = \(\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\)dx dy,  the line integral being taken along the entire boundary C of S such that S is on the left as one advance along C

Proof: Let any line parallel to either co-ordinate axes cut C in at most two points.

Let S  lie between the lines x=a, x=b,and y=c y=d.

Let y=f(x) be the curve C₁ (AEB) and y=g(x)  be the curve C₂ (ADB) where f(x)≤ g(x)

Consider \(\iint_S \frac{\partial P}{\partial y} d x d y=\int_{x=a}^{x=b} \int_{y=f(x)}^{y=g(x)}\left(\frac{\partial P}{\partial y} d y\right) d x\)

= \(=\int_{x=a}^{x=b}[P(x, y)]_{y=f(x)}^{y=g(x)} d x=\int_a^b[P(x, g(x))-P(x, f(x))] d x\)

= \(\int_a^b P(x, g) d x-\int_a^b P(x, f) d x=-\int_{C_1} P(x, y) d x-\int_{C_2} P(x, y) d x=-\int_{c_3} P d x\)

Similarly , we can prove that \(\iint_S\)\(\left(\frac{\partial Q}{\partial x}\right)\)dx dy=\(\int_C\)dy.

∴\(\int_C\) P dx + Q dy = \(\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\)dx dy

27. Evaluate \(\oint_C\)(x dy-y dx) around the circle C where C is x² +y²=1.

Solution:

By Green’s theorem, \(\int_c P d x+Q d y=\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\)

Put P = -y, Q = x. Then \(\frac{\partial Q}{\partial x}=1, \frac{\partial P}{\partial y}=-1\)

∴ \(\int_C x d y-y d x=\iint_S[1-(-1)] d x d y=2 \iint_S d x d y\)

= \(2 \text { (Area of the surface } S \text { ) }=2 \pi(1)^2=2 \pi\)

28. If f and g are two continuous and differentiable scalar point functions over the region V enclosed by the surface S, then prove that

1. \(\int_V\left[f \nabla^2 g+\nabla f \cdot \nabla g\right] d V\)=\(=\int_S(f \nabla g) \cdot N d S\)
2. \(\int_V\left(f \nabla^2 g-g \nabla^2 f\right) d V\)=\(=\int_S(f \nabla g-g \nabla f) \cdot \mathbf{N} d S\)

Solution:

1. Let F = f ∇g.

Then \(\nabla \cdot \mathbf{F}=\nabla \cdot(f \nabla g)=f(\nabla \cdot \nabla g)+\nabla f \cdot \nabla g=f \nabla^2 g+\nabla f \cdot \nabla g\)

By Gauss’s divergence theorem, \(\int_V \nabla \cdot \mathbf{F} d V=\int_S \mathbf{F} \cdot \mathbf{N} d S\)

⇒ \(\int_V\left[f \nabla^2 g+\nabla f \cdot \nabla g\right] d V=\int_S(f \nabla g) \cdot \mathbf{N} d S\)

2. From (1); \(\int_V\left(f \nabla^2 g+\nabla f \cdot \nabla g\right) d V=\int_S(f \nabla g) \cdot \mathbf{N} d S\) → (1)

Interchanging f and g in (1), we get \(\int_V\left(g \nabla^2 f+\nabla g \cdot \nabla f\right) d V=\int_s(g \nabla f) \cdot \mathbf{N} d S\) → (2)

(1) – (2) ⇒ \(\int_V\left(f \nabla^2 g-g \nabla^2 f\right) d V=\int_S(f \nabla g-g \nabla f) \cdot \mathbf{N} d S\)

29. Show that the area bounded by a simple closed curve C is given by ½ \(\oint_C\) x dy-y dx and hence find area of ellipse x=a cos θ, y= b sin θ ,0≤θ≤2π.

Solution:

Here P = -y, Q = x. Then \(\frac{\partial Q}{\partial x}=1, \frac{\partial P}{\partial y}=-1\)

By Green’s theorem \(\frac{1}{2} \int_C x d y-y d x=\frac{1}{2} \int_C P d x+Q d y=\frac{1}{2} \iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\)

= \(\frac{1}{2} \iint_S[1-(-1)] d x d y=\iint_S d x d y\) = Area of the surface bounded by the curve C.

Equations of ellipse are x = a cos θ, y = b sin θ, 0 ≤ θ ≤ 2π

∴ \(\frac{d x}{d \theta}=-a \sin \theta, \frac{d y}{d \theta}=b \cos \theta\)

∴ Area of the ellipse = \(\frac{1}{2} \int_c x d y-y d x=\frac{1}{2} \int_0^{2 \pi}[(a \cos \theta)(b \cos \theta) {-}(b \sin \theta)(-a \sin \theta)] d \theta\)

= \(\frac{1}{2} \int_0^{2 \pi}\left(a b \cos ^2 \theta+a b \sin ^2 \theta\right) d \theta=\frac{a b}{2} \int_0^{2 \pi} d \theta=\frac{a b}{2}(2 \pi)=\pi a b \text { sq.unit }\)

30. \(\oint_C\) (cos x sin y- xy) dx + sin x cos y dy, by Green’s theorem, where C is the circle x²+y²=1

Solution:  By Green’s theorem\(\int_C\) P dx + Q dy = \(\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\)dx dy

Let P = cos x sin y -xy, Q = sin x cos y. \(\frac{\partial P}{\partial y}=\cos x \cos y-x, \frac{\partial Q}{\partial x}=\cos x \cos y\).

The limits of the surface of the circle x² + y² = 1 are x = -1 to x = 1 and \(y=-\sqrt{1-x^2} \text { to } y=\sqrt{1-x^2} \text {. }\)

∴\(\int_c(\cos x \sin y-x y) d x+\sin x \cos y d y=\iint_s(\cos x \cos y-\cos x \cos y+x) d x d y\)

= \(\iint_S x d x d y=\int_{x=-1}^{x=1} \int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}} x d x d y=\int_{x=-1}^{x=1} 2 x d x \int_0^{y=\sqrt{1-x^2}} d y=\int_{x=-1}^{x=1} 2 x \sqrt{1-x^2} d x=0\)

31. Evaluate by Green’s theorem \(\oint_C\)(x²– cosh y) + (y + sinx) dy, where C is the rectangle with vertices (0, 0), (π, 0), (π, 1), (0, 1).

Solution:

By theorem \(\int_c P d x+Q d y=\iint_s\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\)

Let \(P=x^2-\cosh y, Q=y+\sin x \text {.. Then } \frac{\partial P}{\partial y}=-\sinh y, \frac{\partial Q}{\partial x}=\cos x\)

The limits of the surface of integration are x = 0 to x = π and y = 0 to y = 1.

∴ \(\int_C\left(x^2-\cosh y\right) d x+(y+\sin x) d y=\int P d x+Q d y\)

= \(\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y=\int_{x=0}^{x=\pi} \int_{y=0}^{y=1}(\cos x+\sinh y) d x d y\)

= \(\int_{x=0}^{x=\pi}[y \cos x+\cosh y]_{y=0}^{y=1} d x=\int_0^{\pi}(\cos x+\cosh 1-1) d x\)

= \([\sin x+x \cosh 1-x]_{x=0}^{x=\pi}=\pi \cosh 1 – \pi=\pi(\cosh 1-1)=\pi\left(\frac{e+e^{-1}}{2}-1\right)\)

32. Using Green’s theorem, evaluate \(\oint_C\)(x² +y²)dx + 3xy² dy where C is the circle x²+y²=4

Solution:

Here \(P=x^2+y^2, Q=3 x y^2\) ∴ \(\frac{\partial P}{\partial y}=2 y, \frac{\partial Q}{\partial x}=3 y^2\). The limits of the surface of integration are x = -2 to x = 2 and \(y=-\sqrt{4-x^2} \text { to } y=\sqrt{4-x^2}\)

By Green’s theorem, \(\int_C\left(x^2+y^2\right) d x+3 x y^2 d y=\iint_S\left(3 y^2-2 y\right) d x d y\)

= \(\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}\left(3 y^2-2 y\right) d x d y=\int_{x=-2}^{x=2}\left[y^3-y^2\right]_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} d x=\int_{x=-2}^{x=2} 2\left(4-x^2\right)^{3 / 2} d x\)

= \(4 \int_0^2\left(4-x^2\right)^{3 / 2} d x=4 \int_0^{\pi / 2}\left(4-4 \sin ^2 \theta\right)^{3 / 2} 2 \cos \theta d \theta \text { where } x=2 \sin \theta\)

= \(64 \int_0^{\pi / 2} \cos ^4 \theta d \theta=64 \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}=12 \pi\)

33. Evaluate \(\oint_C\) (3x + 4y) dx + (2x− 3y) dy where ‘C’ is the circle  x²+y²= 4.

Solution:

Here P = 3x + 4y, Q = 2x – 3y. \(\frac{\partial P}{\partial y}=4, \frac{\partial Q}{\partial x}=2\). The limits of the surface of integration are x = -2 to x = 2 and \(y=-\sqrt{4-x^2} \text { to } y=\sqrt{4-x^2}\).

By Green’s theorem, \(\int_C(3 x+4 y) d x+(2 x-3 y) d y\)

= \(\iint_y(2-4) d x d y=\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}(-2) d x d y=\int_{x=-2}^{x=2}[-2 y]_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}\)

= \(\int_{-2}^2-4 \sqrt{4-x^2} d x=-8 \int_0^2 \sqrt{4-x^2} d x=-8\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \text{Sin}^{-1} \frac{x}{2}\right]_{0}^{2}=-8\left[2 \frac{\pi}{2}\right]=-8 \pi\)

34. Evaluate by Green’s theorem \(\oint(y-\sin x) d x\)+ cos x dy where C is the triangle enclosed by the lines x=0, x=π/2,  πy=2x.

Solution:

Here P = y – sin x, Q = cos x. ∴ \(\frac{\partial P}{\partial y}=1, \frac{\partial Q}{\partial x}=-\sin x\)

The limits of the surface of integration are x = 0 to \(x=\frac{\pi}{2} ; y=0 \text { to } y=\frac{2 x}{\pi}\)

By Green’s theorem \(\int_C(y-\sin x) d x+\cos x d y=\int_C P d x+Q d y\)

= \(\iint_s\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y=\int_{x=0}^{x=\pi / 2} \int_{y=0}^{y=2 x / \pi}(-\sin x-1) d x d y\)

=\(\int_{x=0}^{x=\pi / 2}[(-\sin x-1) y]_{y=0}^{y=2 x / \pi} d x\)

= \(\int_0^{\pi / 2}(-\sin x-1) \frac{2 x}{\pi} d x=-\frac{2}{\pi} \int_0^{\pi / 2} x(1+\sin x) d x\)

=\(-\frac{2}{\pi}[x(x-\cos x)]_{0}^{\pi /2}-\int_0^{\pi / 2}(x-\cos x) d x\)

= \(-\frac{2}{\pi}\left[\frac{\pi}{2}\left(\frac{\pi}{2}\right)-\left(\frac{x^2}{2}-\sin x\right)_0^{\pi / 2} \right]=-\frac{2}{\pi}\left[\frac{\pi^2}{4}-\frac{\pi^2}{8}+1\right]=-\frac{\pi}{4}-\frac{2}{\pi}\)

35. Compute \(\oint_C\) (x²− 2xy) dx + (x²y + 3) dy around the boundary C of the region defined by y² = 8x and x = 2 by applying Green’s theorem.

Solution:

Here \(P=x^2-2 x y, Q=x^2 y+3\)  ∴ \(\frac{\partial P}{\partial y}=-2 x, \frac{\partial Q}{\partial x}=2 x y\)

The limits of the surface of integration are x = 0 to x = 2 and y = 0 to \(\sqrt{8 x}\).

By Green’s Theorem, \(\oint_c\left(x^2-2 x y\right) d x+\left(x^2 y+3\right) d y=\int_c P d x+Q d y\)

= \(\iint_S\left[\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right] d x d y=\int_{x=0}^{x=2} \int_{y=-\sqrt{8 x}}^{y=\sqrt{8 x}}(2 x y+2 x) d x d y=\int_{x=0}^{x=2}\left[x y^2+2 x y\right]_{y=-\sqrt{8 x}}^{y=\sqrt{8 x}} d x\)

= \(\int_{x=0}^{x=2}(4 x \sqrt{8 x}) d x=\left[8 \sqrt{2} \frac{x^{5 / 2}}{5 / 2}\right]_{x=0}^{x=2}=\frac{128}{5}\)

36. Find the area bounded by x^2/3 +y^2/3 = a^2/3  using Green’s theorem.

Solution:

The parametric equation of \(x^{2 / 3}+y^{2 / 3}=a^{2 / 3} \text { is } x=a \cos ^3 \theta, y=a \sin ^3\) θ and θ variation from 0 to 2π.

By green’s theorem, Area = \(\frac{1}{2} \int_C(x d y-y d x)\)

= \(\frac{1}{2} \int_{\theta=0}^{\theta=2 \pi} a\cos ^3 \theta 3 a \sin ^2 \theta \cos \theta d \theta-a \sin ^3 \theta\left(3 a \cos ^2 \theta\right)(-\sin \theta) d \theta\)

= \(\frac{1}{2} \int_{\theta=0}^{2 \pi} 3 a^2 \cos ^2 \theta \sin ^2 \theta\left(\cos ^2 \theta+\sin ^2 \theta\right) d \theta=\frac{1}{2} \int_0^{2 \pi} 3 a^2 \cos ^2 \theta \sin ^2 \theta d \theta\)

= \(6 a^2 \int_0^{\pi / 2} \sin ^2 \theta \cos ^2 \theta d \theta=6 a^2 \times \frac{1}{4} \times \frac{1}{2}=\frac{3 a^2}{4} \times \frac{\pi}{2}=\frac{3 \pi a^2}{8} \text { sq.unit }\)

37. Verify Green’s theorem in the plane for \(\oint_C\)(3x²– 8y²) dx + (4y- 6xy) dy where C is the region bounded by y = \(\sqrt{x}\)and y = x².

Solution:

P = \(P=3 x^2-8 y^2, Q=4 y-6 x y . \quad \frac{\partial P}{\partial y}=-16 y ; \frac{\partial Q}{\partial x}=-6 y .\)

⇒ \(\int_C\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y=\iint_s \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\)

= \(\iint_s(-6 y+16 y) d x d y=10 \int_{x=0}^{x=1}\left(\int_{y=x^2}^{y=\sqrt{x}} y d y\right) d x\)

= \(10 \int_{x=0}^{x=1}\left[\frac{y^2}{2}\right]_{y=x^2}^{y=\sqrt{x}} d x \quad=5 \int_0^1\left(x-x^4\right) d x=5\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1=5\left[\frac{1}{2}-\frac{1}{5}\right]=5\left[\frac{5-2}{10}\right]=\frac{3}{2}\)

∴ \(\int_c\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y\)

= \(\int_{C_1}\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y+\int_{C_2}\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y\)

Where c₁ is the curve y = x² from O to A and C₂ is the curve = √x from A to O

= \(\int_0^1\left(3 x^2-8 x^4\right) d x+\left(4 x^2-6 x^3\right) 2 x d x+\int_1^0\left(3 x^2-8 x\right) d x+(4 \sqrt{x}-6 x \sqrt{x}) \frac{d x}{2 \sqrt{x}}\)

= \(\int_0^1\left(3 x^2+8 x^3-20 x^4\right) d x-\int_0^1\left(3 x^2-8 x+2-3 x\right) d x=\int_0^1\left(8 x^3-20 x^4+11 x-2\right) d x\)

= \(\left[2 x^4-4 x^5+\frac{11 x^2}{2}-2 x\right]_0^1=2-4+\frac{11}{2}-2=\frac{3}{2}\)  ∴ Green’s theorem is verified.

38. Verify Green’s theorem in the plane for\(\oint_C\)(xy + y²) dx + x²dy where C is the closed curve of the region bounded by y = x andy = x².

Solution:

Here P = xy + y², Q = x² ∴ \(\frac{\partial P}{\partial y}=x+2 y, \frac{\partial Q}{\partial x}=2 x\)

By Green’s theorem, \(\int_c\left(x y+y^2\right) d x+x^2 d y=\int_c P d x+Q d y=\iint_s\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\)

= \(\iint_s[2 x-(x+2 y)] d x d y=\iint_s(x-2 y) d x d y=\int_{x=0}^{x=1} \int_{y=x^2}^{y=x}(x-2 y) d x d y\)

= \(\int_{x=0}^{x=1}\left[x y-y^2\right]_{y=x^2}^{y=x} d x=\int_0^1\left[\left(x^2-x^2\right)-\left(x^3-x^4\right)\right] d x=\int_0^1\left(x^4-x^3\right) d x\)

= \(\left[\frac{x^5}{5}-\frac{x^4}{4}\right]_0^1=\frac{1}{5}-\frac{1}{4}=-\frac{1}{20}\)

 

\(\int_c\left(x y+y^2\right) d x+x^2 d y\)

= Line integral along y = x² (from O to A) + line integral along y = x (from A to O)

= \(=\int_0^1\left[x\left(x^2\right)+x^4\right] d x+x^2 \cdot 2 x d x+\int_1^0\left(x^2+x^2\right) d x+x^2 d x\)

= \(\int_0^1\left(3 x^3+x^4\right) d x-\int_0^1 3 x^2 d x=\left[3 \frac{x^4}{4}+\frac{x^5}{5}-x^3\right]_{0}^{1}=\frac{3}{4}+\frac{1}{5}-1=\frac{15+4-20}{20}=-\frac{1}{20}\)

39. Verify Green’s theorem in the plane for \(\oint_C\)(2xy − x²)dx + (x +y²) dy where C is the boundary of the region enclosed by y = x² and y²=x described in the positive sense.

Solution:

Here P = 2xy – x², Q = x² + y².

∴ \(\frac{\partial P}{\partial y}=2 x, \frac{\partial Q}{\partial x}=2 x\)

⇒ \(\iint_S\left[\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right] d x d y=\iint_S(2 x-2 x) d x d y=0\)

⇒ \(\int_c P d x+Q d y\) = Line integral along y = x² (from O to A) + Line integral along y² = x (from A to O) = I₁ + I₂

Now \(I_1=\int_{x=0}^{x=1} P d x+Q d y=\int_0^1\left[2 x\left(x^2\right)-x^2\right] d x+\left(x^2+x^4\right) 2 x d x\)

= \(\left.\int_0^1\left(2 x^3-x^2+2 x^3+2 x^5\right) d x=\int_0^1\left(2 x^5+4 x^3-x^2\right) d x=\frac{x^6}{3}+x^4-\frac{x^3}{3}\right]_0^1=1\)

⇒ \(I_2=\int_{x=1}^{x=0} P d x+Q d y=\int_1^0\left(2 x \sqrt{x}-x^2\right) d x+\left(x^2+x\right) \frac{1}{2 \sqrt{x}} d x\)

= \(\int_1^0\left[2 x \sqrt{x}-x^2+x \sqrt{x} / 2+\sqrt{x} / 2\right] d x=\int_1^0\left[5 x^{3 / 2} / 2-x^2+x^{1 / 2} / 2\right] d x\)

= \(\left.\frac{5}{2} \times \frac{x^{5 / 2}}{5 / 2}-\frac{x^3}{3}+\frac{1}{2} \times \frac{x^{3 / 2}}{3 / 2}\right]_1^0=-1+\frac{1}{3}-\frac{1}{3}=-1\)

∴ \(\int_c P d x+Q d y=I_1+I_2=1-1=0\)

∴ Green’s theorem is verified.

40. Verify Green’s theorem, \(\oint_C\)(3x²– 8y²) dx + (4y- 6xy)dy where C is the boundary enclosed by x = 0,y = x+y= 1.

Solution:

Here P = 3x² – 8y², Q = 4y – 6xy. ∴ \(\frac{\partial P}{\partial y}=-16 y, \frac{\partial Q}{\partial x}=-6 y\)

The limits o the surface of integration are x = 0 to x = 1 and y = 0 to y = 1 – x

By Green’s theorem, \(\oint\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y\)

= \(\int_c P d x+Q d y=\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y=\int_{x=0}^{x=1} \int_{y=0}^{y=1-x}(-6 y+16 y) d x d y\)

= \(\int_{x=0}^{x=1} \int_{y=0}^{y=1-x} 10 y d x d y=\int_{x=0}^{x=1}\left(5-10 x+5 x^2\right) d x=\int_{x=0}^{x=1}\left[5 y^2\right]_{x=0}^{y=1-x} d x\)

=\(\int_{x=0}^{x=1} 5(1-x)^2 d x\)

= \(\left[5 x-5 x^2+\frac{5 x^3}{3}\right]_{x=0}^{x=1}=5-5+\frac{5}{3}=\frac{5}{3}\)

Given planes x = 0, y = 0 and x + y = 1 from a triangle in xy-plane with vertices O(0,0), A(1,0) and B(0,1).

∴ \(\int_c\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y\)

= Line integral along \(\overrightarrow{O A}\) + Line along \(\overrightarrow{A B}\) + Line integral along \(\overrightarrow{B O}\).

1. Line integral along \(\left.\overrightarrow{O A}=\int_0^1 3 x^2 d x=x^3\right]_0^1=1\)

2. Line integral along \(\overrightarrow{A B}\). Here x + y = 1 ⇒ y = 1 – x varies from 1 to 0.

Line integral along \(\overrightarrow{A B}=\int_1^0\left[3 x^2-8(1-x)^2\right] d x+[4(1-x)-6 x(1-x)](-d x)\)

= \(\int_1^0\left(3 x^2-8-8 x^2+16 x\right) d x-\left(4-4 x-6 x+6 x^2\right) d x=\int_1^0\left(-11 x^2+26 x-12\right) d x\)

= \(\left[-\frac{11 x^3}{3}+13 x^2-12 x\right]_1^0=\frac{11}{3}-13+12=\frac{8}{3} .\)

3. Line integral along \(\left.\overrightarrow{B O}=\int_1^0 4 y^2 d y=2 y^2\right]_1^0=-2\)

∴ \(\int_C\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y=1+\frac{8}{3}-2=\frac{5}{3}\)

∴ Green’s theorem is verified.

41. State and prove Stake’s theorem.

Solution:  Let S be a surface bounded by a closed non-intersecting curve C. If F is any differentiable vector point function, then

\(\int_C\)F.dr= \(\int_S\) curl F.Nds, where N is the outward drawn unit normal vector to S and C is traversed in the positive direction.

proof: Let S be a surface which is such that its projection on xy,yz,zx or y=h(z,x) where f,g,h are simple valued continuous and differentiable functions. Let F=F₁i+F₂j+F₃k.

Then curl F =∇×F=∇ × (F₁i + F₂j + F₃k)

=∇ × F₁i × F₂j × F₃k

 

Now \(\nabla \times F_1 \mathbf{I}=\left|\begin{array}{ccc}
\mathbf{I} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_1 & 0 & 0
\end{array}\right|=\frac{\partial F_1}{\partial z} \mathbf{J}-\frac{\partial F_1}{\partial y} \mathbf{k}\)

⇒ \(\left(\nabla \times F_1 \mathbf{I}\right) \cdot \mathbf{N}=\left[\frac{\partial F_1}{\partial z}(\mathbf{j} \cdot \mathbf{N})-\frac{\partial F_1}{\partial y}(\mathbf{k} \cdot \mathbf{N})\right]\) (1)

Let z = f(x,y) be the equation of S.

For any point in S, r = xi + yj + zk = xi + yi + f(x,y)k…

∴ \(\frac{\partial \mathbf{r}}{\partial y}=\mathbf{j}+\frac{\partial z}{\partial y} \mathbf{k} \text {. Since } \frac{\partial \mathbf{r}}{\partial y} \text { is the tangent vector to } S, \mathbf{N} \cdot \frac{\partial \mathbf{r}}{\partial y}=0\)

⇒ \(\mathbf{N} \cdot \mathbf{J}+(\mathbf{N} \cdot \mathbf{k}) \frac{\partial z}{\partial y}=0 \Rightarrow \mathbf{N} \cdot \mathbf{j}=-(\mathbf{N} \cdot \mathbf{k}) \frac{\partial z}{\partial y}\)

From (1); \(\left(\nabla \times F_1 \mathbf{i}\right) \cdot \mathbf{N}=-(\mathbf{N} \cdot \mathbf{k}) \frac{\partial F_1}{\partial z} \cdot \frac{\partial z}{\partial y}-(\mathbf{N} \cdot \mathbf{k}) \frac{\partial F_1}{\partial y}\)

∴ \(\left[\left(\nabla \times F_1 \mathbf{i}\right) \cdot \mathbf{N}\right] d S=-\left(\frac{\partial F_1}{\partial y}+\frac{\partial F_1}{\partial z} \cdot \frac{\partial z}{\partial y}\right)(\mathbf{N} \cdot \mathbf{k}) d S\)

= \(-\frac{\partial}{\partial y} F_1(x, y, z) \cos \gamma d S=-\frac{\partial F_1}{\partial y} d x d y\)

Let R be the projection of S on xy-plane. Then

∴ \(\int_S\left(\nabla \times F_1 \mathbf{i}\right) \cdot \mathbf{N} d S=\int_R \int-\frac{\partial F_1}{\partial y} d x d y=\int_C F_1 d x\), by green’s theorem

Similarly \(\int_S\left(\nabla \times F_2 \mathbf{j}\right) \cdot \mathbf{N} d S=\int_c F_2 d y \text { and } \int_S\left(\nabla \times F_3 \mathbf{k}\right) \cdot \mathbf{N} d S=\int_c F_3 d z\)

∴ \(\int_S \nabla \times\left(F_1 \mathbf{i}+F_2 \mathbf{j}+F_3 \mathbf{k}\right) \cdot \mathbf{N} d S=\int_c F_1 d x+F_2 d y+F_3 d z\)

∴ \(\int_S(\nabla \times F) \cdot \mathbf{N} d S=\int_c F \cdot d r\)

42. Prove by Stake’s theorem curl grad φ = 0.

Solution:  Let be a  surface enclosed by a simple closed curve C.

∴ By stoke’s theorem, \(\int_s(\text{curl} \text{grad} \varphi) \cdot \mathbf{N} d S\)

= \(\int_s[\nabla \times(\nabla \varphi)] \cdot \mathbf{N} d S=\int_c \nabla \varphi \cdot d \mathbf{r}=\int_c\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right) \cdot(\mathbf{i} d x+\mathbf{j} d y+\mathbf{k} d y)\)

= \(\left.\int_C\left(\frac{\partial \varphi}{\partial x} d x+\frac{\partial \varphi}{\partial y} d y+\frac{\partial \varphi}{\partial z} d z\right)=\int_C d \varphi=\varphi\right]_P^P=0\)

∴ \(\int_S(\text{curl} \text{grad} \varphi) \cdot \mathbf{N} d S=0 \Rightarrow \text{curl}(\text{grad} \varphi)=0\)

43. Find \(\int_C\)T .dr where T is the unit tangent vector and C is the unit circle in the xy-plane with centre at the origin.

Solution:  By stokes theorem ,\(\int_C\)T.dr=\(\int_S\)(curl T).N dS=\(\int_S\)(∇×T).N dS=\(\int_S\)0 dS=0.

44. Prove that \(\oint_C\)r.dr = 0.

Solution:  By stokes theorem ,\(\int_C\)T.dr=\(\int_S\)(curl r).N dS=\(\int_S\)0. N dS=0.

45. By Stoke’s theorem prove that div curl F = 0.

Solution:  Let S be the surface enclosed by a simple closed curve C

∴ \(\int_s \text{div} \text{curl} \mathbf{F} d S=\int \nabla \cdot \nabla \times \mathbf{F} d S=\int_s \Sigma \mathbf{i} \frac{\partial}{\partial x}(\nabla \times \mathbf{F}) d S=\int_S \Sigma \frac{\partial}{\partial x}[\mathbf{i} \cdot(\nabla \times \mathbf{F})] d S\)

= \(\Sigma \frac{\partial}{\partial x} \int_S(\nabla \times F) \cdot i d S=\Sigma \frac{\partial}{\partial x} \int_C\left(F_2 d y+F_3 d z\right)=0\)

∴ div curl F = 0.

46. Verify Stoke’s theorem to evaluate \(\int_C\)xy dx + xy² dy, where C is the square in the  xy-plane with vertices (1, 0), (- 1, 0), (0, 1), (0,- 1)

.Solution:

Let F = \(x y \mathbf{i}+x y^2 \mathbf{j} . \text{curl} \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x y & x y^2 & 0
\end{array}\right|=\left(y^2-x\right) \mathbf{k}\)

⇒ \(\int_c x y d x+x y^2 d y=\oint_C \mathbf{F} \cdot d \mathbf{r}=\int_s \nabla \times \mathbf{F} \cdot \mathbf{N} d S=\int_S\left(y^2-x\right) \mathbf{k} \cdot \mathbf{N} d S\)

Since k. N ds = dx dy and R is the region ABCD in xy-plane,

We have \(\int_S\left(y^2-x\right) \mathbf{k} \cdot \mathbf{N} d S=\iint_R\left(y^2-x\right) d x d y\)

Equation to \(\stackrel{\leftrightarrow}{A B} \text { is } \frac{x}{1}+\frac{y}{1}=1 \Rightarrow y=1-x\)

Equation to \(\stackrel{\leftrightarrow}{B C} \text { is } \frac{x}{-1}+\frac{y}{1}=1 \Rightarrow y=1+x\)

Equation to \(\stackrel{\leftrightarrow}{C D} \text { is } \frac{x}{-1}+\frac{y}{-1}=1 \Rightarrow y=-1-x\)

Equation to \(\stackrel{\leftrightarrow}{D A} \text { is } \frac{x}{1}+\frac{y}{-1}=1 \Rightarrow y=x-1\)

 

⇒ \(\iint_R\left(y^2-x\right) d x d y=\int_{x=0}^{x=1} \int_{y=0}^{y=1-x}\left(y^2-x\right) d x d y+\int_{x=-1}^{x=0} \int_{y=0}^{y=1+x}\left(y^2-x\right) d x d y\)

+ \(\int_{x=-1}^{x=0} \int_{y=-1-x}^{y=0}\left(y^2-x\right) d x d y+\int_{x=0}^{x=1} \int_{y=x-1}^{y=0}\left(y^2-x\right) d x d y\)

= \(\int_{x=0}^{x=1}\left[\frac{y^3}{3}-x y\right]_{y=0}^{y=1-x} d x+\int_{x=-1}^{x=0}\left[\frac{y^3}{3}-x y\right]_{y=0}^{y=1+x} d x\)

+\(\int_{x=-1}^{x=0}\left[\frac{y^3}{3}-x y\right]_{y=-1-x}^{y=0} d x+\int_{x=0}^{x=1}\left[\frac{y^3}{3}-x y\right]_{y=x-1}^{y=0} d x\)

= \(\int_0^1\left[\frac{(1-x)^3}{3}-x(1-x)\right] d x+\int_{-1}^0\left[\frac{(1+x)^3}{3}-x(1+x)\right] d x\)

–\(\int_{-1}^0\left[\frac{(-1-x)^3}{3}-x(-1-x)\right] d x – \int_0^1\left[\frac{(x-1)^3}{3}-x(x-1)\right] d x\)

= \(\int_0^1\left[\frac{(1-x)^3}{3}-x+x^2\right] d x+\int_{-1}^0\left[\frac{(1+x)^3}{3}-x-x^2\right] d x\)

+\(\int_{-1}^0\left[\frac{(1+x)^3}{3}-x-x^2\right] d x-\int_0^1\left[\frac{(x-1)^3}{3}-x^2+x\right] d x\)

= \(\left[\frac{(1-x)^4}{-12}-\frac{x^2}{2}+\frac{x^3}{3}\right]_0^1+\left[\frac{(1+x)^4}{12}-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-1}^0\)

+\(\left[\frac{(1+x)^4}{12}-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-1}^0-\left[\frac{(x-1)^4}{12}-\frac{x^3}{2}+\frac{x^2}{2}\right]_0^1\)

= \(\left[0-\frac{1}{2}+\frac{1}{3}+\frac{1}{12}\right]+\left[\frac{1}{12}-0-0-0+\frac{1}{2}+\frac{1}{3}\right]\)

+\(\left[\frac{1}{12}-0-0-0+\frac{1}{2}-\frac{1}{3}\right]-\left[0-\frac{1}{3}+\frac{1}{2}-\frac{1}{12}+0-0\right]\)

= \(\frac{-6+4+1}{12}+\frac{1+6-4}{12}+\frac{1+6-4}{12}-\frac{-4+6-1}{12}=-\frac{1}{12}+\frac{3}{12}+\frac{3}{12}-\frac{1}{12}=\frac{4}{12}=\frac{1}{3}\).

Case (1): Line integral along AB: y = 1-x, dy = -dx, x varies from 1 to 0.

∴ \(\int_{C_1} x y d x+x y^2 d y=\int_1^0 x(1-x) d x+x(1-x)^2(-d x)=\int_1^0\left(x-x^2-x+2 x^2-x^3\right) d x\)

= \(\int_1^0\left(x^2-x^3\right) d x=\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_1^0=-\left[\frac{1}{3}-\frac{1}{4}\right]=-\frac{1}{12}\)

Case (2): Line integral along BC: y = 1+x, dy = dx, x varies from 0 to -1.

∴ \(\int_{C_2} x y d x+x y^2 d y=\int_0^{-1} x(1+x) d x+x(1+x)^2 d x=\int_0^{-1}\left(x+x^2+x+2 x^2+x^3\right) d x\)

= \(\int_0^{-1}\left(2 x+3 x^2+x^3\right) d x=\left[x^2+x^3+\frac{x^4}{4}\right]_0^1=1-1+\frac{1}{4}=\frac{1}{4}\)

Case (3): Line integral along CD: y = -1, dy = -dx, x varies from -1 to 0.

∴ \(\int_{C_3} x y d x+x y^2 d y=\int_{-1}^0 x(-1-x) d x+x(-1-x)^2(-d x)\)

=\(\int_{-1}^0\left(-x-x^2-x-2 x^2-x^3\right) d x\)

= \(-\int_{-1}^0\left(2 x+3 x^2+x^3\right) d x=-\left[x^2+x^3+\frac{x^4}{4}\right]_0^{-1}=1-1+\frac{1}{4}=\frac{1}{4}\)

Case (4): Line integral along DA: y = x-1, dy = dx, x varies from 0 to 1.

∴ \(\int_{C_4} x y d x+x y^2 d y=\int_0^{1} x(x-1) d x+x(x-1)^2 d x=\int_0^{1}\left(x^2-x+x^3-2 x^2+x\right) d x\)

= \(\int_0^1\left(x^3-x^2\right) d x=\left[\frac{x^4}{4}-\frac{x^3}{3}\right]_0^1=\frac{1}{4}-\frac{1}{3}=-\frac{1}{12}\)

∴ \(\int_S\)xy dx+xy² dy=1/12+1/4+1/4−1/12=1/3

∴  Stoke’s theorem is verified.

47. Verify Stoke’s theorem for the function F =x²i + xy j integrated round the square in the plane z= 0 whose sides are along the line x = 0, y = 0, x = a, y = a.

Solution:

∴ \({F}=x^2 \mathbf{i}+x y \mathbf{j} . \quad \text{curl} \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x^2 & x y & 0
\end{array}\right|=\mathbf{i}(0-0)-\mathbf{j}(0-0)+\mathbf{k}(y-0)=y \mathbf{k}\)

F. dr = (x2i + xyj) . (dx i + dy j + dz k) = x2 dx + xy dy

Let N be the unit normal vector to the surface of the square.

By Stoke’s theorem \(\int_S \boldsymbol{F} \cdot d \mathbf{r}=\int_S \text{cur} l \boldsymbol{F} \cdot \mathbf{N} d S\)

Since the surface of the square lies in the xy-plane, N = k.

⇒ \(\int_S \text{curl} \mathbf{F} \cdot \mathbf{N}dS =\int_S y k \cdot{k} d S=\int_S y d S=\int_0^a \int_0^a y d x d y\)

=\(\int_0^a\left[\frac{y^2}{2} \right]_0^a d x=\int_0^a \frac{a^2}{2} d x=\left[\frac{a^2 x}{2}\right]_0^a=\frac{a^3}{2}.\)

Case (1): Along the side OA: y = 0 and x varies from 0 to a.

\(\int_{c_1} F \cdot d \mathbf{r}=\int_{c_1} x^2 d x+x y d y=\int_{x=0}^{x=a} x^2 d x=\left[\frac{x^3}{3}\right]_0^a=\frac{a^3}{3}\)

Case (2): Along the side AB: x = a, dx = 0 and y varies from 0 to a.

\(\int_{c_2} F \cdot d \mathbf{r}=\int_{C_2} x^2 d x+x y d y=\int_{y=0}^{y=a} a y d y=\left[\frac{a y^2}{2} \right]_0^a=\frac{a^3}{2}.\)

Case (3): Along the side BC: y = a, dy = 0 and x varies from a to 0.

\(\int_{C3} \mathbf{F} \cdot d \mathbf{r}=\int_{C3} x^2 d x+x y d y=\int_{x=a}^{x=0} x^2 d x=\left[\frac{x^3}{3}\right]_a^0=-\frac{a^3}{3}\)

Case (4): Along the side CO: x = 0, y varies from a to 0.

\(\int_{C_4} \boldsymbol{F} \cdot d \boldsymbol{r}=\int_{C_4} x^2 d x+x y d y=0\)

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_{c_1} \mathbf{F} \cdot d \mathbf{F}+\int_{c_2} \mathbf{F} \cdot d \mathbf{r}+\int_{c_3} \mathbf{F} \cdot d \mathbf{r}+\int_{c_4} \mathbf{F} \cdot d \mathbf{r}=\frac{a^3}{3}+\frac{a^3}{2}-\frac{a^3}{3}+0=\frac{a^3}{2}\)

=\(\int_S\) curl F.NdS

∴ Stoke’s theorem is verified

48. Verify Stoke’s theorem to evaluate \(\oint_C\) F .dr where F =y²i + x² j- (x+ z)k and C is the boundary of the triangle with vertices (0, 0, 0), (1, 0, 0), (1, 1, 0).

Solution:

F = y² i + x² j – (x + z) k.

∴ \(\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y^2 & x^2 & -x-z
\end{array}\right|=\mathbf{i}(0-0)-\mathbf{j}(-1-0)+\mathbf{k}(2 x-2 y)=\mathbf{j}+(2 x-2 y) \mathbf{k}\)

Let N be the unit normal vector to the surface of the triangle.

Since the triangle lies in xy-plane, N=k, and dS=dx dy.

In xy -plane, the vertices of the triangle are O(0,0), A(1,0), and B(1,1).

By Stokes theorem,

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S=\int_S[\mathbf{j}+(2 x-2 y) \mathbf{k}] \cdot \mathbf{k} d S\)

= \(\iint_R(2 x-2 y) d x d y=\int_{x=0}^{x=1} \int_{y=0}^{y=x}(2 x-2 y) d x d y\) [∵ Along OB, x = y]

= \(\left.=\int_{x=0}^{x=1}\left[2 x y-y^2\right]{ }_{y=0}^{y=x} d x=\int_0^1\left(2 x^2-x^2\right) d x=\int_0^1 x^2 d x=\frac{x^3}{3}\right]_0^1=\frac{1}{3}\)

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_{O A} \mathbf{F} \cdot d \mathbf{r}+\int_{A B} \mathbf{F} \cdot d \mathbf{r}+\int_{B O} \mathbf{F} \cdot d \mathbf{r}\)

1. Along OA, y = 0, z = 0, dy = 0, dz = 0, x varies from 0 to 1

∴ \(\int_{O A} \mathbf{F} \cdot d \mathbf{r}=\int_0^{1} y^2 d x=0\)

2. Along AB, x = 1, z = 0, dx = 0, dz = 0, y varies from 0 to 1

∴ \(\left.\int_{A B} \mathbf{F} \cdot d \mathbf{r}=\int_0^1 x^2 d y=\int_0^1 d y=y\right]_0^1=1\)

3. Along BO, x = y, z = 0, dx = dy and x varies from 1 to 0.

∴ \(\left.\int_{B O} \mathbf{F} \cdot d \mathbf{r}=\int_1^0 y^2 d x+x^2 d y=\int_1 x^2 d x+x^2 d x=\int_1 2 x^2 d x=\frac{2 x^{30}}{3}\right]_1=-\frac{2}{3}\)

∴ \(\int_C F \cdot d \mathbf{F}=0+1-\frac{2}{3}=\frac{1}{3}\)

∴ Stoke’s theorem is verified.

49. Verify Stoke’s theorem for \(\oint_C\) F =- y³i + x³j,  where S is the circular disc x² +y² ≤ 1, z= 0.

Solution:  F=−y³i+x³j.

The boundary of C of S is a circle in xy -plane, x² +y²= 1, z=0

The parametric equations are x=cos θ,y=sin θ,z=0 where 0≤ θ≤2π.

∴ \(\int_c \mathbf{F} \cdot d \mathbf{r}=\int_c-y^3 d x+x^3 d y=\int_{\theta=0}^{\theta=2 \pi}-\sin ^3 \theta(-a \sin \theta) d \theta+\cos ^3 \theta(a \cos \theta) d \theta\)

Put x = sin θ

∴ dx = cos θ dθ

x = 0 ⇒ θ = 0

x = 1 ⇒ θ = π/2

= \(\int_0^{2 \pi}\left(\cos ^4 \theta+\sin ^4 \theta\right) d \theta=\int_0^{2 \pi}\left[\left(\cos ^2 \theta+\sin ^2 \theta\right)^2-2 \cos ^2 \theta \sin ^2 \theta\right] d \theta\)

= \(\int_0^{2 \pi}\left[1-\frac{1}{2} \sin ^2 2 \theta\right] d \theta=\int_0^{2 \pi}\left[1-\frac{1-\cos 4 \theta}{4}\right] d \theta\)

= \(\int_0^{2 \pi}\left[\frac{3+\cos 4 \theta}{4}\right] d \theta=\left[\frac{3 \theta}{4}+\frac{\sin 4 \theta}{16}\right]_0^{2 \pi}=\left[\frac{3 \pi}{2}+0\right]=\frac{3 \pi}{2}\)

∴ \(\nabla \times \mathbf{F}=\left[\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
-y^3 & x^3 & 0
\end{array}\right]=\mathbf{i}(0-0)-\mathbf{j}(0+0)+\mathbf{k}\left(3 x^2+3 y^2\right)=3\left(x^2+y^2\right) \mathbf{k}\)

Let R be the projection of S in the xy-plane.

∴ \(\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S=\iint_R 3\left(x^2+y^2\right) \mathbf{k} \cdot \mathbf{k} d x d y=3 \iint_R\left(x^2+y^2\right) d x d y\)

= \(3 \int_{x=-1}^{x=1} \int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}}\left(x^2+y^2\right) d x d y=12 \int_{x=0}^{x=1} \int_{y=0}^{y=\sqrt{1-x^2}}\left(x^2+y^2\right) d x d y\)

= \(12 \int_{x=0}^{x=1}\left[x^2 y+\frac{y^3}{3}\right]_0^{\sqrt{1-x^2}} d x=12 \int_0^1\left[x^2 \sqrt{1-x^2}+\frac{1-x^2}{3} \sqrt{1-x^2}\right] d x\)

= \(4 \int_0^1\left(1+2 x^2\right) \sqrt{1-x^2} d x=4 \int_0^{\pi / 2}\left(1+2 \sin ^2 \theta\right) \cos \theta \cos \theta d \theta\)

= \(4 \int_0^{\pi / 2}\left(\cos ^2 \theta+2 \sin ^2 \theta \cos ^2 \theta\right) d \theta=4\left[\frac{1}{2} \cdot \frac{\pi}{2}+2 \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right]=4\left[\frac{\pi}{4}+\frac{\pi}{8}\right]=\frac{3 \pi}{2}\)

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S\)

∴ Stoke’s theorem is verified.

50. Verify Stoke’s theorem for F = (2x-y) i -yz² j -y²zk, where S is the upper half surface of the sphere x² +y² +z²= 1 and C is its boundary.

Solution: The boundary C os S is a circle in xy-plane, i.e…,  x² +y² +z²= 1 z=0.

The parametric equations are x=cos t,y= sin t, 0≤t≤2π

F=(2x-y)i-yz²j+y²zk

r=xi+yj+zk⇒dr=dxi+dyj+dzk

F.dr=(2x-y)dx-yz² dy+y²z dz=(2cos t-sin t) (-sin t)(dt)

=(sin² t-2 cos t sin t)dt=(sin² t-sin 2t)dt.

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_0^{2 \pi}\left(\sin ^2 t-\sin 2 t\right) d t=4 \int_0^{\pi / 2} \sin ^2 t d t+\left[\frac{\cos 2 t}{2}\right]_0^{2 \pi}\)

= \(4 \cdot \frac{1}{2} \cdot \frac{\pi}{2}+\frac{1}{2}[\cos 4 \pi-\cos 0]=\pi+\frac{1}{2}(1-1)=\pi\)

Also \(\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x-y & -y z^2 & -y^2 z
\end{array}\right|=\mathbf{i}(0-0)-\mathbf{j}(0-0)+\mathbf{k}(0+1)=\mathbf{k}\)

Let R be the projection of S in the xy-plane. Then k. N dS = dx dy.

∴ \(\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S=\int_S \mathbf{k} \cdot \mathbf{N} d S=\iint_R d x d y\)

= \(4 \int_{x=0}^{x=1} \int_{y=0}^{y=\sqrt{1-x^2}} d x d y=4 \int_{x=0}^{x=1}[y]{ }_{y=0}^{y=\sqrt{1-x^2}} d x=4 \int_0^1 \sqrt{\left(1-x^2\right)} d x\)

= \(4\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \text{Sin}^{-1} x\right]_0^1=4\left[\frac{1}{2} \times \frac{\pi}{2}\right]=\pi\)

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S\)

∴ Stoke’s theorem is verified.

51. Verify Stoke’s theorem for A = 2 yi + 3xj− z² k where S is the upper half surface of the sphere x²+y² + z² = 9 and C is its boundary.

Solution: The boundary Cof S is a circle in xy-plane  is the circle  x²+y²=9,z=0

The parametric equations are x=3 cos θ,y=3 sin θ , z=0

⇒dx =− 3 sin θ dθ ,dy =3 cos θ dθ,dz=0

∴ \(\int_C\)A.dr=\(\int_C\)(2yi+3xj−z²k).(dxi+dyj+dzk)

= \(\int_c 2 y d x+3 x d y-z d z=\int_0^{2 \pi} 2(3 \sin \theta)(-3 \sin \theta d \theta)+3(3 \cos \theta)(3 \cos \theta d \theta)-0\)

= \(\int_0^{2 \pi}\left[-18 \sin ^2 \theta+27 \cos ^2 \theta\right] d \theta=\int_0^{2 \pi}\left(27-45 \sin ^2 \theta\right) d \theta\)

= \(27(2 \pi-0)-45 \times 4 \times \frac{1}{2} \times \frac{\pi}{2}=54 \pi-45 \pi=9 \pi\)

∴ \(\nabla \times \mathbf{A}=\left|\begin{array}{lll}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 y & 3 x & -z^2
\end{array}\right|=\mathbf{i}(0-0)-\mathbf{j}(0-0)+\mathbf{k}(3-2)=\mathbf{k}\)

Let R be the projection of S in the xy-plane.

∴ \(\int_S(\nabla \times \mathbf{A}) \cdot \mathbf{N} d S=\iint_R(\mathbf{k} \cdot \mathbf{N}) d S=\iint_R d x d y\)

= \(\int_{x=-3}^{x=3} \int_{y=-\sqrt{9-x^2}}^{y=\sqrt{9-x^2}} d x d y=4 \int_{x=0}^{x=3} \int_{y=0}^{y=\sqrt{9-x^2}} d x d y\)

= \(4 \int_{x=0}^{x=3}\left[y\right]_{y=0}^{\sqrt{=9-x}} d x=4 \int_{0}^3 \sqrt{9-x^2} d x=4\left[\frac{x}{2} \sqrt{9-x^2}+\frac{9}{2} \text{Sin}^{-1} \frac{x}{3}\right]_0^3=4\left[\frac{9}{2} \times \frac{\pi}{2}\right]=9 \pi\)

∴ \(\int_S(\nabla \times \mathbf{A}) \cdot \mathbf{N} d S=\int_c \mathbf{A} \cdot d \mathbf{r}\)

∴ Stoke’s theorem is verified.

52. Apply Stoke’s theorem to evaluate\(\int_C\) (y dx + z dy+x dz) where C is the curve of intersection of x²+ y² + z²= a² and x + z =a.

Solution:  

For the sphere x² + y² + z² = a², Centre O = (0, 0, 0), radius = a.

Let A be the centre and r be the radius of the circle of intersection of x² + y² + z² = a² and x + z = a.

∴ \(O A=\left|\frac{0+0-a}{\sqrt{2}}\right|=\frac{a}{\sqrt{2}}\)

⇒ \(r^2=a^2-O A^2=a^2-\frac{a^2}{2}=\frac{a^2}{2} \Rightarrow r=\frac{a}{\sqrt{2}}\)

Since the plane of the circle is perpendicular to the y-axis, the vector normal to the plane is j.  ∴ N = j.

Let F = yi + zj + xk

∴ \(\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y & z & x
\end{array}\right|=\mathbf{i}(0-1)-\mathbf{j}(1-0)+\mathbf{k}(0-1)\)

= – i – j – k

By stoke’s theorem, \(\int_C(y d x+z d y+x d z)=\int_C \mathbf{F} \cdot d \mathbf{r}=\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S\)

= \(\int_S(-\mathbf{i}-\mathbf{j}-\mathbf{k}) \cdot \mathbf{j} d S=\int_S(-1) d S=-S=-(\text { Area of the circle })={-}\left(\pi a^2 / 2\right)\)

53.Evaluate \(\int_S\)∇xF.N dS using Stoke’s theorem, where F = (2x -y) i -yz²-y²z k and S is tlie upper halfof the sphere x² +y² +z² = 1 and C is its boundary.

Solution:  The boundary C of S is a circle in xy-plane, i.e x² +y² =1,z=0.

The parametric equations are x=cos t, y=sint, 0≤t ≤2π.

F = (2x – y)i – y z j + y z k.

r = xi + yj + zk ⇒ dr = dxi + dyj + dzk.

= \(\mathbf{F} \cdot d \mathbf{r}=(2 x-y) d x-y z^2 d y+y^2 z d z=(2 \cos t-\sin t)(-\sin t)(d t)\)

= \(\left(\sin ^2 t-2 \cos t \sin t\right) d t=\left(\sin ^2 t-\sin 2 t\right) d t\)

By Stoke’s Theorem,

∴ \(\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d s=\int_C \mathbf{F} \cdot d \mathbf{r}=\int_0^{2 \pi}\left(\sin ^2 t-\sin 2 t\right) d t=4 \int_0^{\pi / 2} \sin ^2 t d t+\left[\frac{\cos 2 t}{2}\right]_0^{2 \pi}\)

= \(4 \cdot \frac{1}{2} \cdot \frac{\pi}{2}+\frac{1}{2}[\cos 4 \pi-\cos 0]=\pi+\frac{1}{2}(1-1)=\pi\)

54. Evaluate  \(\int_S\)(∇x F). N dS where F =yi + (x- 2xz) j -xy k and S is the surface S  of the sphere x² +y² + z² = a², above the xy-plane.

Solution:  The boundary C of S is a circle in xy-plane, i.e. x² +y² =1,z=0.

The parametric equations are x=a cos t, y= a sin t,z=0, 0≤t ≤2π. And dx =-a isn’t dt, dy= a cos t dt, dz=0.

∴ \(\int_S \nabla \times \mathbf{F} \cdot \mathbf{N} d S=\int_C \mathbf{F} \cdot d \mathbf{r}=\int_C[y \mathbf{i}+(x-2 x z) \mathbf{j}-x y \mathbf{k}] \cdot d \mathbf{r}\)

= \(\int_C y d x+(x-2 x z) d y-x y d z=\int_{t=0}^{t=2 \pi} a \sin t(-a \sin t) d t+(a \cos t-0) a \cos t d t\)

= \(a^2 \int_{t=0}^{t=2 \pi}\left(\cos ^2 t-\sin ^2 t\right) d t\)

= \(a^2 \int_{t=0}^{t=2 \pi} \cos 2 t d t=\left[\frac{a^2}{2} \sin 2 t\right]_{t=0}^{t=2 \pi}=0\)

55. Prove that \(\oint_C\) f∇g .dr  \(\oint_C\)(∇fx ∇g) .N dS.

Solution: By Stoke’s theorem,

∴ \(\oint_c(f \nabla g) \cdot d \mathbf{r}=\int_S[\nabla \times(f \nabla g)] \cdot \mathbf{N} d S=\int_s[\nabla f \times \nabla g+f \text { curl } \text{grad} g] \cdot \mathbf{N} d S\)

= \(\int_S(\nabla f \times \nabla g) \cdot \mathbf{N} d S\) ∵ curl grad g = 0.

56. Prove that \(\int_S\) curlφ f .dS = \(\int_C\) φ f .dr− \(\int_S\) ∇φ x f.dS.

Solution: Applying Stoke’s theorem to the function φ f.

∴ \(\oint_c(f \nabla g) \cdot d \mathbf{r}=\int_s[\nabla \times(f \nabla g)] \cdot \mathbf{N} d S=\int_s[\nabla f \times \nabla g+f \mathrm{curl} \text{grad} g] \cdot \mathbf{N} d S\)

∴ \(\int_s(\nabla f \times \nabla g) \cdot \mathbf{N} d S .\)

∵ curl grad g = 0.

57.Prove that \(\oint_C\)f∇f.dr = 0.

Solution:

By Stoke’s theorem, \(\int_C(f \nabla f) \cdot d \mathbf{r}=\int_S(c u r l f \nabla f) \cdot \mathbf{N} d S\)

= \(\int_S \phi \text{curl} \mathbf{f} \cdot d \mathbf{S}=\int_C \phi \mathbf{f} \cdot d \mathbf{r}-\int_S \nabla \phi \times \mathbf{f} \cdot d \mathbf{S} .\)

Vector Integration- 4  Exercise−(4)

 

1. Evaluate \(\int_0^1\)(e^t + e^-2t j + Z k) dt.

Solution: 

⇒ \(\left.\left.\left.\int_0^1\left(e^t \mathbf{i}+e^{-2 t} \mathbf{j}+t \mathbf{k}\right) d t=\mathbf{i} \int_0^1 e^t d t+\mathbf{j} \int_0^1 e^{-2 t} d t+\mathbf{k} \int_0^1 t d t=\mathbf{i} e^t\right]_0^1+\mathbf{j} \frac{e^{-2 t}}{-2}\right]_0^1+\mathbf{k} \frac{t^2}{2}\right]_0^1\)

= \(\mathbf{i}(e-1)+\mathbf{j}\left(\frac{e^{-2}}{-2}+\frac{1}{2}\right)+\frac{1}{2} \mathbf{k}=\mathbf{i}(e-1)+\mathbf{j}\left(\frac{1-e^{-2}}{2}\right)+\frac{1}{2} \mathbf{k}\)

2. IfF(Z) = ti + (t²– 2t) j + (3t² + 3t³) k then find \(\int_0^1\)f(t) dt.

Solution:

⇒ \(\int_0^1 \mathbf{f}(t) d t=\int_0^1\left[t \mathbf{i}+\left(t^2-2 t\right) \mathbf{j}+\left(3 t^2+3 t^3\right) \mathbf{k}\right] d t\)

= \(\int_0^1 t d t+\mathbf{j} \int_0^1\left(t^2-2 t\right) d t+\mathbf{k} \int_0^1\left(3 t^2+3 t^3\right) d t\)

= \(i\left[\frac{t^2}{2}\right]_0^1+\mathbf{j}\left[\frac{t^3}{3}-t^2\right]_0^1+\mathbf{k}\left[t^3+\frac{3 t^4}{4}\right]_0^1=\frac{1}{2} \mathbf{i}-\frac{2}{3} \mathbf{j}+\frac{7}{4} \mathbf{k} . .\)

3. If f(t) = (t−t²) i + 2t³ j- 3k, then find \(\int_1^2\)f(t) dt.

Solution:

⇒ \(\int_1^2 \mathbf{f}(t) d t=\int_1^2\left[\left(t-t^2\right) \mathbf{i}+2 t^3 \mathbf{j}-3 \mathbf{k}\right] d t=\mathbf{i} \int_1^2\left(t-t^2\right) d t+\mathbf{j} \int_1^2 2 t^3 d t-\mathbf{k} \int_1^2 3 a\)

= \(\mathbf{i}\left[\frac{t^2}{2}-\frac{t^3}{3}\right]_1^2+\mathbf{j}\left[\frac{t^4}{2}\right]_1^2-\mathbf{k}[3 t]=\mathbf{i}\left[2-\frac{8}{3}-\frac{1}{2}+\frac{1}{3}\right]+\mathbf{j}\left[8-\frac{1}{2}\right]-\mathbf{k}[6-3]\)

= \(-\frac{5}{6} i+\frac{15}{2} j-3 k\)

4. If f (t) = 2i- J + 2k and  f(3) = 4i- 2j + 3k then find\(\int_2^ 3\)f.\(\frac{d f}{d t}\) dt.

Solution:

\(\frac{d}{d t}\left\{f(t)^2\right\}=2 \mathbf{f}(t) \cdot \frac{d \mathbf{f}}{d t} \Rightarrow \int \mathbf{f}(t) \cdot \frac{d \mathbf{f}}{d t} d t=\frac{1}{2} \mathbf{f}(t)^2\)

∴ \(\left.\int_2^3\left(\mathbf{f} \cdot \frac{d \mathbf{f}}{d t}\right) d t=\frac{1}{2} \mathbf{f}(t)^2\right]_2^3=\frac{1}{2}\left[\mathbf{f}(3)^2-\mathbf{f}(2)^2\right]=\frac{1}{2}\left[(4 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k})^2-(2 \mathbf{i}-\mathbf{j}+2 \mathbf{k})^2\right]\)

= \(\frac{1}{2}[(16+4+9)-(4+1+4)]=\frac{1}{2}(29-9)=10\)

5. lf f(t) = 5t²i + tj-t³k. find\(\left(f \times \frac{d^2 \mathbf{f}}{d t^2}\right)\)dt.

Solution:

∴ \(\int_1^2\left(\mathbf{f} \times \frac{d^2 \mathbf{f}}{d t^2}\right) d t=\left[\mathbf{f} \times \frac{d \mathbf{f}}{d t}\right]_1^2=\left[-2 t^3 \mathbf{i}+5 t^4 \mathbf{j}-5 t^2 \mathbf{k}\right]_1^2=-14 \mathbf{i}+75 \mathbf{j}-15 \mathbf{k} .\)

6. If \(\frac{d^2 \mathbf{r}}{d t^2}\)= 6ti- 24t² + 4 sin t k, find r given that r = 2i + j and  \(\frac{d r}{d t}\)=−i−3k at t=0

Solution:

\(\frac{d \mathbf{r}}{d t}=\int \frac{d^2 \mathbf{r}}{d t^2} d t=\int\left[6 t \mathbf{i}-24 t^2 \mathbf{j}+4 \sin t \mathbf{k}\right] d t=3 t^2 \mathbf{i}-8 t^3 \mathbf{j}-4 \cos t \mathbf{k}+\mathbf{c}_1\)

At \(t=0, \frac{d \mathbf{r}}{d t}=-\mathbf{i}-3 \mathbf{k} \Rightarrow-4 \mathbf{k}+\mathbf{c}_1=-\mathbf{i}-3 \mathbf{k} \Rightarrow \mathbf{c}_1=-\mathbf{i}+\mathbf{k}\)

∴ \(\frac{d \mathbf{r}}{d t}=3 t^2 \mathbf{i}-8 t^3 \mathbf{j}-4 \cos t \mathbf{k}-\mathbf{i}+\mathbf{k}=\left(3 t^2-1\right) \mathbf{i}-8 t^3 \mathbf{j}+(1-4 \cos t)\)

r = \(\left.\int \frac{d \mathbf{r}}{d t} d t=\int\left[\left(3 t^2-1\right) \mathbf{i}-8 t^3 \mathbf{j}+(1-4 \cos t) \mathbf{k}\right)\right] d t=\left(t^3-t\right) \mathbf{i}-2 t^4 \mathbf{j}+(t-4 \sin t) \mathbf{k}+\mathbf{c}_2\)

At \(t=0, \mathbf{r}=2 \mathbf{i}+\mathbf{j} \Rightarrow \mathbf{c}_2=2 \mathbf{i}+\mathbf{j}\)

∴ \(r=\left(t^3-t\right) \mathbf{i}-2 t^4 \mathbf{j}+(t-4 \sin t) \mathbf{k}+2 \mathbf{i}+\mathbf{j}=\left(t^3-t+2\right) \mathbf{i}+\left(1-2 t^4\right) \mathbf{j}+(t-4 \sin t) \mathbf{k}\)

7. lf\(\frac{d^2 \mathbf{r}}{d t^2}\)=−k²r, show that \(\left(\frac{d r}{d t}\right)^2\) = c −k²r².

Solution:

Given that \(\frac{d^2 \mathbf{r}}{d t^2}=-k^2 \mathbf{r} \Rightarrow 2 \frac{d \mathbf{r}}{d t} \cdot \frac{d^2 \mathbf{r}}{d t^2}=- k^2\left(2 \mathbf{r} \cdot \frac{d \mathbf{r}}{d t}\right)\)

⇒ \(\int\left(2 \frac{d \mathbf{r}}{d t} \cdot \frac{d^2 \mathbf{r}}{d t^2}\right) d t=-k^2 \int\left(2 \mathbf{r} \cdot \frac{d \mathbf{r}}{d t}\right) d t \Rightarrow\left(\frac{d \mathbf{r}}{d t}\right)^2=-k^2 \mathbf{r}^2+\mathbf{c} \Rightarrow\left(\frac{d \mathbf{r}}{d t}\right)^2=\mathbf{c}-k^2 \mathbf{r}^2\)

8. IfA = ti− t²j+ (t- 1)k, B = 2 t² + 6tk, find (a)\(\int_1^2\)(A.B) dt   (b) \(\int_0^2\) (A x B) dt

Solution:

1. \(\mathbf{A} \cdot \mathbf{B}=\left[t \mathbf{i}-t^2 \mathbf{j}+(t-1) \mathbf{k}\right] \cdot\left[2 t^2 \mathbf{i}+6 t \mathbf{k}\right]=2 t^3+6 t(t-1)=2 t^3+6 t^2-6 t^2\)

∴ \(\int_0^2(\mathbf{A} \cdot \mathbf{B}) d t=\int_0^2\left(2 t^3+6 t^2-6 t\right) d t=\left[\frac{4 t^4}{4}+\frac{6 t^3}{3}-\frac{6 t^2}{2}\right]_0^2=16+16-12=20\)

2. \(\mathbf{A} \times \mathbf{B}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
t & -t^2 & t-1 \\
2 t^2 & 0 & 6 t
\end{array}\right|=\mathbf{i}\left(-6 t^3-0\right)-\mathbf{j}\left(6 t^2-2 t^3+2 t^2\right)+\mathbf{k}\left(0+2 t^4\right)\)

= \(-6 t^3 \mathbf{i}+\left(2 t^3-8 t^2\right) \mathbf{j}+2 t^4 \mathbf{k}\)

⇒ \(\int_0^2(\mathbf{A} \times \mathbf{B}) d t=\int_0^2\left[-6 t^3 \mathbf{i}+\left(2 t^3-8 t^2\right) \mathbf{j}+2 t^4 \mathbf{k}\right] d t=\left[-\frac{6 t^4}{4} \mathbf{i}+\left(\frac{2 t^4}{4}-\frac{8 t^3}{3}\right) \mathbf{J}+\frac{2 t^5}{5} \mathbf{k}\right]_0^2\)

= \(-24 \mathbf{i}+\left(8-\frac{64}{3}\right) \mathbf{j}+\frac{64}{5} \mathbf{k}=-24 \mathbf{i}-\frac{40}{3} \mathbf{j}+\frac{64}{5} \mathbf{k}\)

9. Evaluate \(\int_c\)(x² +y²) dx, where C is the arc of the parabola y²= 4ax between (0, 0)and (a, 2a).

Solution:

⇒ \(\int_C\left(x^2+y^2\right) d x\)

⇒ \(=\int_{x=0}^{x=a}\left(x^2+4 a x\right) d x\)

⇒ \(=\left[\frac{x^3}{3}+2 a x^2\right]_{x=0}^{x=a}\)

⇒ \(=\frac{a^3}{3}+2 a^3\)

∴ \(\frac{7 a^3}{3}\)

10. Evaluate \(\int_c \frac{d x}{(x+y)}\) , where  C is the curve x=at²,y=2at,0≤t≥2

Solution:

⇒ \(\int_C \frac{d x}{x+y}\)

⇒ \(=\int_{t=0}^{t=2} \frac{d\left(a t^2\right)}{a t^2+2 a t}\)

⇒ \(=\int_{t=0}^{t=2} \frac{2 a t d t}{a t^2+2 a t}\)

∴ \(=2 \int_0^2 \frac{d t}{t+2}\)=\(2 \log (t+2)]_0^2\)

11. Show that \(\int_c\)[(x-y)³ dx + (x-y)³ dy] = 3πa⁴ taken along the circle x²+y² = a² in    the counter clockwise sense.

Solution:

The parametric equations circle x+y = a are x=a cos θ, y= a sin θ.

dx= -a sin θ dθ, dy= a cos θ dθ and θ varies from 0 to 2π.

⇒ \(\int_C\left[(x-y)^3 d x+(x-y)^3 d y\right]\)

⇒ \(=\int_0^{2 \pi}(a \cos \theta-a \sin \theta)^3(-a \sin \theta d \theta)\)+ \((a \cos \theta-a \sin \theta)^3(a \cos \theta d \theta)\)

⇒ \(a^4 \int_0^{2 \pi}(\cos \theta-\sin \theta)^4 d \theta=a^4 \int_0^{2 \pi}\left(\cos ^4 \theta-4 \cos ^3 \theta \sin \theta+6 \cos ^2 \theta \sin ^2 \theta-4 \cos \theta \sin ^4 \theta+\sin ^4 \theta\right) d \theta\)

⇒ \(a^4 4 \int_0^{\pi / 2}\left(\cos ^4 \theta-0+6 \cos ^2 \theta \sin ^2 \theta-0+\sin ^4 \theta\right) d \theta\)

⇒ \(4 a^4 \int_0^{\pi / 2}\left[\left(\cos ^2 \theta+\sin ^2 \theta\right)^2+4 \cos ^2 \theta \sin ^2 \theta\right] d \theta\)

∴ \(4 a^4 \int_0^{\pi / 2}\left(1+4 \cos ^2 \theta \sin ^2 \theta\right) d \theta=4 a^4\left[\frac{\pi}{2}+4 \times \frac{1}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right]=3 \pi a^4\).

12. Define line integral and explain the Cartesian form of line integral.

An integral which is to be evaluated along a curve is called a ” Line Integral. Suppose r=xi+yj+zk defines a piecewise smooth curve C joining two points A and B. Suppose F is a vector point function defined and continuous along C. If s denotes the arc length of the curve C then \(\frac{d r}{d s}\)= T is a unit vector along the tangent to the curve C at the point r.

The component of the vector F along the tangent is F.\(\frac{d r}{d s}\). The integral of F.\(\frac{d r}{d s}\) along from A to B written as \(\int_A^B\left(F \cdot \frac{d r}{d s}\right)\) ds \(=\int_A^B \mathbf{F} \cdot d \boldsymbol{r}\)  \(=\int_C F \cdot d r\) . is an example of a line integral. it is called tangent line integral of F along C.

Cartesian form: If F = F₁i +F₂ j + F₃k then

F.dr =(F₁i +F₂ j + F₃k). (dxi+dyj+dzk)= F₁dx+F₂dy+F₃dz

∴ \(\int_C\)F.dr=\(\int_C\)F₁dx+F₂dy+F₃dz

If the parametric equation of the curve C are x=x(t),y(t), z=z(t) and if t₁ at A, t=t₂ at  B then \(\int_C\)F.dr=\(=\int_{t_1}^{t_2}\left[F_1 \frac{d x}{d t}+F_2 \frac{d y}{d t}+F_3 \frac{d z}{d t}\right]\)dt

13. If F- 3xyi- 5zj + 10xk, evaluate \(\int_c\)F.dr along the curve, x = t² + 1, y = 2t², z= t³  from t = 1 to t= 2.

Solution:

Let \(\mathbf{r}=x \mathbf{i}+y \mathbf{J}+z \mathbf{k}=\left(t^2+1\right) \mathbf{i}+2 t^2 \mathbf{j}+t^3 \mathbf{k} \Rightarrow \frac{d \mathbf{r}}{d t}=2 t \mathbf{l}+4 t \mathbf{j}+3 t^2 \mathbf{k} …\)

⇒ \(\mathbf{F}=3 x y \mathbf{i}-5 z \mathbf{j}+10 x \mathbf{k}=3\left(t^2+1\right) 2 t^2 \mathbf{i}-5 t^3 \mathbf{j}+10\left(t^2+1\right) \mathbf{k}\)

⇒ \(6 t^2\left(t^2+1\right) \mathbf{i}-5 t^3 \mathbf{J}+10\left(t^2+1\right) \mathbf{k}\)

⇒ \(\mathbf{F} \cdot \frac{d \mathbf{r}}{d t}=6 t^2\left(t^2+1\right) \cdot 2 t-5 t^3 \cdot 4 t+10\left(t^2+1\right) \cdot 3 t^2\)

⇒ \(12 t^5+12 t^3-20 t^4+30 t^4+30 t^2=12 t^5+10 t^4+12 t^3+30 t^2\)

∴ \(\left.\int \mathbf{F} \cdot d \mathbf{r}=\int\left(\mathbf{F} \cdot \frac{d \mathbf{r}}{d t}\right) d t=\int_1^2\left(12 t^5+10 t^4+12 t^3+30 t^2\right) d t=12 t^6+2 t^5+3 t^4+10 t^3\right]_1^2\)

= 138 + 64 + 48 + 80 – 2 – 2 – 3 – 10 = 320 – 17 = 303.

14. If F- 3xyi- 5zj + l0xk, evaluate \(\int_c\)F.dr along the curve, x =t² , y = 2t², from z = t³  to t= 1 t=2

Solution: Let r=xi+yj+zk=t²i+2t²j+t³k ⇒\(\frac{d r}{d t}\)=2ti+4tj+3t²k

F=3xyi-5zj+10xk=3t²(2t²)i-5t³j+10t²k= 6t⁴i-5t³j+10t²k

⇒ \(\mathbf{F} \cdot \frac{d \mathbf{r}}{d t}=\left(6 t^4 \mathbf{i}-5 t^3 \mathbf{j}+10 t^2 \mathbf{k}\right) \cdot\left(2 t \mathbf{i}+4 t \mathbf{j}+3 t^2 \mathbf{k}\right)\)

⇒ \(\left(6 t^4\right)(2 t)+\left(-5 t^3\right)(4 t)+\left(10 t^2\right)\left(3 t^2\right)=12 t^5-20 t^4+30 t^4=12 t^5+10 t^4\)

∴ \(\left.\int \mathbf{F} \cdot d \mathbf{r}=\int \mathbf{F} \cdot \frac{d \mathbf{r}}{d t}\right) d t=\int\left(12 t^5+10 t^4\right) d t=\left[2 t^6+2 t^5\right]_1^2=(128+64)-(2+2)=188\).

15. Find ∫F .dr where F = xyi+yzj + zxk over the curve Curve r = ti + t²j + t³k, t varying from −1 to 1 .

Solution:

⇒ \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+3 \mathbf{k}=t \mathbf{i}+t^2 \mathbf{j}+t^3 \mathbf{k} \Rightarrow x=t, y=t^2, z=t^3 \text { and } \frac{d \mathbf{r}}{d t}=\mathbf{i}+2 t \mathbf{j}+3 t^2 \mathbf{k}\)

⇒ \(\mathbf{F}=x y \mathbf{i}+y z \mathbf{j}+z x \mathbf{k}=t^3 \mathbf{i}+t^{\mathbf{5}} \mathbf{j}+t^4 \mathbf{k}\)

∴ \(\int \mathbf{F} \cdot d \mathbf{r}=\int\left(\mathbf{F} \cdot \frac{d \mathbf{r}}{d t}\right) d t=\int_{-1}^1\left(t^3+2 t^5+3 t^5\right) d t=\int_{-1}^1\left(t^3+5 t^5\right) d t=\left[\frac{t^4}{4}+5 \frac{t^7}{7}\right]_{-1}^1\)

⇒ \(\left(\frac{1}{4}+\frac{5}{7}\right)-\left(\frac{1}{4}-\frac{5}{7}\right)=\frac{10}{7}\)

16. Find \(\int_c\)F.dr where C is the arc of y- x² in xy-plane from (0, 0) to (1, 1) and F=x²i +y²j.

Solution:  The equation of the given curve is y= x²⇒ 2x dx. Given F=xi+yj+zk the curve lies in xy plane from (0,0) t0 (1,2) the limits of integration are  x=0 to  x=1

∴ \(\int_c F \cdot d r\)

⇒ \(=\int_C x^2 d x+y^2 d y\)

⇒ \(=\int_0^1 x^2 d x+x^4(2 x d x)\)

⇒ \(=\int_0^1\left(x^2+2 x^3\right) d x\) \(=\left[\frac{x^3}{3}+\frac{2 x^6}{6}\right]_0^1\)

∴ \(=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\)

17. If F = 3xy i -y² j, evaluate  \(\int_c\)Fdr where C is the curve y = 2x² in the xy-plane from (0, 0) to (1, 2).

Solution:

The equation of given curve C is y= x²⇒ 4x dx. The integration was performed in xy- plane along C from (0,0) to (1,2).

∴ x varies from 0 to 1.

⇒ \(\int_C F \cdot d r\)

⇒ \(\int_C\left(3 x y \mathbf{i}-y^2 \mathbf{j}\right) \cdot(d x \mathbf{i}+d y \mathbf{j}+d z \mathbf{k})\)

⇒ \(=\int_C\left(3 x y d x-y^2 d y\right)\)

⇒ \(=\int_{x=0}^{x=1} 3 x \cdot\left(2 x^2\right) d x-4 x^4 \cdot 4 x d x\)

⇒ \(=\int_0^1\left(6 x^3-16 x^3\right) d x\) \(\left.=6 \frac{x^4}{4}-16 \frac{x^6}{6}\right]_0^1\)

⇒ \(=\frac{3}{2}-\frac{8}{3}=\frac{9-16}{6}\)

∴ \(-\frac{7}{6}\).

18. Evaluate   \(\int_c\)F.dr, where F = x²y² i +yj and the curve C is y= 4x in the xy-plane c from (0, 0) to (4, 4).

Solution:

The equation of curve C is y²=4x⇒2y dy =4dx⇒ ydy =2dx.

∴ x varies from 0 to 4.

⇒ \(\int_C F \cdot d r\)

⇒ \(\int_C x^2 y^2 d x+y d y\)

⇒ \(\int_0^4 x^2 \cdot 4 x d x+2 d x\)

⇒ \(\int_0^4\left(4 x^3+2\right) d x\) \(\left.=x^4+2 x\right]_0^4\)

=256+8=264.

19. Evaluate \(\int_c\)F.dr where F = 3x²i + (2xz -y)j + zk along the straight line C from(0,0,0) to (1,2).

Solution:

Equation of the joining (0,0,0) to (2,1,3) are  \(\frac{x}{2}=\frac{y}{1}=\frac{z}{3}=t\) (say).

∴ x = 2t, y = t, z = 3t, t varies from 0 to 1. Also \(\frac{d x}{d t}=2, \frac{d y}{d t}=1, \frac{d z}{d t}=3\)

⇒ \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \Rightarrow d \mathbf{r}=d x \mathbf{i}+d y \mathbf{j}+d z \mathbf{k}=\left(\frac{d x}{d t} \mathbf{i}+\frac{d y}{d t} \mathbf{j}+\frac{d z}{d t} \mathbf{k}\right) d t=(2 \mathbf{i}+\mathbf{j}+3 \mathbf{k}) d t\)

⇒ \(\mathbf{F}=3 x^2 \mathbf{i}+(2 x z-y) \mathbf{j}+z \mathbf{k}=12 t^2 \mathbf{i}+\left(12 t^2-t\right) \mathbf{j}+3 t \mathbf{k}\)

⇒ \(\int_C \mathbf{F}. d \mathbf{r}=\int_C\left[12 t^2 \mathbf{i}+\left(12 t^2-t\right) \mathbf{j}+3 t \mathbf{k}\right] \cdot(2 \mathbf{i}+\mathbf{j}+3 \mathbf{k}) d t=\int_C\left(24 t^2+12 t^2-t+9 t\right) d t\)

∴ \(\left.\int_0^1\left(36 t^2+8 t\right) d t=12 t^3+4 t^2\right]_0^1=12+4=16\)

20. Evaluate \(\int_c\)A.dr where C is the line joining (0,0,0) and (2,1,1), given A = (2y + 3)i + xzj + (yz−x)k.

Solution:

The equations of the line joining (0,0,0) and (2,1,1) are  \(\frac{x}{2}=\frac{y}{1}=\frac{z}{3}\) =t.

Then along the line C,x=2t,y=t,z=t.

∴ At (0,0,0,) , t=0 and at (2,1,1,) , t=1.

r=xi+yj+zk=2ti+rj=tk.

∴ dr=(2i+j+k)dt.

∫A.dr=\(\int_C[2(2 y+3)+1(x z)+1(y z-x)] d t\)

=\(\int_0^1[2(2 t+3)+(2 t \cdot t)+(t \cdot t-2 t)] d t\)

∴ \(\left.=\int_0^1\left(2 t^2+2 t+6\right) d t=t^3+t^2+6 t\right]_0^1\)=8.

21. Evaluate \(\oint_c\)F .dr where C is the circle x²+y² = 1, z = 0 and F =yi + zj +xk.

Solution:

The equation of the circle is x+y+=1,z=0

∴ dz=0.

In parametric form, x=cos θ,y=sin θ, z=0 and θ varies from 0 to 2π.

∴ \(\int_C F \cdot d r\)=\(\int_C(y \mathbf{i}+z \mathbf{j}+x \mathbf{k}) \cdot(\mathbf{i} d x+\mathbf{j} d y+\mathbf{k} d z)\)

⇒ \(\int_C(y d x+z d y+x d z)\)=\(\int_C y d x\)

⇒ \(=\int_{\theta=0}^{2 \pi} \sin \theta(-\sin \theta) d \theta\)

⇒ \(\theta=-4 \int_0^{\pi / 2} \sin ^2 \theta d \theta\)

=4(1/2)(π/2)=−π

22. If F = (3x² + 6y)i- 14yzi + 20xz² k, evaluate \(\int_c\)F.dralong the straight line joining (0, 0, 0) to (1, 0, 0) to(l, 1, 0) to (1, 1, 1)5

Solution:

(1) Line integral along the line from (0, 0, 0) to (1, 0, 0):  Here y = 0, z = 0, x varies from 0 to 1, dy = 0, dz = 0.

\(\int_{c_1} \mathbf{F} \cdot d r\)=\(\int_0^1 3 x^2 d x\)

=1.

(2) Line integral along the line from (1,0,0) to (1,1,0):

Here x=1, y varies from 0 to 1, z=0; dx=0, dz=0.

∴ \(\int_{c_2} \mathbf{F} \cdot d r\)=\(\int_0^1\)=0

(3) Line integral along the line from (1,1,0) t0 (1,1,1):

Here x=1,y=1, and z varies from 0 to 1. dx=0, dy=0.

⇒ \(\int_{c_3} \mathbf{F} \cdot d r\)=\(\int_0^1 20 z^2 d z\)=\(\frac{20 z^3}{3}\)\(]_0^1\)

⇒ \(\frac{20}{3}\)

∴ \(\int_{c} \mathbf{F} \cdot d r\)\(\)=\(\int_{c_1} \mathbf{F} \cdot d r\)+\(\int_{c_2} \mathbf{F} \cdot d r\)+\(\int_{c_3} \mathbf{F} \cdot d r\)=1+0+\(\frac{20}{3}\)=\(\frac{23}{3}\).

23. If F =(x² +y²)i− 2xy j, evaluate \(\oint_c\)F .dr where the curve C is the rectangle in the  xy-plane bounded by y = 0,y = b, x = 0, x = a.

Solution:

Since the integration takes place in xy -plane (z=0),

∴ \(\oint_c\)F.dr= \(\oint_c\)F₁dx+F₂ dy=\(\oint_c\)(x²+y²)dx-2xy dy

(1) Line integral along OP: Here y=0, dy=0 and x varies from o to a. \(\int_{O P} F \cdot d r\)=\(\int_0^a x^2 d x\)=\(\frac{a^3}{3}\)

(2)  Line integral along PQ: Here x=a, dx=0, and y changes from 0 to b.

∴ \(\int_{P Q} F \cdot d r\)=\(\int_a^b(-2 a y) d y\)=\(\left.-a y^2\right]_0^b\)=−ab²

(3) Line integral along QR: Here y=b, dy=0, and x changes from a to 0.

∴\(\int_{Q R} F \cdot d r\)=\(\int_b^0 0\)=0

∴\(=\int_a^0\left(x^2+b^2\right) d x\)

=\(\left[\frac{x^3}{3}+b^2 x\right]_a^0\)

=\(-\frac{1}{3} a^3-a b^2\)

(4) Line integral along RO: Here x=0, dx=0, and y varies from b to 0.

∴ \(\int_{R O} F \cdot d r\)=\(\int_b^0 0\)

∴ \(\oint_c \boldsymbol{F} \cdot d \boldsymbol{r}\)=\(\int_{O P} F \cdot d r\)+ \(\int_{P Q} F \cdot d r\)+\(\int_{Q R} F \cdot d r\)+\(\int_{R O} F \cdot d r\)=\(\frac{a^3}{3}-a b^2-\frac{1}{3} a^3-a b^2\)=-2ab²

24. Find \(\int_c\)y²dx−x²dy, where C the curve represents sides of ΔABC, where A =(1, 0), B = (0, 1), C = (- 1, 0).

Solution:

Equation of \(\overleftrightarrow{A B}\) is \(\frac{y-0}{1-0}\)=\(=\frac{x-1}{0-1}\)⇒ y=1−x

Equation  of \(\overleftrightarrow{B C}\) is \(\frac{y-1}{0-1}\)\(=\frac{x-0}{-1-0}\) ⇒ y=1+x.

Equation of \(\overleftrightarrow{C A}\) is y=0.

Case(1): Line integral along \(\overleftrightarrow{A B}\): y=1-x, dy =−dx, x varies from 1 to 0

∴ \(\int_{c_1} y^2 d x-x^2 d y=\int_1^0(1-x)^2 d x-x^2(-d x)=\int_1^0\left(1+x^2-2 x+x^2\right) d x\)

= \(\int_1^0\left(1-2 x+2 x^2\right) d x=\left[x-x^2+\frac{2 x^2}{3}\right]_1^0=-\left(1-1+\frac{2}{3}\right)=-\frac{2}{3}\)

Case(2): Line integral aling \(\overleftrightarrow{B C}\): y=1+x,dy=dx, x varies from 0 to −1

∴ \(\int_{c_2} y^2 d x-x^2 d y=\int_0^{-1}(1+x)^2 d x-x^2 d x=\int_0^{-1}\left(1+2 x+x^2-x^2\right) d x\)

= \(\int_0^{-1}(1+2 x) d x=\left[x+x^2\right]_0^{-1}=-1+1=0\)

Case(3): Line integral along \(\overleftrightarrow{C A}\): y=0,dy=0, x varies from −1 to 1.

∴ \(\int_{c_3} y^2 d x-x^2 d y=\int_{-1}^1 0=0\)

∴ \(\int_C y^2 d x-x^2 d y=\int_{c_1} y^2 d x-x^2 d y+\int_{c_2} y^2 d x-x^2 d y+\int_{c_3} y^2 d x-x^2 d y\)

= \(-\frac{2}{3}+0+0=-\frac{2}{3}\)

25. Find the work done when a force F = (x² − y² + x) i− (2xy+y)j moves a particle in xy-plane from (0, 0) to (1, 1) along the parabola y²= x.

Solution:

Given curve is y²=x⇒ 2y dy=dx. Work done by F is \(\int_C F \cdot d r\). the integration is performed in xy-plane and y varies from 0 to 1.

⇒ \(\int_c \mathbf{F} \cdot d \mathbf{r}=\int_c\left(x^2-y^2+x\right) d x-(2 x y+y) d y=\int_0^1\left(y^4-y^2+y^2\right) 2 y d y-\left(2 y^3+y\right) d y\)

⇒ \(\int_0^1\left(2 y^5-2 y^3-y\right) d y=\left[\frac{y^6}{3}-\frac{y^4}{2}-\frac{y^2}{2}\right]_0^1=\frac{1}{3}-\frac{1}{2}-\frac{1}{2}=\frac{2-3-3}{6}=\frac{-4}{6}=\frac{-2}{3}\)

26. If F = (x +y²) i−2xj+ 2yzk, evaluate \(\int_S\) F . N dS where S is the surface of plane 2x +y + 2z = 6 in the first octant.

Solution: Let φ =2x+y=2z-6.

The vector normal to the surfaces S is ∇φ=\(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)=2i+j+2k.

Unit normal vector, N \(=\frac{2 i+j+2 k}{\sqrt{4+1+4}}\)=\(=\frac{1}{3}(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})\).

Let R be the projection of S on xy-plane.

Now R is bounded by the x-axis, y-axis, and the line 2x+y=6z= 0

⇒ \(\mathbf{F} \cdot \mathbf{N}=\left[\left(x+y^2\right) \mathbf{i}-2 x \mathbf{j}+2 y z \mathbf{k}\right] \cdot \frac{1}{3} (2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})=\frac{1}{3}\left[2 x+2 y^2-2 x+4 y z\right]=\frac{2}{3}\left(y^2+2 y z\right)\)

⇒ \(\mathbf{N} \cdot \mathbf{k}=\frac{1}{3}(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k}) \cdot \mathbf{k}=\frac{2}{3}\)

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}=\iint_R \frac{2}{3}\left(y^2+2 y z\right) \frac{d x d y}{(2 / 3)}=\iint_R\left(y^2+2 y z\right) d x d y\)

⇒ \(\iint_R\left[y^2+y(6-2 x-y)\right] d x d y=2 \int_{x=0}^{x=3} \int_{y=0}^{y=6-2 x} y(3-x) d x d y\)

⇒ \(2 \int_{x=0}^{x=3}\left[\frac{y^2}{2}(3-x)\right]_{y=0}^{y=6-2 x} d x=\int_{x=0}^{x=3}(3-x)(6-2 x)^2 d x\)

⇒ \(\int_0^3(3-x)\left(36-24 x+4 x^2\right) d x=\int_0^3\left(108-108 x+36 x^2-4 x^3\right) d x\)

∴ \(\left[108 x-54 x^2+12 x^3-x^4\right]_0^3=324-486+324-81=81\)

27. Evaluate \(\int_S\)F.N dS where F = xy i- x²j + (x + z) k and S is the surface of the planes 2x + 2y + z=6 in the first octant.

Solution:

Let φ =2x+2y=z-6.

The vector normal to the surfaces S is  ∇φ=\(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)=2i+j+2k.

Unit normal vector, N \(=\frac{2 i+2 j+k}{\sqrt{4+4+1}}\)=\(\frac{1}{3}\)(2i+2j+k).

Let R be the projection of S on xy-plane.

Now R is bounded by x-axis, y-axis, and the line 2x+y=6,z= 0

F.N =[(xyi-x²j+(x+z)K].1/3(2i+2j+k)=\(=\frac{1}{3}\left(2 x y-2 x^2+x+z\right)\).

⇒ \(\mathbf{N} \cdot \mathbf{k}=\frac{1}{3}(2 \mathbf{i}+2 \mathbf{j}+\mathbf{k}) \cdot \mathbf{k}=\frac{1}{3}\)

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}=\iint_R \frac{1}{3}\left(2 x y-2 x^2+x+z\right) \frac{d x d y}{(1 / 3)}\)

⇒ \(\iint_R\left(2 x y-2 x^2+x+6-2 x-2 y\right) d x d y=\int_{x=0}^{x=3} \int_{y=0}^{y=3-x}\left(2 x y-2 x^2-x-2 y+6\right) d x d y\)

⇒ \(\int_{x=0}^{x=3}\left[x y^2-2 x^2 y-x y-y^2+6 y\right]_{y=0}^{y=3-x} d x\)

⇒ \(\int_{x=0}^{x=3}\left[x(3-x)^2-2 x^2(3-x)-x(3-x)-(3-x)^2+6(3-x)\right] d x\)

⇒ \(\int_0^3\left(9 x-6 x^2+x^3-6 x^2+2 x^3-3 x+x^2-9+6 x-x^2+18-6 x\right) d x\)

⇒ \(\int_0^3\left(3 x^3-12 x^2+6 x+9\right) d x=\left[\frac{3 x^4}{4}-4 x^3+3 x^2+9 x\right]_0^3=\frac{243}{4}-108+27+27\)

∴ \(\frac{243}{4}-54=\frac{27}{4}\)

28.Evaluate\(\int_S\) F. N dS where F= 18zi- 12J + 3yk and S is the part of the planes 2x + 3y + 6z=12 located in the first octant.

Solution: Let φ =2x+3y+6z-12. The vector normal to the plane is ∇φ=2i+3j+6k.

Unit normal vector, N=\(\frac{2 i+3 j+6 k}{\sqrt{4+9+36}}\)=\(\frac{2 \mathbf{i}+3 \mathbf{j}+6 \mathbf{k}}{7}\).

Let R be the projection of S on xy-plane.

∴ \(\mathbf{F} \cdot \mathbf{N}=(18 z \mathbf{i}-12 \mathbf{j}+3 y \mathbf{k}) \cdot\left(\frac{2 \mathbf{i}+3 \mathbf{j}+6 \mathbf{k}}{7}\right)=\frac{1}{7}(36 z-36+18 y)=\frac{6}{7}(6 z-6+3 y)\)

⇒ \(\mathbf{N} \cdot \mathbf{k}=\frac{1}{7}(2 \mathbf{i}+3 \mathbf{j}+6 \mathbf{k}) \cdot \mathbf{k}=\frac{6}{7} \cdot \quad d S=\frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}=\frac{d x d y}{6 / 7}\)

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}=\iint_R \frac{6}{7}(6 z-6+3 y) \frac{d x d y}{(6 / 7)}=\iint_R(6 z-6+3 y) d x d y\)

⇒ \(\iint_R(12-2 x-3 y-6+3 y) d x d y=\int_{x=0}^{x=6} \int_{y=0}^{y=(12-2 x) / 3}(6-2 x) d x d y\)

⇒ \(\int_{x=0}^{x=6}[(6-2 x) y]_{y=0}^{y=(12-2 x) / 3} d x=\int_{x=0}^{x=6} \frac{4}{3}(3-x)(6-x) d x=\frac{4}{3} \int_0^6\left(x^2-9 x+18\right) d x\)

∴ \(\frac{4}{3}\left[\frac{x^3}{3}-1 \frac{9 x^2}{2}+18 x\right]_0^6=\frac{4}{3}[72-162+108]=24\)

29. Evaluate \( \int_S\)F.N dS where F =yi + 2xj -zk and S is the surface of the plane s 2x+y = 6 in the first octant, cut of f by the plane z = 4.

Solution: Let φ =2x+y-6.

The vector to the surfaces S is ∇φ =\(i \frac{\partial \varphi}{\partial x}+j \frac{\partial \varphi}{\partial y}+k \frac{\partial \varphi}{\partial z}=\)=2i+j

Unit normal vector  to th surface is N \(=\frac{\nabla \varphi}{|\nabla \varphi|}\)=\(\frac{2 \mathbf{i}+\mathbf{j}}{\sqrt{5}}\).

Let r be the projection of S over xz plane.

The boundaries of R is x=0 to x=3 and z-0 to z=4.

∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R(y \mathbf{i}+2 x \mathbf{j}-z \mathbf{k}) \cdot \frac{2 \mathbf{i}+\mathbf{j}}{\sqrt{5}} \frac{d x d z}{|\mathbf{j} \cdot \mathbf{n}|}\)

⇒ \(\iint_R \frac{2 y+2 x}{\sqrt{5}} \frac{d x d z}{1 / \sqrt{5}}=\iint_R[2(6-2 x)+2 x] d x d z=\int_{x=0}^{x=3} \int_{z=0}^{z=4}(12-2 x) d x d z\)

⇒ \(\left.\int_{x=0}^{x=3}(12-2 x) z\right]{ }_{z=0}^{z=4} d x=\int_{x=0}^{x=3}(12-2 x) 4 d x=4\left[12 x-x^2\right]_0^3=4[36-9]=108\)

30. Evaluate Evaluate \(\int_S\)F.N dS where F = zi +xj−  3y² zk and S is the surfaces x² +y² = 16 included in the first octant between z = 0 and z = 5.

Solution: Let φ = x²+y²-16

The normal to the surfaces S is grad  φ = ∇φ=\(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)=2xi+2yj

Unit normal vector, N=\(\frac{2 x i+2 y i}{\sqrt{4 x^2+4 y^2}}\)

=\(\frac{x \mathbf{i}+y \mathbf{j}}{\sqrt{x^2+y^2}}\)

=\(\frac{x \mathbf{i}+y \mathbf{j}}{4}\)

Let R be the projection of S on xy-plane .

In yz- plane, for the surface y varies from 0 to 4 and z varies from 0 to 5.

Then \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{i}|}=\int_{y=0}^{y=4} \int_{z=0}^{z=5} \frac{x z+x y}{4} \cdot \frac{d y d z}{x / 4}\)

⇒ \(\int_{y=0}^{y=4} \int_{z=0}^{z=5}(z+y) d y d z=\int_{y=0}^{y=4}\left[z^2 / 2+y z\right]{ }_{z=0}^{z=5} d y=\int_0^4\left[\frac{25}{2}+5 y\right] d y=\left[\frac{25 y}{2}+\frac{5 y^2}{2}\right]_0^4\)

⇒ 50 + 40 = 90.

31. Evaluate \(\int_S\)F.N dS where F = 6z i + (2x +y) j- x k and S is the surface of the region bounded by x² + z² = 9, x == 0,y = 0, z = 0 and y = 8.

Solution: Let φ x² +y²+ z² − 9

The normal to the surfaces S is ∇φ \(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)=2xi+2zk

Unit normal vector, N=\(\frac{2 x i+2 y i}{\sqrt{4 x^2+4 y^2}}\)

=\(\frac{x \mathbf{i}+z \mathbf{k}}{\sqrt{x^2+z^2}}\)

=\(=\frac{1}{3}(x \mathbf{i}+z \mathbf{k})\)

F.N =(6zi+(2x+y)j-xk).\(\frac{1}{3}\) (xi+zk)=1/3 (6xz-xz)=\(\frac{5}{3}\) xz

N.K=\(\frac{1}{3}\) (xi+zk)k. =z/3. Let R be the projection of S on xy-plane.

In xy-plane, for the surfaces x varies from 0 to 3 and y varies from 0 to 8.

∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{k}|}=\int_{x=0}^{x=3} \int_{y=0}^{y=8}\left(\frac{5}{3} x z\right) \frac{d x d y}{(z / 3)}\)

\(=\int_{x=0}^{x=3} \int_{y=0}^{y=8} 5 x d x d y\)=\(\int_{x=0}^{x=3} 40 x d x=\left[20 x^{2}\right]_{0}^{3}=180\)

32. Evaluate \(\int_S\)F.N dS, where F =yzi + zxj + xyk. and S is the part ofthe sphere x² +y² + z² = 1 which lies in the first octant.

Solution: Let φ =x² +y²+ z²−1

Normal vector to the surfaces is ∇φ=2(xi+yj+zk)

Unit normal vector, N=\(\frac{2(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})}{\sqrt{4 x^2+4 y^2+4 z^2}}\)

=\(\frac{x \mathbf{i}+y \mathbf{j}+\mathbf{k}}{\sqrt{x^2+y^2+z^2}}\)

=xi+yj+zk.

F.N (yzi+zxj+xyk).(xi+yj+zk)=xyz+xyz+xyz=3xyz, N.i=x.

Let R be the projection of S on yz-plane.

Then \(\int_S F \cdot N d S\)

=\(\iint_R F \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \boldsymbol{i}|}\)

=\(\iint_R 3 x y z d y d z / x\)

=\(3 \iint_R y z d y d z\)

In yz-plane x=0, the equation of the surface becomes y²+ z²=1

∴ y varies from o to 1 and z varies from o to \(\sqrt{1-y^2}\)

∴ \(\int_s \mathbf{F} \cdot \mathbf{N} d S=3 \int_{y=0}^{y=1} \int_{z=0}^{z=\sqrt{1-y^2}} y z d y d z=3 \int_{y=0}^{y=1}\left[\frac{z^2}{2}\right]_{z=0}^{z=\sqrt{1-y^2}} y d y\)

⇒ \(\frac{3}{2} \int_0^1 y\left(1-y^2\right) d y=\frac{3}{2}\left[\frac{y^2}{2}-\frac{y^4}{4}\right]=\frac{3}{2}\left[\frac{1}{2}-\frac{1}{4}\right]=\frac{3}{8}\)

33. Evaluate \(\int_S\)F.N dS where F =y²z² i + z²x² j + x²y² k and the surfaces x² +y² + z² = 1 above xy-plane.

Solution: Let φ= x² +y²+ z²−1

Normal vector to the surfaces is ∇φ=2(xi+yj+zk)

Unit normal vector, N=\(=\frac{2(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})}{\sqrt{4 x^2+4 y^2+4 z^2}}\)

=\(\frac{x+y+z k}{\sqrt{x^2+y^2+z^2}}=\)

=xi+yj+zk.

F.N =( y² z² i+z² x² j+x² y² k).(xi+yj+zk)=xy² z² +x² yz² +x² y² z

Let R be the projection of S in xy−plane.

Then \(\int_S F \cdot N d S\)

=\(\iint_R F \cdot \mathbf{N} \frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}\)

=\(\iint_R \frac{\left(x y^2 z^2+x^2 y z^2+x^2 y^2 z\right)}{z} d x d y\)

=\(\iint_R\left(x y^2 z+x^2 y z+x^2 y^2\right) d x d y\)

In xy-plane , z=0 and the equation of the surfaces becomes x+y=1.

x varies from -1 to 1 and z varies from

=\(\sqrt{1-x^2}\)  to \(\sqrt{1-x^2}\)

∴ \(\int_S F \cdot N d S\)

=\(\iint_R\left(x y^2 z+x^2 y z+x^2 y^2\right) d x d y\)

=\(\iint_R x^2 y^2 d x d y\)

=\(y\int_{x=-1}^{x=1} \int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}} x^2 y^2 d x d y\)

 

⇒ \(4 \int_{x=0}^{x=1} \int_{y=0}^{y=\sqrt{1-x^2}} x^2 y^2 d x d y=4 \int_{x=0}^{x=1}\left[\frac{x^2 y^3}{3}\right]_0^{\sqrt{1-x^2}} d x=\frac{4}{3} \int_{x=0}^{x=1} x^2\left(1-x^2\right)^{3 / 2} d x\)

Put x = sin θ.

Then dx = cos θ dθ

x = 0, 1 ⇒ θ = 0, π/2

⇒ \(\frac{4}{3} \int_{\theta=0}^{0=\pi / 2} \sin ^2 \theta\left(1-\sin ^2 \theta\right)^{3 / 2} \cos \theta d \theta\)

⇒ \(\frac{4}{3} \int_0^{\pi / 2} \sin ^2 \theta \cos ^4 \theta d \theta=\frac{4}{3} \times \frac{1}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}=\frac{\pi}{24}\)

34. Find the surface area of the sphere given by x = a sin θ cos φ, y = a sin θ sin φ , z= a cos θ,0≤θ≤π,0≤φ≤2π.

Solution: x² +y² +z² = a² sin² θ cos² φ+a² sin² θ sin² φ +a² cos²  θ

= a² sin² θ (cos² φ+sin² φ)a² cos²  θ=a² sin² θ  + a² cos²  θ= a²( sin² θ + cos²  θ) =a²

35. If F = 4xzi -y² j+yz k, evaluate ∫F . N dS where S is the surface of the cube bounded by x = 0, x = a ,y = 0, y=a, z = 0, z = a.

Solution:

Consider the cube OABCPQRS surrounded by the following faces:

(1) For the faces PQAR, i is the outward normal:

∴ N=i, x=a, dS=dy dz.

∴ \(\int_{R_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{y=0}^{y=a} \int_{z=0}^{z=a}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot \mathbf{i} d y d z\)

⇒ \(\int_{y=0}^{y=a} \int_{z=0}^{z=a} 4 x z d y d z=\int_{y=0}^{y=a} \int_{z=0}^{z=a} 4 a z d y d z\)

⇒ \(\int_{y=0}^{y=a}\left[2 a z^2\right]_{z=0}^{z=a} d y=\int_{y=0}^{y=a} 2 a^3 d y=\left[2 a^3 y\right]_0^a=2 a^4\)

(2) For the faces OBSC,−i is the outward normal:

∴ N=-i,x=0 , and dS =dy dz

∴ \(\int_{R_2} \mathbf{F} \cdot \mathbf{N} d S\)

=\(\iint_{R_2}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot(-\mathbf{i}) d y d z\)

=\(-\iint_{R_2} 4 x z d y d z\)=0

(3) For the face BQPS, j is the outward normal:

∴N=j,y=a, and dS=dx dz

∴\(\int_{R_3} \mathbf{F} \cdot \mathbf{N} d S\)

=\(\iint_{R_3}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right)\).j dx dz

=\(-\iint_{R_3} y^2 d x d z\)

=\(-a^2 \int_{x=0}^{x=a} \int_{z=0}^{z=a} d x d z\)

=\(-a^2[x]_0^a[z]_0^a\)

=-a⁴

(4) For the OARC, -j is the  outward normal:

∴ N=-j, y=0 and dS =dx dz

∴ \(\int_{R_4} \mathbf{F} \cdot \mathbf{N} d S\)

=\(\int_{R_4}\)(4xzi-y²j+yzk).(-j) dS

=\(\int_{R_4}\)∫y² dx dz=0

(5) For the face PRCS, k is the outward normal:

∴ N=k, z=a, and dS= dx dy

∴ \(\int_{R_5} \mathbf{F} \cdot \mathbf{N} d S\)

=\(\int_{R_5}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot \mathbf{k} d S\)

=\(\iint_{R_5} y z \cdot d x d y\)

=\(\int_{x=0}^{x=a} \int_{y=0}^{y=a} a y d x d y\)

=\(a[x]_0^a \cdot\left[\frac{y^2}{2}\right]_0^a\)

∴ \(=\frac{a^4}{2}\)

(6) For the face OAQB, -k is the outward normal:

∴ N=-k, z=0 and dS= dx dy

∴ \(\int_{R_6} \mathbf{F} \cdot \mathbf{N} d S\)

=\(\int_{R_6}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot(-\mathbf{k}) d S\)

=\(-\iint_{R_6} y z d x d y\)=0

∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d S\)=2a⁴+0-a⁴+0+½ a⁴+0=3/2a⁴.

36. If F = 2xzi−xj+yk, evaluate ∫F dV where V is the region bounded by the surfaces x = 0, x = 2, y = 0, y = 6, z=x², z = 4.

Solution:

⇒ \(\int_V \mathbf{F} d V=\int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4}\left(2 x z \mathbf{i}-x \mathbf{j}+y^2 \mathbf{k}\right) d x d y d z\)

⇒ \(i\int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} 2 x z d x d y d z-\mathbf{j} \int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} x d x d y d z+\mathbf{k} \int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} y^2 d x d y d z\)

⇒ \(\left.\left.\left.=\mathbf{i} \int_{x=0}^{x=2} \int_{y=0}^{y=6} x z^2\right]_{z=x^2}^{z=4} d x d y-\mathbf{j} \int_{x=0}^{x=2} \int_{y=0}^{y=6} x z\right]_{z=x^2}^{z=4} d x d y+\mathbf{k} \int_{x=0}^{x=2} \int_{y=0}^{y=6} y^2 z\right]_{z=x^2}^{z=4} d x d y\)

⇒ \(\text { i } \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left(16 x-x^5\right) d x d y-\mathbf{j} \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left(4 x-x^3\right) d x d y+\mathbf{k} \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left(4 y^2-x^2 y^2\right) d x d y\)

⇒ \(\left.\left.=\mathbf{i} \int_{x=0}^{x=2}\left(16 x-x^5\right) y\right]_{y=0}^{y=6} d x-\mathbf{j} \int_{x=0}^{x=2}\left(4 x-x^3\right) y\right]_{y=0}^{y=6} d x+\mathbf{k} \int_{x=0}^{x=2}\left[4 y^3 / 3-x^2 y^3 / 3\right]_{y=0}^{y=6} d x\)

⇒ \(\mathbf{i} \int_0^2\left(96 x-6 x^5\right) d x-\mathbf{j} \int_0^2\left(24 x-6 x^3\right) d x+\mathbf{k} \int_0^2\left(288-72 x^2\right) d x\)

∴ \(\mathbf{i}\left[48 x^2-x^6\right]_0^2-\mathbf{j}\left[12 x^2-3 x^4 / 2\right]_0^2+\mathbf{k}\left[288 x-24 x^3\right]_0^2=128 \mathbf{i}-24 \mathbf{j}+384 \mathbf{k}\)

37. Evaluate \(\int_V\)F dV where F =xi +yj + zk and V is the region bounded by x = 0, x =2,y = 0, y = 6, z = 4 and z=x² .

Solution:

⇒ \(\int_V F d V=\int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4}(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) d x d y d z\)

⇒ \(i\int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} x d x d y d z+j\int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} y d x d y d z+k \int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} z d x d y d z\)

⇒ \(i \int_{x=0}^{x=2} \int_{y=0}^{y=6}[x z]_{z=x^2}^4 d x d y+\mathbf{j} \int_{x=0}^{x=2} \int_{y=0}^{y=6}[y z]_{z=x^2}^{z=4} d x d y+\mathbf{k} \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left[\frac{z^2}{2}\right]_{z=4}^{x^2} d x d y\)

⇒ \(i \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left(4 x-x^3\right) d x d y+\mathbf{j} \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left(4-x^2\right) y d x d y+\mathbf{k} \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left[8-\frac{x^4}{2}\right] d x d y\)

⇒ \(\mathbf{i} \int_{x=0}^{x=2}\left[\left(4 x-{x}^3\right) y\right]_{y=0}^{y=6} d x+\mathbf{j} \int_{x=0}^{x=2}\left[\left(4 x-x^2\right) \frac{y^2}{2}\right]_{y=0}^{y=6} d x+\mathbf{k} \int_{x=0}^{x=2}\left[\left(8-\frac{x^4}{2}\right) y\right]_{y=0}^{y=6} d x\)

⇒ \(\mathbf{i} \int_{x=0}^{x=2} 6\left(4 x-x^3\right) d x+\mathbf{j} \int_{x=0}^{x=2} 18\left(4-x^2\right) d x+\mathbf{k} \int_{x=0}^{x=2} 6\left(8-\frac{x^4}{2}\right) d x\)

⇒ \(\mathbf{i}\left[12 x^2-\frac{6 x^4}{4}\right]_{x=0}^{x=2}+\mathbf{j}\left[18\left(4 x-\frac{x^3}{3}\right)\right]_{x=0}^{x=2}+\mathbf{k}\left[6\left(8 x-\frac{x^5}{10}\right)\right]_{x=0}^{x=2}\)

⇒ \(\mathbf{i}(48-24)+\mathbf{j}\left[18\left(8-\frac{8}{3}\right)\right]+\mathbf{k}\left[6\left(16-\frac{32}{10}\right)\right]=24 \mathbf{i}+96 \mathbf{j}+\frac{384}{5} \mathbf{k}\)

38. If F = (2x²– 3z) i- 2xy j- 4x k, then evaluate ∫\(\int_V\)∫∇.F dV where V is the closed region bounded by the planes x = 0,y = 0, z = 0 and 2x + 2y + z = 4. Also, Evaluate ∫\(\int_V\)∫∇×F dV.

Solution:

∴ \(\nabla \cdot \mathbf{F}=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right) \cdot\left[\left(2 x^2-3 z\right) \mathbf{i}-2 x y \mathbf{j}-4 x \mathbf{k}\right]\)

⇒ \(\frac{\partial}{\partial x}\left(2 x^2-3 z\right)+\frac{\partial}{\partial y}(-2 x y)+\frac{\partial}{\partial z}(-4 x)=4 x-2 x=2 x\)

∴ \(\iiint_V \nabla \cdot \mathbf{F} d V=\iiint_V 2 x d x d y d z=2 \int_{x=0}^{x=2} \int_{y=0}^{y=2-x} \int_{z=0}^{z=4-2 x-2 x} x d x d y d z\)

⇒ \(\left.2 \int_{x=0}^{x=2} \int_{y=0}^{y=2-x} x[z]\right]_{z=0}^{z=4-2 x-2 y} d x d y=2 \int_{x=0}^{x=2} \int_{y=0}^{y=2-x} x(4-2 x-2 y) d x d y\)

⇒ \(\left.2 \int_{x=0}^{x=2}\left[4 x y-2 x^2 y-x y^2\right]\right]_{y=0}^{y=2-x} d x=2 \int_0^2\left[4 x(2-x)-2 x^2(2-x)-x(2-x)^2\right] d x\)

⇒ \(2 \int_0^2\left[x^3-4 x^2+4 x\right] d x=2\left[\frac{1}{4} x^4-\frac{4}{3} x^3+2 x^2\right]_0^2=2\left[4-\frac{32}{3}+8\right]=\frac{8}{3}\)

We have \(\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x^2-3 z & -2 x y & -4 x
\end{array}\right|\)

⇒ \(\left[\frac{\partial}{\partial y}(-4 x)-\frac{\partial}{\partial z}(-2 x y)\right] \mathbf{i}-\left[\frac{\partial}{\partial x}(-4 x)-\frac{\partial}{\partial z}\left(2 x^2-3 z\right)\right] \mathbf{j}+\left[\frac{\partial}{\partial x}(-2 x y)-\frac{\partial}{\partial y}\left(2 x^2-3 z\right)\right] \mathbf{k}\)

⇒ \(0 \mathbf{i}-(-4+3) \mathbf{j}+(-2 y) \mathbf{k}=\mathbf{j}-2 y \mathbf{k}\)

∴ \(\iiint_V \nabla \times \mathbf{F} d V=\iiint_V(\mathbf{j}-2 y \mathbf{k}) d x d y d z\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=2-x z=4} \int_{z=0}^{-2 x-2 y}(\mathbf{j}-2 y \mathbf{k})dx dy dz =\int_{x=0}^{x=2} \int_{y=0}^{y=2-x}(\mathbf{j}-2 y \mathbf{k})(4-2 x-2 y) d x d y.\)

⇒ \(\int_{x=0}^{x=2}\left[\mathbf{j}\left(4 y-2 x y-y^2\right)-2 \mathbf{k}\left(2 y^2-x y^2-\frac{2}{3} y^3\right)\right]_{y=0}^{y=2-x} d x\)

⇒ \(=\int_{x=0}^{x=2}\left[\mathbf{j}(2-x)(4-2 x-2+x)-2 \mathbf{k}(2-x)^2\left\{2-x-\frac{2}{3}(2-x)\right\}\right] d x\)

⇒ \(\int_0^2\left[(2-x)^2 \mathbf{j}-\frac{2}{3}(2-x)^3 \mathbf{k}\right] d x=\int_0^2\left[(x-2)^2 \mathbf{j}+\frac{2}{3}(x-2)^3 \mathbf{k}\right] d x\)

⇒ \(\left[\frac{(x-2)^3}{3}\right]_0^2 \mathbf{j}+\left[\frac{2}{3} \frac{(x-2)^4}{4}\right]_0^2 \mathbf{k}=\frac{8}{3} \mathbf{j}-\frac{8}{3} \mathbf{k}=\frac{8}{3}(\mathbf{j}-\mathbf{k})\)

39. Evaluate ∫∫∫(2x+y)dV where V is closed region bounded by the cylinder z = 4−x² and the planes x = 0,y = 0 and y = 2, z = 0.

Solution:

∴ \(\iiint_V(2 x+y) d V=\int_{x=0}^{x=2} \int_{y=0}^{y=2} \int_{z=0}^{z=4-x^2}(2 x+y) d x d y d z=\int_{x=0}^{x=2} \int_{y=0}^{y=2}[(2 x+y) z]\int_{z=0}^{z=4-x^2} d x d y\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=2}(2 x+y)\left(4-x^2\right) d x d y=\int_{x=0}^{x=2}\left[\left(4-x^2\right)\left(2 x y+\frac{y^2}{2}\right)\right]_{y=0}^{y=2} d x\)

⇒ \(\int_{x=0}^{x=2}\left(4-x^2\right)(4 x+2) d x=\int_0^2\left(8+16 x-2 x^2-4 x^3\right) d x\)

⇒ \(\left[8 x+8 x^2-\frac{2 x^3}{3}-x^4\right]_0^2=16+32-\frac{16}{3}-16=\frac{80}{3}\)

40. If φ = 45x²y, evaluate ∫\(\int_V\)∫φ dV where V is the closed region bounded by the plane 4x + 2y + z = 8,x = 0,y = 0,z = 0.

Solution:

∴ \(\iiint_V \varphi d V=\iiint_V 45 x^2 y d x d y d z=\int_{x=0}^{x=2} \int_{y=0}^{y=4-2 x} \int_{z=0}^{z=8-4 x-2 y} 45 x^2 y d x d y d z\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=4-2 x}\left[45 x^2 y z\right]_{z=0}^{z=8-4 x-2 y} d x d y=\int_{x=0}^{x=2} \int_{y=0}^{y=4-2 x}\left[45 x^2 y(8-4 x-2 y)\right] d x d y\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=4-2 x} 45 x^2\left(8 y-4 x y-2 y^2\right) d x d y=\int_{x=0}^{x=2} 45 x^2\left[4 y^2-2 x y^2-\frac{2}{3} y^3\right]_{y=0}^{y=4-2 x} d x\)

⇒ \(\int_{x=0}^{x=2}\left[720 x^2(2-x)^2-360 x^3(2-x)^2-240 x^2(2-x)^3\right] d x\)

⇒ \(120 \int_{x=0}^{x=2}\left[6 x^2\left(4-4 x+x^2\right)-3 x^3\left(4-4 x+x^2\right)-2 x^2\left(8-12 x+6 x^2-x^3\right)\right] d x\)

⇒ \(120 \int_{x=0}^{x=2}\left(24 x^2-24 x^3+6 x^4-12 x^3+12 x^4-3 x^5-16 x^2+24 x^3-12 x^4+2 x^5\right) d x\)

⇒ \(120 \int_{x=0}^{x=2}\left(8 x^2-12 x^3+6 x^4-x^5\right) d x=120\left[\frac{8 x^3}{3}-3 x^4+\frac{6 x^5}{5}-\frac{x^6}{6}\right]_0^2\)

= 2560 – 5760 + 4608 – 1280 = 128.

41. Evaluate I=∫∫∫\sqrt{\left(a^2 b^2 c^2-b^2 c^2 x^2-c^2 a^2 y^2-a^2 b^2 z^2\right)} dx dy dz taken throughout the domain \(\left\{(x, y, z): \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1\right\}\)

Solution: The limits of integration are ;x varies from -a to a; y varies from -b\(\sqrt{1-x^2 / a^2}\) to b\(\sqrt{1-x^2 / a^2}\) and z varies from -c \(\sqrt{1-x^2 / a^2-y^2 / b^2}\) to c \(\sqrt{1-x^2 / a^2-y^2 / b^2}\)

∴ \(I=\iiint \sqrt{\left(a^2 b^2 c^2-b^2 c^2 x^2-c^2 a^2 y^2-a^2 b^2 z^2\right)} d x d y d z\)

⇒ \(a b c \int_{x=-a}^{x=a} \int_{y=-b \sqrt{1-x^2 / a^2}}^{y=b \sqrt{1-x^2 / a^2}} \quad \int_{z=-c \sqrt{1-x^2 / a^2-y^2 / b^2}}^{z=c \sqrt{1-x^2 / a^2-y^2 / b^2}} \sqrt{\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)} d x d y d z\)

⇒ \(8a b c \int_{x=-a}^{x=a} \int_{y=0}^{y=b \sqrt{y=1-x^2 / a^2}} \quad \int_{z=0}^{z=c \sqrt{z=1-x^2 / a^2-y^2 / b^2}} \sqrt{\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)} d x d y d z\)

⇒ \(=8 a b c^2 \int_{x=0}^{x=a} \int_{y=0}^{y=b \sqrt{y=1-x^2 / a^2}}\left[0+\frac{1}{2}\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)\left(\frac{\pi}{2}-0\right)\right] d x d y\)

⇒ \(8 a b c^2 \frac{\pi}{4} \int_{x=0}^{x=a} \int_{y=0}^{y=b \sqrt{1-x^2 / a^2}}\left[\left(1-\frac{x^2}{a^2}\right)-\frac{y^2}{b^2}\right] d x d y\)

⇒ \(2 a b c^2 \pi \int_{x=0}^{x=a}\left[\left(1-\frac{x^2}{a^2}\right) y-\frac{y^3}{3 b^2}\right]_{y=0}^{y=b \sqrt{1-x^2 / a^2}} d x\)

⇒ \(2 a b c^2 \pi \int_0^a b\left(1-\frac{x^2}{a^2}\right)^{3 / 2}-\frac{1}{3 b^2} b^3\left(1-\frac{x^2}{a^2}\right)^{3 / 2} d x\)

Put x = a sin θ.

Then dx = a cos θ dθ.

x = 0, a ⇒ θ = 0, π/2

⇒ \(2 a b c^2 \pi \int_0^a \frac{2 b}{3}\left(1-\frac{x^2}{a^2}\right)^{3 / 2} d x=\frac{4 a b^2 c^2 \pi}{3} \int_0^{\pi / 2}\left(1-\sin ^2 \theta\right)^{3 / 2}(a \cos \theta d \theta)\)

∴ \(\frac{4 a^2 b^2 c^2 \pi}{3} \int_0^{\pi / 2} \cos ^4 \theta d \theta=\frac{4 a^2 b^2 c^2 \pi}{3} \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}=\frac{a^2 b^2 c^2 \pi^2}{4}\)

42. Find the volume of a sphere of radius ‘a’.

Solution: The equation of the sphere with the center origin and radius a is x²+y²+z²=a²

The volume of the sphere V \(=\int_V d V\)

The limits of integration are x=± a,y=±\(\sqrt{a^2-x^2}\),z=±\(\sqrt{a^2-x^2-y^2}\)

∴ \(V=\int_{v}dv=\int_{x=-a}^{x=a} \int_{y=- \sqrt{a^2 – x^2}}^{y= \sqrt{a^2+x^2}} \int_{z=- \sqrt{a^2-x^2-y^2}}^{z= \sqrt{a^2-x^2-y^2}}d x d y d z\)

⇒ \(8 \int_{x=0}^{x=a} \int_{y=0}^{y=\sqrt{a^2-x^2}} \int_{z=0}^{z=\sqrt{a^2-x^2-y^2}} d x d y d z=8 \int_{x=0}^{x=a} \int_{y=0}^{y=\sqrt{a^2-x^2}}[z]_{z=0}^{z=\sqrt{a^2-x^2-y^2}} d x d y\)

⇒ \(8 \int_{x=0}^{x=a} \int_{y=0}^{y=\sqrt{a^2-x^2}} \sqrt{a^2-x^2-y^2} d x d y\)

⇒ \(8 \int_{x=0}^{x=a}\left[\frac{y}{2} \sqrt{a^2-x^2-y^2}+\frac{a^2-x^2}{2} \text{Sin}^{-1} \frac{y}{\sqrt{a^2-x^2}}\right]_{y=0}^{y=\sqrt{a^2-x^2}} d x\)

⇒ \(8 \int_{x=0}^{x=a}\left(\frac{a^2-x^2}{2}\right) \frac{\pi}{2} d x=2 \pi\left[a^2 x-\frac{x^3}{3}\right]_0^a=2 \pi\left[a^3-\frac{a^3}{3}\right]=\frac{4 \pi a^3}{3}\)

Vector Differentiation- 3 Exercise 3 Solved Problems (Contd)

70. Find the angle between the surfaces x²yz = 3x + z² and 3x²-y + 2z=1 at (1,-2, 1).

Solution:

A vector normal to the surfaces f=xy²z-3x-z² is grad (xy²z-3x-z²)

=(y²z-3)i+2xyzj+(xy²+2z)k

A vector normal to g=3x-y+2z-1 is 6xi-2yj+2k

At(1,-2,1), grad g=6i+4j+2k:

If θ is the angle between the surfaces, then

cos θ \(=\frac{1(6)+(-4) 4+2 \cdot 2}{\sqrt{1+16+4} \sqrt{36+16+4}}\)

= \(\frac{-6}{\sqrt{21} \sqrt{56}}\)

 ⇒ θ =Cos^-1\(\left(\frac{-3}{7 \sqrt{6}}\right)\)

71. Find the cosine of the angle between the surfaces x²y + z = 3, x log z-y² = 4 at P (- 1, 2, 1)The normal to the surfaces xy+z=3 is 2xyi+xj+k

Solution:

At (-1,2,1), the normal is 4i+j+k

The normal to the surface x log z-y=4 is log zi-2yjj+(x/z)k

At(-1,2,1) the normal is -4j-k

The angle between the surfaces is equal to the angle between normal to the surfaces.

∴ cos θ \(=\left|\frac{4 \cdot 0+1(-4)+1(-1)}{\sqrt{4^2+1^2+1^2} \sqrt{(-4)^2+(-1)^2}}\right|\)

= \(\frac{5}{\sqrt{18} \sqrt{17}}\)

⇒ θ  = cos^-1\(\frac{5}{\sqrt{306}}\)

72. Define divergence of a vector point function.

Divergence: If F is continuously differentiable vector point function then \(\mathbf{i} \cdot \frac{\partial \boldsymbol{F}}{\partial x}+\mathbf{j} \cdot \frac{\partial \boldsymbol{F}}{\partial y}+\mathbf{k} \cdot \frac{\partial \boldsymbol{F}}{\partial z}\) is called divergence of F and it is denoted by div F or . F

73. If F₁i+F₂j+F₃k then prove that div F = \(=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}\).

Solution:

⇒ \({div} \mathbf{F}=\mathbf{i} \cdot \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{F}}{\partial z}\)

⇒ \(\mathbf{i} \cdot\left(\frac{\partial F_1}{\partial x} \mathbf{i}+\frac{\partial F_2}{\partial x} \mathbf{j}+\frac{\partial F_3}{\partial x} \mathbf{k}\right)+\mathbf{j} \cdot\left(\frac{\partial F_1}{\partial y} \mathbf{i}+\frac{\partial F_2}{\partial y} \mathbf{j}+\frac{\partial F_3}{\partial y} \mathbf{k}\right)+\mathbf{k} \cdot\left(\frac{\partial F_1}{\partial z} \mathbf{i}+\frac{\partial F_2}{\partial z} \mathbf{j}+\frac{\partial F_3}{\partial z} \mathbf{k}\right)\)

⇒ \(\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}\)

= \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_1 & F_2 & F_3
\end{array}\right|\)

74. If F and G are two vector point functions then prove that div (F ± G) = div F ± div G.

Solution:

⇒ \({div}(\mathbf{F}+\mathbf{G})=\mathbf{i} \cdot \frac{\partial}{\partial x}(\mathbf{F}+\mathbf{G})+\mathbf{j} \cdot \frac{\partial}{\partial y}(\mathbf{F}+\mathbf{G})+\mathbf{k} \cdot \frac{\partial}{\partial z}(\mathbf{F}+\mathbf{G})\)

⇒ \(\mathbf{i} \cdot\left(\frac{\partial \mathbf{F}}{\partial x}+\frac{\partial \mathbf{G}}{\partial x}\right)+\mathbf{j} \cdot\left(\frac{\partial \mathbf{F}}{\partial y}+\frac{\partial \mathbf{G}}{\partial y}\right)+\mathbf{k} \cdot\left(\frac{\partial \mathbf{F}}{\partial z}+\frac{\partial \mathbf{G}}{\partial z}\right)\)

⇒ \(\mathbf{i} \cdot \frac{\partial \mathbf{F}}{\partial x}+\mathbf{i} \cdot \frac{\partial \mathbf{G}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{F}}{\partial y}+\mathbf{j} \cdot \frac{\partial \mathbf{G}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{F}}{\partial z}+\mathbf{k} \cdot \frac{\partial \mathbf{G}}{\partial z}\)

⇒ \(\left(\mathbf{i} \cdot \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{F}}{\partial z}\right)+\left(\mathbf{i} \cdot \frac{\partial \mathbf{G}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{G}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{G}}{\partial z}\right)={div} \mathbf{F}+{div} \mathbf{G}\)

Similarly, we can prove that div(F-G)= div F-div G

75. If F = xyz i + x²y²z j + xyz³ k then find div F at (2, 1,- 3)

Solution: div F =\(\frac{\partial}{\partial x}\)(xyz)+\(\frac{\partial}{\partial y}\)(x²y²z)+\(\frac{\partial}{\partial z}\)(xyz³)

=yz+2x²yz+3xyz² At (2,1,-3) , div F=-3-24+54=27.

76. Show that div r = 3

Solution: Let r=xi=yj+zk. Then \(\frac{\partial r}{\partial x}\)=i, \(\frac{\partial r}{\partial y}\)=j, \(\frac{\partial r}{\partial z}\)=k

div r=∇.r\(=\mathbf{i} \cdot \frac{\partial \mathbf{r}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{r}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{r}}{\partial z}\)=i.i+j.j+k.k= 1+1+1=3.

77. Show that div (r x a) = 0

Solution:

Let a = \(a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}, \mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

⇒ \(\mathbf{r} \times \mathbf{a}=\left|\begin{array}{ccc}
\mathbf{i} &\mathbf{j}& \mathbf{k} \\
x & y & z \\
a_1 & a_ 2 & a_3
\end{array}\right|=\mathbf{i}\left(a_3 y-a_2 z\right)-\mathbf{j}\left(a_3 x-a_1 z\right)+\mathbf{k}\left(a_2 x+a_1 y\right)\)

⇒ \({div}(\mathbf{r} \times \mathbf{a})=\nabla \cdot(\mathbf{r} \times \mathbf{a})=\frac{\partial}{\partial x}\left(a_3 y-a_2 y\right)+\frac{\partial}{\partial y}\left(a_1 z-a_3 x\right)+\frac{\partial}{\partial z}\left(a_2 x-a_1 y\right)\)

= 0+0+0 = 0

78. Show that div\(\frac{\underline{r}}{r}\)=\(\frac{2}{r}\)

Solution:

Let a = \(a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}, \mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} .\)

⇒ \(\mathbf{r} \times \mathbf{a}=\left|\begin{array}{lll}
\mathbf{i} & \mathbf{j}&  \mathbf{k} \\
x &  y & z \\
a_ 1& a_ 2& a_ 3
\end{array}\right|=\mathbf{i}\left(a_3 y-a_2 z\right)-\mathbf{j}\left(a_3 x-a_1 z\right)+\mathbf{k}\left(a_2 x+a_1 y\right)\)

⇒ \({div}(\mathbf{r} \times \mathbf{a})=\nabla \cdot(\mathbf{r} \times \mathbf{a})=\frac{\partial}{\partial x}\left(a_3 y-a_2 y\right)+\frac{\partial}{\partial y}\left(a_1 z-a_3 x\right)+\frac{\partial}{\partial z}\left(a_2 x-a_1 y\right)\)

= 0+0+0 =0

79. Define solenoidal vector point function.

Solenoidal: A vector point function F is said to be solenoidal if div F=0

80. Show that F = 3y⁴ z²+ 4x³z² j- 3x²y² k is solenoidal.

Solution:

div F =\(\frac{\partial}{\partial x}\left(3 y^4 z^2\right)+\frac{\partial}{\partial y}\left(4 x^3 z^2\right)+\frac{\partial}{\partial z}\left(-3 x^2 y^2\right)\)=0

∴ F is solenoidal.

81. Prove that F =y³ z² i-3x²z⁵ J- 15x⁵y⁴ k is solenoidal vector.

Solution:

div F \(=\frac{\partial}{\partial x}\left(3 y^4 z^2\right)+\frac{\partial}{\partial y}\left(4 x^3 z^2\right)+\frac{\partial}{\partial z}\left(-3 x^2 y^2\right)\)

= 0+0+0=0

∴ F is solenoidal.

82. If F = (x + 3y) i + (y-2z)y + (x+pz) k is solenoidal, find p.

Solution:

F is solenoidal div F =0 ⇒ ∇.F=0

⇒  \(\frac{\partial}{\partial x}\{x+3 y\}+\frac{\partial}{\partial y}\{y-2 z\}+\frac{\partial}{\partial z}\{x+p z\}\)=0 ⇒ 1+1+p=0 ⇒ p-2.

83. Define the curl of a vector point function.

 Curl:   If F is a continuously differentiable vector point function then \(\mathbf{i} \times \frac{\partial F}{\partial x}+j \times \frac{\partial F}{\partial y}+\mathbf{k} \times \frac{\partial F}{\partial z}\) is called curl of F. It is denoted by curl F or ∇×F.

84. If F = F₁i+F₂j+F₃k then prove that curl F \(=\left|\begin{array}{ccc}\mathbf{1} & \mathbf{j} & \mathbf{k} \\\frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\F_1 & F_2 & F_3\end{array}\right|\)

Solution:

⇒ \({curl} \mathbf{F}=\mathbf{i} \times \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{F}}{\partial z}\)

⇒ \(\mathbf{i} \times\left(\frac{\partial F_1}{\partial x} \mathbf{i}+\frac{\partial F_2}{\partial x} \mathbf{j}+\frac{\partial F_3}{\partial x} \mathbf{k}\right)+\mathbf{j} \times\left(\frac{\partial F_1}{\partial y} \mathbf{i}+\frac{\partial F_2}{\partial y} \mathbf{j}+\frac{\partial F_3}{\partial y} \mathbf{k}\right)+\mathbf{k} \times\left(\frac{\partial F_1}{\partial z} \mathbf{i}+\frac{\partial F_2}{\partial z} \mathbf{j}+\frac{\partial F_3}{\partial z} \mathbf{k}\right)\)

⇒ \(\frac{\partial F_2}{\partial x} \mathbf{k}-\frac{\partial F_3}{\partial x} \mathbf{j}-\frac{\partial F_1}{\partial y} \mathbf{k}+\frac{\partial F_3}{\partial y} \mathbf{i}+\frac{\partial F_1}{\partial z} \mathbf{j}-\frac{\partial F_2}{\partial z} \mathbf{i}\)

⇒ \(\mathbf{i}\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)-\mathbf{j}\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right)+\mathbf{k}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_1 & F_2 & F_3
\end{array}\right|\)

85. If F and G are two vector point functions then prove that curl (F ± G) = curl F ± curl G.

Solution:

⇒ \({curl}(\mathbf{F}+\mathbf{G})=\mathbf{i} \times \frac{\partial}{\partial x}(\mathbf{F}+\mathbf{G})+\mathbf{j} \times \frac{\partial}{\partial y}(\mathbf{F}+\mathbf{G})+\mathbf{k} \times \frac{\partial}{\partial z}(\mathbf{F}+\mathbf{G})\)

⇒ \(\mathbf{i} \times\left(\frac{\partial \mathbf{F}}{\partial x}+\frac{\partial \mathbf{G}}{\partial x}\right)+\mathbf{j} \times\left(\frac{\partial \mathbf{F}}{\partial y}+\frac{\partial \mathbf{G}}{\partial y}\right)+\mathbf{k} \times\left(\frac{\partial \mathbf{F}}{\partial z}+\frac{\partial \mathbf{G}}{\partial z}\right)\)

⇒ \(\mathbf{i} \times \frac{\partial \mathbf{F}}{\partial x}+\mathbf{i} \times \frac{\partial \mathbf{G}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{F}}{\partial y}+\mathbf{j} \times \frac{\partial \mathbf{G}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{F}}{\partial z}+\mathbf{k} \times \frac{\partial \mathbf{G}}{\partial z}\)

⇒ \(\left(\mathbf{i} \times \frac{\partial \mathbf{F}}{\partial x} + \mathbf{j} \times \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{F}}{\partial z}\right)+\left(\mathbf{i} \times \frac{\partial \mathbf{G}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{G}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{G}}{\partial z}\right)={curl} \mathbf{F}+{curl} \mathbf{G}\)

Similarly, we can prove that curl (F-G) = curl G

86. If F = xyz i + zx² j+ xy² z k then find curl F at (1, 2,- 1).

Solution:

⇒ \({curl} \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x y z & z x^2 & x y^2 z
\end{array}\right|=\mathbf{i}\left(2 x y z-x^2\right)-\mathbf{j}\left(y^2 z-x y\right)+\mathbf{k}(2 x z-x z)\)

At (1,2,-1), curl F = -5i+6j-k

87. If F=x²yi- 2xzj + 2yz k, find curl F at (1, 1, 1)

Solution:

⇒ \({curl} \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x^2 y & -2 x z & 2 y z
\end{array}\right|=\mathbf{i}(2 z+2 x)-\mathbf{j}(0-0)+\mathbf{k}\left(-2 z-x^2\right)\)

At (1,1,1), curl F = 4i-3k

88. Find div F and curl F where F = xy² i + 2x²yzj- 3yz²k at (1,-1, 1).

Solution:

⇒ \({div} \mathbf{F}=\frac{\partial}{\partial x}\left(x y^2\right)+\frac{\partial}{\partial y}\left(2 x^2 y z\right)+\frac{\partial}{\partial z}\left(-3 y z^2\right)=y^2+2 x^2 z-6 y z\)

⇒ \({At}(1,-1,1), {div} \mathbf{F}=(-1)^2+2(1)^2(1)-6(-1)(1)=1+2+6=9\)

⇒ \({curl} \mathbf{F}=\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x y^2 & 2 x^2 y z & -3 y z
\end{array}\right|=\mathbf{i}\left(-3 z^2-2 x^2 y\right)-\mathbf{j}(0-0)+\mathbf{k}(4 x y z-2 x y)\)

⇒ At (1,-1,1), curl F – i (-3+2) + k (-4+2) = -i-2k = -6i

89. Iff =x²yi-2xzj + 2yzk, find (1) div f (2) curl f.

Solution: Given f=x²yi-2xyzj+2yzk

⇒ div f=\(\frac{\partial}{\partial x}\left\{x^2 y\right\}+\frac{\partial}{\partial y}\{-2 x z\}+\frac{\partial}{\partial z}\{2 y z\}\)=2xy+2y.

⇒ curl f=\(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x^2 y & -2 x z & 2 y z
\end{array}\right|\)

=i(2z+2x)-j(0-0)+k(-2z-x²)

=(2x+2z)i-(x²+2z)k.

90. If F =x²zi-2y³ z² j+xy²zk find div F and curl F at (1,- 1, 1).

Solution:

⇒ \({div} \mathbf{F}=\frac{\partial}{\partial x}\left\{x^2 z\right\}+\frac{\partial}{\partial y}\left\{-2 y^3 z^2\right\}+\frac{\partial}{\partial z}\left\{x y^2 z\right\}=2 x z-6 y^2 z^2+x y^2\)

At (1,-1,1), div F = 2-6+1 -3

⇒ \({curl} \mathbf{F}=\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x^2 z & -2 y^3 z^2 & x y^2 z
\end{array}\right|=\mathbf{i}\left[2 x y z+4 y^3 z\right]-\mathbf{j}\left[y^2 z-x^2\right]+\mathbf{k}[0-0]\)

⇒ \({At}(1,-1,1),{curl} \mathbf{F}=\mathbf{i}\left[2(1)(-1)(1)+4(-1)^3(1)\right]-\mathbf{j}\left[(-1)^2(1)-(1)^2\right]=-6 \mathbf{i}\)

91. Show that curl r = 0

Solution:

Let r=xi+yj+zk.Then \(\frac{\partial \mathbf{r}}{\partial x}\)=i, \(\frac{\partial \mathbf{r}}{\partial y}\)=j, \(\frac{\partial \mathbf{r}}{\partial z}\)=k

⇒ curl r=∇× r=\(=\mathbf{i} \times \frac{\partial \mathbf{r}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{r}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{r}}{\partial z}\)=

i×i+j×j+k×k

=0+0+0=0.

92. Show that curl (r x a) =- 2a

Solution:

⇒ \({curl}(\mathbf{r} \times \mathbf{a})=\nabla \times(\mathbf{r} \times \mathbf{a})=\mathbf{I} \times \frac{\partial}{\partial x}(\mathbf{r} \times \mathbf{a})+\mathbf{j} \times \frac{\partial}{\partial y}(\mathbf{r} \times \mathbf{a})+\mathbf{k} \times \frac{\partial}{\partial z}(\mathbf{r} \times \mathbf{a})\)

⇒ \(\mathbf{i} \times\left(\frac{\partial \mathbf{r}}{\partial x} \times \mathbf{a}\right)+\mathbf{j} \times\left(\frac{\partial \mathbf{r}}{\partial y} \times \mathbf{a}\right)+\mathbf{k} \times\left(\frac{\partial \mathbf{r}}{\partial z} \times \mathbf{a}\right)=\mathbf{i} \times(\mathbf{i} \times \mathbf{a})+\mathbf{j} \times(\mathbf{j} \times \mathbf{a})+\mathbf{k} \times(\mathbf{k} \times \mathbf{a})\)

= (i.a) i – (i.i) a + (j.a) j – (j.j) a+ (k.a) k – (k.k) a

= (i.a) i +(j.a) j + (k.a) K -3a = a-3a = -2a

93. If a and b are constant vectors then show that

1. div {(r x a) x b} =- 2 (b . a)
2. curl {(r x a) x b} = b x a .

Solution:

Let a \( =a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}, \mathbf{b}=b_1 \mathbf{i}+b_2 \mathbf{j}+b_3 \mathbf{k} \text {. }\)

⇒ \(\mathbf{r} \times \mathbf{a}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
x & y & z \\
a_1 & a_2 & a_3
\end{array}\right|=\mathbf{i}\left(a_3 y-a_2 z\right)-\mathbf{j}\left(a_3 x-a_1 z\right)+\mathbf{k}\left(a_2 x-a_1 y\right)\)

⇒ \((\mathbf{r} \times \mathbf{a}) \times \mathbf{b}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
a_3 y-a_2 z & a_1 z-a_3 x & a_2 x-a_1 y \\
b_1 & b_2 & b_3
\end{array}\right|\)

⇒ \(\mathbf{i}\left(a_1 b_3 z-a_3 b_3 x-a_2 b_2 x+a_1 b_2 y\right)-\mathbf{J}\left(a_3 b_3 y-a_2 b_3 z-a_2 b_1 x+a_1 b_1 y\right)\)

+ \(\mathbf{k}\left(a_3 b_2 y-a_2 b_2 z-a_1 b_1 z-a_3 b_1 x\right)\)

1. \({div}[(\mathbf{r} \times \mathbf{a}) \times \mathbf{b}]=\frac{\partial}{\partial x}\left(a_1 b_3 z-a_3 b_3 x-a_2 b_2 x+a_1 b_2 y\right)\)

⇒ \(+\frac{\partial}{\partial y}\left(-a_3 b_3 y+a_2 b_3 z+a_2 b_1 x-a_1 b_1 y\right)+\frac{\partial}{\partial z}\left(a_3 b_2 y-a_2 b_2 z-a_1 b_1 z+a_3 b_1 x\right)\)

⇒ \(-a_3 b_3-a_2 b_2-a_3 b_3-a_1 b_1-a_2 b_2-a_1 b_1\)

⇒ \(-2\left(a_1 b_1+a_2 b_2+a_3 b_3\right)=-2(a \cdot b)=-2(b \cdot a)\)

2. curl [(r×a)×b]

⇒ \(i\left(a_3 b_2-a_2 b_3\right)-\mathbf{j}\left(a_3 b_1-a_1 b_3\right)+\mathbf{k}\left(a_2 b_1-a_1 b_2\right)=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
b_1 & b_2 & b_3 \\
a_1 & a_2 & a_3
\end{array}\right|=\mathbf{b} \times \mathbf{a}\)

94. Define the irrotational vector point function.
Irrotational: A vector point function F is said to be irrotational if curl F =0.

95. Show that F -yz i +zx j + xy k is irrotational

Solution:

curl F= \(=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y z & z x & x y
\end{array}\right|\) =i(x-x)j(y-y)+k(z-z)=0

∴ F is irrotational.

96. Show that F = (sin y + z)i + (x cosy-z)j + (x – y)k is irrotational.

curl F\(=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\sin y+z & x \cos y-z & x-y
\end{array}\right|\)=i(-1+1)-j(1-1)+k(cosy-cosy)=0

∴ F is irrotational.

97. Show that rⁿ r  is irrotational

Solution:

Let r=xi+yj+zk. Then \(\frac{\partial \mathbf{r}}{\partial x}\)=i, \(\frac{\partial \mathbf{r}}{\partial y}\)=j,\(\frac{\partial \mathbf{r}}{\partial z}\)= k

⇒ curl r=∇×r=\(\mathbf{i} \times \frac{\partial \mathbf{r}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{r}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{r}}{\partial z}=\)=i×i+j×j+k×k=

⇒ 0+0+0=0

98. Show that f(r) r  is irrotational. Find when it is solenoidal.

Solution:

⇒ \(\mathbf{F}=r^n \mathbf{r}=r^n x \mathbf{i}+r^n y \mathbf{j}+r^n z \mathbf{k}\)

⇒ \({curl} \mathbf{F}=\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{J} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
r^n x & r^n y & r^n z
\end{array}\right|\)

⇒ \(\mathbf{i}\left[\frac{\partial}{\partial y}\left(r^n z\right)-\frac{\partial}{\partial z}\left(r^n y\right)\right]-\mathbf{j}\left[\frac{\partial}{\partial x}\left(r^n z\right)-\frac{\partial}{\partial z}\left(r^n x\right)\right]+\mathbf{k}\left[\frac{\partial}{\partial x}\left(r^n y\right)-\frac{\partial}{\partial y}\left(r^n x\right)\right]\)

⇒ \(i\left[n r^{n-1} z \cdot \frac{y}{r}-n r^{n-1} \cdot y \cdot \frac{z}{r}\right]-j\left[n r^{n-1} z \cdot \frac{x}{r}-n r^{n-1} x \cdot \frac{z}{r}\right]+k\left[n r^{n-1} y \cdot \frac{x}{r}-n r^{n-1} x \cdot \frac{y}{r}\right]\)

= 0

∴ F is irrational

⇒ \(r^n \mathbf{r}=r^n x \mathbf{I}+r^n y \mathbf{j}+r^n z \mathbf{k} \text { is solenoidal }\)

⇒ \(r^n+x n r^{n-1} \frac{\partial r}{\partial x}+r^n+y n r^{n-1} \frac{\partial r}{\partial y}+r^n+z n r^{n-1} \frac{\partial r}{\partial z}=0\)

⇒ \(r^n+x n r^{n-1} \frac{\partial r}{\partial x}+r^n+y n r^{n-1} \frac{\partial r}{\partial y}+r^n+z n r^{n-1} \frac{\partial r}{\partial z}=0\)

⇒ \(3 r^n+n r^{n-1} \frac{x^2}{r}+n r^{n-1} \frac{y^2}{r}+n r^{n-1} \frac{z^2}{r}=0 \Rightarrow 3 r^n+n r^{n-2}\left(x^2+y^2+z^2\right)=0\)

⇒ \(3 r^n+n r^n=0 \Rightarrow n+3=0 \Rightarrow n=-3 .\)

99. Show that f(r) r is irrotational.

Solution:

⇒ \(\nabla \times \frac{\mathbf{r}}{r^2}=\nabla \times\left(\frac{x}{r^2} \mathbf{i}+\frac{y}{r^2} \mathbf{j}+\frac{z}{r^2} \mathbf{k}\right)=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x / r^2 & y / r^2 & z / r^2
\end{array}\right|\)

⇒ \(i\left[\frac{\partial}{\partial y}\left(\frac{z}{r^2}\right)-\frac{\partial}{\partial z}\left(\frac{y}{r^2}\right)\right]-\mathbf{j}\left[\frac{\partial}{\partial x}\left(\frac{z}{r^2}\right)=\frac{\partial}{\partial z}\left(\frac{x}{r^2}\right)\right]+\mathbf{k}\left[\frac{\partial}{\partial x}\left(\frac{y}{r^2}\right)-\frac{\partial}{\partial y}\left(\frac{x}{r^2}\right)\right]\)

⇒ \(i\left[-2 r^{-3} \frac{\partial r}{\partial y} \cdot z+2 r^{-3} \frac{\partial r}{\partial z} y\right]-\mathbf{j}\left[-2 r^{-3} \frac{\partial r}{\partial x} z+2 r^{-3} \frac{\partial r}{\partial z} x\right]+\mathbf{k}\left[-2 r^{-3} \frac{\partial r}{\partial x} y+2 r^{-3} \frac{\partial r}{\partial y} x\right]\)

⇒ \(\mathbf{i}\left(-2 r^{-4} y z+2 r^{-4} z y\right)-\mathbf{j}\left(-2 r^{-4} x z+2 r^{-4} z x\right)+\mathbf{k}\left(-2 r^{-4} x y+2 r^{-4} y z\right)=\mathbf{0}\)

∴ \(r / r^2\) is irrational

100. Show that r/r² is always irrotational

Solution:

⇒ \(\nabla \times \frac{\mathbf{r}}{r^2}=\nabla \times\left(\frac{x}{r^2} \mathbf{i}+\frac{y}{r^2} \mathbf{j}+\frac{z}{r^2} \cdot \mathbf{k}\right)=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x / r^2 & y / r^2 & z / r^2
\end{array}\right|\)

⇒ \(\mathbf{i}\left[\frac{\partial}{\partial y}\left(\frac{z}{r^2}\right)-\frac{\partial}{\partial z}\left(\frac{y}{r^2}\right)\right]-\mathbf{j}\left[\frac{\partial}{\partial x}\left(\frac{z}{r^2}\right)=\frac{\partial}{\partial z}\left(\frac{x}{r^2}\right)\right]+\mathbf{k}\left[\frac{\partial}{\partial x}\left(\frac{y}{r^2}\right)-\frac{\partial}{\partial y}\left(\frac{x}{r^2}\right)\right]\)

⇒ \(i\left[-2 r^{-3} \frac{\partial r}{\partial y} \cdot z+2 r^{-3} \frac{\partial r}{\partial z} y\right]-\mathbf{j}\left[-2 r^{-3} \frac{\partial r}{\partial x} z+2 r^{-3} \frac{\partial r}{\partial z} x\right]+\mathbf{k}\left[-2 r^{-3} \frac{\partial r}{\partial x} y+2 r^{-3} \frac{\partial r}{\partial y} x\right]\)

⇒ \(\mathbf{i}\left(-2 r^{-4} y z+2 r^{-4} z y\right)-\mathbf{j}\left(-2 r^{-4} x z+2 r^{-4} z x\right)+\mathbf{k}\left(-2 r^{-4} x y+2 r^{-4} y z\right)=\mathbf{0}\)

∴ \(\mathbf{r} / \boldsymbol{r}^2\) is irrational

101. Find the constants a, b, c so that (x + 2y + az)i + (bx −3y-z)j + (4x + cy + 2z) k is irrotational.

Solution:

⇒ \((x+2 y+a z) \mathbf{i}+(b x-3 y-z) \mathbf{j}+(4 x+c y+2 z) \mathbf{k} \text { is irrotational }\)

⇒ \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x+2 y+a z & b x-3 y-z & 4 x+c y+2 z
\end{array}\right|=0\)

⇒ \(\mathbf{i}(c+1)-\mathbf{j}(4-a)+\mathbf{k}(b-2)=\mathbf{0} \Rightarrow c+1=0,4-a=0, b-2=0\)

⇒ a = 4, b = 2, c = 1

102. If A is a differentiable vector point function and φ is a differentiable scalar point function then prove that div (φA) = ( gradφ ).A + φ (div A).

Solution:

curl (φA)=∇×(φA) =i×\(\frac{\partial}{\partial x}\)(φA)+j×\(\frac{\partial}{\partial y}\)(φA)+k× \(\frac{\partial}{\partial y}\) (φA)

⇒ \({div}(\varphi \mathbf{A})=\nabla \cdot(\varphi \mathbf{A})=\mathbf{i} \cdot \frac{\partial}{\partial x}(\varphi \mathbf{A})+\mathbf{j} \cdot \frac{\partial}{\partial y}(\varphi \mathbf{A})+\mathbf{k} \cdot \frac{\partial}{\partial z}(\varphi \mathbf{A})\)

⇒ \(\mathbf{i} \cdot\left[\frac{\partial \varphi}{\partial x} \mathbf{A}+\varphi \frac{\partial \mathbf{A}}{\partial x}\right]+\mathbf{j} \cdot\left[\frac{\partial \varphi}{\partial y} \mathbf{A}+\varphi \frac{\partial \mathbf{A}}{\partial y}\right]+\mathbf{k} \cdot\left[\frac{\partial \varphi}{\partial z} \mathbf{A}+\varphi \frac{\partial \mathbf{A}}{\partial z}\right]\)

⇒ \(\mathbf{i} \frac{\partial \varphi}{\partial x} \cdot \mathbf{A}+\varphi\left(\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x}\right)+\mathbf{j} \frac{\partial \varphi}{\partial y} \cdot \mathbf{A}+\varphi\left(\mathbf{j} \cdot \frac{\partial \mathbf{A}}{\partial y}\right)+\mathbf{k} \frac{\partial \varphi}{\partial z} \cdot \mathbf{A}+\varphi\left(\mathbf{k} \cdot \frac{\partial \mathbf{A}}{\partial z}\right)\)

⇒ \(\left[\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{J} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right] \cdot \mathbf{A}+\varphi\left[\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{A}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{A}}{\partial z}\right]\)

⇒ \((\nabla \varphi) \cdot \mathbf{A}+\varphi(\nabla \cdot \mathbf{A})=({grad} \varphi) \cdot \mathbf{A}+\varphi({div} \mathbf{A})\)

103. If A is a differentiable vector point function and φ is a differentiable scalar point function then prove that curl (φA) = (grad φ)x A + φ ( curl A).

Solution:

⇒ curl (φA)=∇ × (φA)\(=\mathbf{i} \times \frac{\partial}{\partial x}(\varphi \mathbf{A})+\mathbf{j} \times \frac{\partial}{\partial y}(\varphi \mathbf{A})+\mathbf{k} \times \frac{\partial}{\partial z}(\varphi \mathbf{A})\)

⇒ \(\mathbf{i} \times\left[\frac{\partial \varphi}{\partial x} \mathbf{A}+\varphi \frac{\partial \mathbf{A}}{\partial x}\right]+\mathbf{j} \times\left[\frac{\partial \varphi}{\partial y} \mathbf{A}+\varphi \frac{\partial \mathbf{A}}{\partial y}\right]+\mathbf{k} \times\left[\frac{\partial \varphi}{\partial z} \mathbf{A}+\varphi \frac{\partial \mathbf{A}}{\partial z}\right]\)

⇒ \(\mathbf{i} \frac{\partial \varphi}{\partial x} \times \mathbf{A}+\varphi\left(\mathbf{i} \times \frac{\partial \mathbf{A}}{\partial x}\right)+\mathbf{j} \frac{\partial \varphi}{\partial y} \times \mathbf{A}+\varphi\left(\mathbf{j} \times \frac{\partial \mathbf{A}}{\partial y}\right)+\mathbf{k} \frac{\partial \varphi}{\partial z} \times \mathbf{A}+\varphi\left(\mathbf{k} \times \frac{\partial \mathbf{A}}{\partial z}\right)\)

⇒ \(\left[\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right] \times \mathbf{A}+\varphi\left[\mathbf{i} \times \frac{\partial \mathbf{A}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{A}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{A}}{\partial z}\right]\)

⇒ \((\nabla \varphi) \times \mathbf{A}+\varphi(\nabla \times \mathbf{A})=({grad} \varphi) \times \mathbf{A}+\varphi({curl} \mathbf{A})\)

104. If A and B are two differentiable vector point functions then prove that

Solution:
grad (A.B) = (B.∇) A + (A.∇) B + B x (curl A) + A x (curl B)

⇒ \(\mathbf{A} \times({curl} \mathbf{B})=\mathbf{A} \times(\nabla \times \mathbf{B})=\mathbf{A} \times\left(\mathbf{i} \times \frac{\partial \mathbf{B}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{B}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{B}}{\partial z}\right)\)

⇒ \(\mathbf{A} \times\left(\mathbf{i} \times \frac{\partial \mathbf{B}}{\partial x}\right)+\mathbf{A} \times\left(\mathbf{j} \times \frac{\partial \mathbf{B}}{\partial y}\right)+\mathbf{A} \times\left(\mathbf{k} \times \frac{\partial \mathbf{B}}{\partial z}\right)\)

⇒ \(\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right) \mathbf{i}-(\mathbf{A} \cdot \mathbf{i}) \frac{\partial \mathbf{B}}{\partial x}+\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial y}\right) \mathbf{j}-(\mathbf{A} \cdot \mathbf{j}) \frac{\partial \mathbf{B}}{\partial y}+\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial z}\right) \mathbf{k}-(\mathbf{A} \cdot \mathbf{k}) \frac{\partial \mathbf{B}}{\partial z}\)

⇒ \(\mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right)+\mathbf{j}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial y}\right)+\mathbf{k}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial z}\right)-\left[(\mathbf{A} \cdot \mathbf{i}) \frac{\partial \mathbf{B}}{\partial x}+(\mathbf{A} \cdot \mathbf{j}) \frac{\partial \mathbf{B}}{\partial y}+(\mathbf{A} \cdot \mathbf{k}) \frac{\partial \mathbf{B}}{\partial z}\right]\)

⇒ \(\mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right)+\mathbf{j}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial y}\right)+\mathbf{k}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial z}\right)-(\mathbf{A} \cdot \nabla) \mathbf{B}\)

⇒ \(\text { Similarly } \mathbf{B} \times({curl} \mathbf{A})=\mathbf{i}\left(\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial x}\right)+\mathbf{j}\left(\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial y}\right)+\mathbf{k}\left(\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial z}\right)-(\mathbf{B} \cdot \nabla) \mathbf{A}\)

∴ \(\mathbf{A} \times({curl} \mathbf{B})+\mathbf{B} \times({curl} \mathbf{A})\)

⇒ \(\mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}+\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial x}\right)+\mathbf{j}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial y}+\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial y}\right)+\mathbf{k}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial z}+\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial z}\right)-(\mathbf{A} \cdot \nabla) \mathbf{B}-(\mathbf{B} \cdot \nabla) \mathbf{A}\)

⇒ \(\mathbf{A} \times({curl} \mathbf{B})+\mathbf{B} \times({curl} \mathbf{A})+(\mathbf{A} \cdot \nabla) \mathbf{B}+(\mathbf{B} \cdot \nabla) \mathbf{A}\)

⇒ \(\mathbf{i} \frac{\partial}{\partial x}(\mathbf{A} \cdot \mathbf{B})+\mathbf{j} \frac{\partial}{\partial y}(\mathbf{A} \cdot \mathbf{B})+\mathbf{k} \frac{\partial}{\partial z}(\mathbf{A} \cdot \mathbf{B})=\nabla(\mathbf{A} \cdot \mathbf{B})={grad}(\mathbf{A} \cdot \mathbf{B})\)

105. If A and B are two differential vector point functions then prove that div(A×B)= B.(curl A)− A.(curl B)

Solution:

div (A×B) = \(\nabla \cdot(\mathbf{A} \times \mathbf{B})=\mathbf{i} \cdot \frac{\partial}{\partial x}(\mathbf{A} \times \mathbf{B})+\mathbf{j} \cdot \frac{\partial}{\partial y}(\mathbf{A} \times \mathbf{B})+\mathbf{k} \cdot \frac{\partial}{\partial z}(\mathbf{A} \times \mathbf{B})\)

⇒ \(\mathbf{i} \cdot\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)+\mathbf{j} \cdot\left(\frac{\partial \mathbf{A}}{\partial y} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial y}\right)+\mathbf{k} \cdot\left(\frac{\partial \mathbf{A}}{\partial z} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial z}\right)\)

⇒ \(\left(\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}\right)+\mathbf{i} \cdot\left(\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)+\left(\mathbf{j} \cdot \frac{\partial \mathbf{A}}{\partial y} \times \mathbf{B}\right)+\mathbf{j} \cdot\left(\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial y}\right)+\left(\mathbf{k} \cdot \frac{\partial \mathbf{A}}{\partial z} \times \mathbf{B}\right)+\mathbf{k} \cdot\left(\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial z}\right)\)

⇒ \(\left(\mathbf{i} \times \frac{\partial \mathbf{A}}{\partial x}\right) \cdot \mathbf{B}+\left(\mathbf{j} \times \frac{\partial \mathbf{A}}{\partial y}\right) \cdot \mathbf{B}+\left(\mathbf{k} \times \frac{\partial \mathbf{A}}{\partial z}\right) \cdot \mathbf{B}-\mathbf{i} \cdot\left(\frac{\partial \mathbf{B}}{\partial x} \times \mathbf{A}\right)-\mathbf{j} \cdot\left(\frac{\partial \mathbf{B}}{\partial y} \times \mathbf{A}\right)-\mathbf{k} \cdot\left(\frac{\partial \mathbf{B}}{\partial z} \times \mathbf{A}\right)\)

⇒ \(\left(\mathbf{i} \times \frac{\partial \mathbf{A}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{A}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{A}}{\partial z}\right) \cdot \mathbf{B}-\left[\left(\mathbf{I} \times \frac{\partial \mathbf{B}}{\partial x}\right) \cdot \mathbf{A}+\left(\mathbf{j} \times \frac{\partial \mathbf{B}}{\partial y}\right) \cdot \mathbf{A}+\left(\mathbf{k} \times \frac{\partial \mathbf{B}}{\partial z}\right) \cdot \mathbf{A}\right]\)

⇒ \((\nabla \times \mathbf{A}) \cdot \mathbf{B}-\left[\mathbf{i} \times \frac{\partial \mathbf{B}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{B}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{B}}{\partial z}\right] \cdot \mathbf{A}=(\nabla \times \mathbf{A}) \cdot \mathbf{B}-(\nabla \times \mathbf{B}) \cdot \mathbf{A}\)

= B. (curl A) – A. (curl B).

106. If A and B are irrotational vector point functions then A×B is solenoidal.

Solution:

A, B are irrotational curl A=0, curl B

div(A×B)=B. (curl A)-A. (curl B)=B.0-A.0.

∴ A×B is solenoidal.

107. If A and B are two differentiable vector point functions then prove that.

Solution:

⇒ \({curl}(\mathbf{A} \times \mathbf{B})=\nabla \times(\mathbf{A} \times \mathbf{B})=\mathbf{i} \times \frac{\partial}{\partial x}(\mathbf{A} \times \mathbf{B})+\mathbf{j} \times \frac{\partial}{\partial y}(\mathbf{A} \times \mathbf{B})+\mathbf{k} \times \frac{\partial}{\partial z}(\mathbf{A} \times \mathbf{B})\)

⇒ \(\mathbf{i} \times\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)+\mathbf{j} \times\left(\frac{\partial \mathbf{A}}{\partial y} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial y}\right)+\mathbf{k} \times\left(\frac{\partial \mathbf{A}}{\partial z} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial z}\right)\)

⇒ \(\mathbf{i} \times\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}\right)+\mathbf{i} \times\left(\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)+\mathbf{j} \times\left(\frac{\partial \mathbf{A}}{\partial y} \times \mathbf{B}\right)\)

= \(+\mathbf{j} \times\left(\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial y}\right)+\mathbf{k} \times\left(\frac{\partial \mathbf{A}}{\partial z} \times \mathbf{B}\right)+\mathbf{k} \times\left(\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial z}\right)\)

⇒ \((\mathbf{i} \cdot \mathbf{B}) \frac{\partial \mathbf{A}}{\partial x}-\left(\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x}\right) \mathbf{B}+\left(\mathbf{i} \cdot \frac{\partial \mathbf{B}}{\partial x}\right) \mathbf{A}-(\mathbf{i} \cdot \mathbf{A}) \frac{\partial \mathbf{B}}{\partial x}+(\mathbf{j} \cdot \mathbf{B}) \frac{\partial \mathbf{A}}{\partial y}-\left(\mathbf{j} \cdot \frac{\partial \mathbf{A}}{\partial y}\right) \mathbf{B}\)

⇒ \(+\left(\mathbf{j} \cdot \frac{\partial \mathbf{B}}{\partial y}\right) \mathbf{A}-(\mathbf{j} \cdot \mathbf{A}) \frac{\partial \mathbf{B}}{\partial y}+(\mathbf{k} \cdot \mathbf{B}) \frac{\partial \mathbf{A}}{\partial z}-\left(\mathbf{k} \cdot \frac{\partial \mathbf{A}}{\partial z}\right) \mathbf{B}+\left(\mathbf{k} \cdot \frac{\partial \mathbf{B}}{\partial z}\right) \mathbf{A}-(\mathbf{k} \cdot \mathbf{A}) \frac{\partial \mathbf{B}}{\partial z}\)

⇒ \(\left[(\mathbf{B} \cdot \mathbf{i}) \frac{\partial \mathbf{A}}{\partial x}+(\mathbf{B} \cdot \mathbf{j}) \frac{\partial \mathbf{A}}{\partial y}+(\mathbf{B} \cdot \mathbf{k}) \frac{\partial \mathbf{A}}{\partial z}\right]-\mathbf{B}\left[\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{A}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{A}}{\partial z}\right]\)

⇒ \(+\mathbf{A}\left[\mathbf{i} \cdot \frac{\partial \mathbf{B}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{B}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{B}}{\partial z}\right]-\left[(\mathbf{A} \cdot \mathbf{i}) \frac{\partial \mathbf{B}}{\partial x}+(\mathbf{A} \cdot \mathbf{j}) \frac{\partial \mathbf{B}}{\partial y}+(\mathbf{A} \cdot \mathbf{k}) \frac{\partial \mathbf{B}}{\partial z}\right]\)

⇒ \((\mathbf{B} \cdot \nabla) \mathbf{A}-\mathbf{B}(\nabla \cdot \mathbf{A})+\mathbf{A}(\nabla \cdot \mathbf{B})-(\mathbf{A} \cdot \nabla) \mathbf{B}\)

⇒ \(\mathbf{A}({div} \mathbf{B})-\mathbf{B}({div} \mathbf{A})+(\mathbf{B} \cdot \nabla) \mathbf{A}-(\mathbf{A} \cdot \nabla) \mathbf{B}\)

108. If  φ is a differentiable scalar function then prove that

Solution:

div grad φ=∇.∇φ\(=\frac{\partial^2 \varphi}{\partial x^2}+\frac{\partial^2 \varphi}{\partial y^2}+\frac{\partial^2 \varphi}{\partial z^2}\)

∇.∇φ=∇.\(\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right)\)

⇒ \(\frac{\partial}{\partial x}\left(\frac{\partial \varphi}{\partial x}\right)+\frac{\partial}{\partial y}\left(\frac{\partial \varphi}{\partial y}\right)+\frac{\partial}{\partial z}\left(\frac{\partial \varphi}{\partial z}\right)\)

= \(\frac{\partial^2 \varphi}{\partial x^2}+\frac{\partial^2 \varphi}{\partial y^2}+\frac{\partial^2 \varphi}{\partial z^2}\)

109. If  φ is a differentiable scalar point function then the prove that curl (grad φ)=0

Solution:

⇒ \({grad} \varphi=\nabla \varphi=\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)

⇒ \({curl}({grad} \varphi)=\nabla \times(\nabla \varphi)=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\frac{\partial \varphi}{\partial x} & \frac{\partial \varphi}{\partial y} & \frac{\partial \varphi}{\partial z}
\end{array}\right|\)

⇒ \(\mathbf{I}\left(\frac{\partial^2 \varphi}{\partial y \partial z}-\frac{\partial^2 \varphi}{\partial y \partial z}\right)-\mathbf{J}\left(\frac{\partial^2 \varphi}{\partial x \partial z}-\frac{\partial^2 \varphi}{\partial z \partial x}\right)+\mathbf{k}\left(\frac{\partial^2 \varphi}{\partial x \partial y}-\frac{\partial^2 \varphi}{\partial y \partial x}\right)=\mathbf{0}\)

110. If φ  is a differentiable scalar point function then this proves that grad φ  is irrotational.

Solution:

φ  is a differentiable scalar point function ⇒ curl (grad φ)=0

⇒ grad φ is irrotational.

111. If F is a differentiable vector point function then the prove that div (curl f)=0

Solution:

Let F = \(F_1 \mathbf{i}+F_2 \mathbf{j}+F_3 \mathbf{k}\)

⇒ \({curl} \mathbf{F}=\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_1 & F_2 & F_3
\end{array}\right|=\mathbf{i}\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)-\mathbf{J}\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right)+\mathbf{k}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\)

⇒ div (curl F) = \(\frac{\partial}{\partial x} \cdot\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)-\frac{\partial}{\partial y}\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right)+\frac{\partial}{\partial z}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\)

⇒ \(\frac{\partial^2 F_3}{\partial x \partial y}-\frac{\partial^2 F_2}{\partial x \partial z}-\frac{\partial^2 F_3}{\partial y \partial x}+\frac{\partial^2 F_1}{\partial y \partial z}+\frac{\partial^2 F_2}{\partial z \partial x}-\frac{\partial^2 F_1}{\partial z \partial y}=0\)

112. If F is a differentiable vector point function then prove that curl F is solenoidal.

Solution: Fis differentiable   vector point function ⇒  div (curl F)=0

⇒  curl F is solenoidal.

113. If F is a differentiable vector point function then prove that curl (curl F)=grad (div F)-∇²F.

Solution:

Curl (curl F) = \(\nabla \times(\nabla \times \mathbf{F})=\mathbf{i} \times \frac{\partial}{\partial x}(\nabla \times \mathbf{F})+\mathbf{j} \times \frac{\partial}{\partial y}(\nabla \times \mathbf{F})+\mathbf{k} \times \frac{\partial}{\partial z}(\nabla \times \mathbf{F})\)

⇒ \(\mathbf{i} \times \frac{\partial}{\partial x}\left(\mathbf{i} \times \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{F}}{\partial z}\right)\)

+\(\mathbf{j} \times \frac{\partial}{\partial y}\left(\mathbf{i} \times \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{F}}{\partial z}\right)\)

⇒ \(+\mathbf{k} \times \frac{\partial}{\partial z}\left(\mathbf{i} \times \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{F}}{\partial z}\right)\)

⇒ \(\mathbf{i} \times\left(\mathbf{i} \times \frac{\partial^2 \mathbf{F}}{\partial x^2}+\mathbf{j} \times \frac{\partial^2 \mathbf{F}}{\partial x \partial y}+\mathbf{k} \times \frac{\partial^2 \mathbf{F}}{\partial x \partial z}\right)\)

+ \(\mathbf{j} \times\left(\mathbf{i} \times \frac{\partial^2 \mathbf{F}}{\partial y \partial x}+\mathbf{j} \times \frac{\partial^2 \mathbf{F}}{\partial y^2}+\mathbf{k} \times \frac{\partial^2 \mathbf{F}}{\partial y \partial z}\right)\)

⇒ \(+\mathbf{k} \times\left(\mathbf{i} \times \frac{\partial^2 \mathbf{F}}{\partial z \partial x}+\mathbf{j} \times \frac{\partial^2 \mathbf{F}}{\partial z \partial y}+\mathbf{k} \times \frac{\partial^2 \mathbf{F}}{\partial z^2}\right)\)

⇒ \(\left(\mathbf{i} \cdot \frac{\partial^2 \mathbf{F}}{\partial x^2}\right) \mathbf{i}-(\mathbf{i} \cdot \mathbf{i}) \frac{\partial^2 \mathbf{F}}{\partial x}+\left(\mathbf{i} \cdot \frac{\partial^2 \mathbf{F}}{\partial x \partial y}\right) \mathbf{j}-(\mathbf{i} \cdot \mathbf{j}) \frac{\partial^2 \mathbf{F}}{\partial x \partial y}+\left(\mathbf{i} \cdot \frac{\partial^2 \mathbf{F}}{\partial x \partial z}\right) \mathbf{k}-(\mathbf{i} \cdot \mathbf{k}) \frac{\partial^2 \mathbf{F}}{\partial x \partial z}\)

⇒ \(+\left(\mathbf{j} \cdot \frac{\partial^2 \mathbf{F}}{\partial y \partial x}\right) \mathbf{i}-(\mathbf{j} \cdot \mathbf{i}) \frac{\partial^2 \mathbf{F}}{\partial y \partial x}+\left(\mathbf{j} \cdot \frac{\partial^2 \mathbf{F}}{\partial y^2}\right) \mathbf{j}-(\mathbf{j} \cdot \mathbf{j}) \frac{\partial^2 \mathbf{F}}{\partial y^2}+\left(\mathbf{j} \cdot \frac{\partial^2 \mathbf{F}}{\partial y \partial z}\right) \mathbf{k}\)

⇒ \(-(\mathbf{J} \cdot \mathbf{k}) \frac{\partial^2 \mathbf{F}}{\partial y \partial z}+\left(\mathbf{k} \cdot \frac{\partial^2 \mathbf{F}}{\partial z \partial x}\right) \mathbf{i}-(\mathbf{k} \cdot \mathbf{i}) \frac{\partial^2 \mathbf{F}}{\partial z \partial x}+\left(\mathbf{k} \cdot \frac{\partial^2 \mathbf{F}}{\partial z \partial y}\right) \mathbf{j}-(\mathbf{k} \cdot \mathbf{j}) \frac{\partial^2 \mathbf{F}}{\partial z \partial y}+\left(\mathbf{k} \cdot \frac{\partial^2 \mathbf{F}}{\partial z^2}\right) \mathbf{k}-(\mathbf{k} \cdot \mathbf{k}) \frac{\partial^2 \mathbf{F}}{\partial z^2}\)

⇒ \(\mathbf{i} \frac{\partial}{\partial x}\left(\mathbf{i} \cdot \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{F}}{\partial z}\right)+\mathbf{j} \frac{\partial}{\partial y}\left(\mathbf{i} \cdot \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{F}}{\partial z}\right)\)

⇒ \(+\mathbf{k} \frac{\partial}{\partial z}\left(\mathbf{i} \cdot \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{F}}{\partial z}\right)-\left(\frac{\partial^2 \mathbf{F}}{\partial x^2}+\frac{\partial^2 \mathbf{F}}{\partial y^2}+\frac{\partial^2 \mathbf{F}}{\partial z^2}\right)\)

⇒ \(\mathbf{i} \frac{\partial}{\partial x}(\nabla \cdot \mathbf{F})+\mathbf{j} \frac{\partial}{\partial x}(\nabla \cdot \mathbf{F})+\mathbf{k} \frac{\partial}{\partial z}(\nabla \cdot \mathbf{F})-\nabla^2 \mathbf{F}=\nabla(\nabla \cdot \mathbf{F})-\nabla^2 \mathbf{F}={grad}({div} \mathbf{F})-\nabla^2 \mathbf{F}\)

114.Show that ∇² (rⁿ)=n(n+1)r^n-2

Solution:

⇒ \(\nabla^2\left(r^n\right)=\frac{\partial^2}{\partial x^2}\left(r^n\right)+\frac{\partial^2}{\partial y^2}\left(r^n\right)+\frac{\partial^2}{\partial z^2}\left(r^n\right)\)

⇒ \(\frac{\partial}{\partial x}\left[n r^{n-1} x / r\right]+\frac{\partial}{\partial y}\left[n r^{n-1} y / r\right]+\frac{\partial}{\partial z}\left[n r^{n-1} z / r\right]\)

⇒ \(n r^{n-2}+n(n-2) x \cdot r^{n-3} x / r+n r^{n-2}+n(n-2) y r^{n-3} y / r+n r^{n-2}\)

⇒ \(+n(n-2) z \cdot r^{n-3} z / r\)

⇒ \(3 n r^{n-2}+n(n-2) r^{n-4}\left(x^2+y^2+z^2\right)=3 n r^{n-2}+n(n-2) r^{n-2}\)

⇒ \(n(3+n-2) r^{n-2}=n(n+1) r^{n-2}\)

115. Show that ∇² \(\left(\frac{1}{r}\right)\)=0

Solution:

⇒ \(\nabla^2\left(\frac{1}{r}\right)=\frac{\partial^2}{\partial x^2}\left(\frac{1}{r}\right)+\frac{\partial^2}{\partial y^2}\left(\frac{1}{r}\right)+\frac{\partial^2}{\partial z^2}\left(\frac{1}{r}\right)=\frac{\partial}{\partial x}\left\{-\frac{1}{r^2} \frac{x}{r}\right\}+\frac{\partial}{\partial y}\left\{-\frac{1}{r^2} \frac{y}{r}\right\}+\frac{\partial}{\partial z}\left\{-\frac{1}{r^2} \frac{z}{r}\right\}\)

⇒ \(\frac{r^3(-1)+x 3 r^2\left(\frac{x}{r}\right)}{r^6}+\frac{r^3(-1)+y 3 r^2\left(\frac{y}{r}\right)}{r^6}+\frac{r^3(-1)+z 3 r^2\left(\frac{z}{r}\right)}{r^6}\)

⇒ \(-\frac{1}{r^3}+\frac{3 x^2}{r^5}-\frac{1}{r^3}+\frac{3 y^2}{r^5}-\frac{1}{r^3}+\frac{3 z^2}{r^5}=-\frac{3}{r^3}+\frac{3}{r^5}\left(x^2+y^2+z^2\right)=-\frac{3}{r^3}+\frac{3}{r^3}=0\)

116. Prove that ∇² (log r)=\(\frac{1}{r^2}\)

Solution:

⇒ \(\nabla^2(\log r)=\frac{\partial^2}{\partial x^2}(\log r)+\frac{\partial^2}{\partial y^2}(\log r)+\frac{\partial^2}{\partial z^2}(\log r)\)

⇒ \(\frac{\partial}{\partial x}\left\{\frac{1}{r} \cdot \frac{x}{r}\right\}+\frac{\partial}{\partial y}\left\{\frac{1}{r} \cdot \frac{y}{r}\right\}+\frac{\partial}{\partial z}\left\{\frac{1}{2} \cdot \frac{z}{r}\right\}\)

⇒ \(\frac{r^2 1-x 2 r\left(\frac{x}{r}\right)}{r^4}+\frac{r^2-y 2 r\left(\frac{y}{r}\right)}{r^4}+\frac{r^2-z 2 r\left(\frac{z}{x}\right)}{r^4}=\frac{3 r^2-2 x^2-2 y^2-2 z^2}{r^4}=\frac{3 r^2-2 r^2}{r^4}=\frac{1}{r^2}\)

117. Show that ∇² \(\left(\frac{x}{r^3}\right)\)=0.

Solution:

⇒ \(\nabla^2\left(\frac{x}{r^3}\right)=\frac{\partial^2}{\partial x^2}\left(\frac{x}{r^3}\right)+\frac{\partial^2}{\partial y^2}\left(\frac{x}{r^3}\right)+\frac{\partial^2}{\partial z^2}\left(\frac{x}{r^3}\right)\)

⇒ \(\frac{\partial}{\partial x}\left[\frac{r^3-x \cdot 3 r^2 \cdot x / r}{r^6}\right]+\frac{\partial}{\partial y}\left[\frac{-3 x}{r^4} \cdot \frac{y}{r}\right]+\frac{\partial}{\partial z}\left[\frac{-3 x}{r^4} \cdot \frac{z}{r}\right]\)

⇒ \(\frac{\partial}{\partial x}\left[\frac{1}{r^3}-\frac{3 x^2}{r^5}\right]+\frac{\partial}{\partial y}\left[\frac{-3 x y}{r^5}\right]+\frac{\partial}{\partial z}\left[\frac{-3 x z}{r^5}\right]\)

⇒ \(-\frac{3}{r^4} \cdot \frac{x}{r}-\frac{r^5(6 x)-3 x^2 \cdot 5 r^4 x / r}{r^{10}}-\frac{r^5(3 x)-3 x y \cdot 5 r^4 \cdot y / r}{r^{10}}-\frac{r^5(3 x)-3 x z \cdot 5 r^4 \cdot z / r}{r^{10}}\)

⇒ \(-\frac{3 x}{r^5}-\frac{6 x}{r^5}+\frac{15 x^3}{r^7}-\frac{3 x}{r^5}+\frac{15 x y^2}{r^7}-\frac{3 x}{r^5}+\frac{15 x z^2}{r^7}\)

⇒ \(\frac{-15 x}{r^5}+\frac{15 x\left(x^2+y^2+z^2\right)}{r^7}=\frac{-15 x}{r^5}+\frac{15 x}{r^5}=0\)

118. Show that  ∇²  \(\left(\frac{r}{r^3}\right)\)

Solution:

⇒ \(\nabla^2\left(\frac{\boldsymbol{r}}{r^3}\right)=\frac{\partial^2}{\partial x^2}\left(\frac{x}{r^3}\right)+\frac{\partial^2}{\partial y^2}\left(\frac{y}{r^3}\right)+\frac{\partial^2}{\partial z^2}\left(\frac{z}{r^3}\right)\)

⇒ \(\frac{r^3-x \cdot 3 r^2 \cdot x / r}{r^6}+\frac{r^3-y \cdot 3 r^2 \cdot y / r}{r^6}+\frac{r^3-z \cdot 3 r^2 \cdot z / r}{r^6}\)

⇒ \(\frac{3 r^3-3 r\left(x^2+y^2+z^2\right)}{r^6}=\frac{3 r^3-3 r^3}{r^6}=0\)

119. Show that ∇²  f(r) =\(=\frac{d^2 f}{\partial r^2}+\frac{2}{r} \frac{d f}{d r}\)

Solution:

⇒ \(\nabla^2 f(r)=\nabla \cdot\{\nabla f(r)\}={div}\{{grad} f(r)\}={div} .\left\{f^{\prime}(r){grad} r\right\}={div}\left\{\frac{1}{r} f^{\prime}(r) \mathbf{r}\right\}\)

⇒ \(\frac{1}{r} f^{\prime}(r){div} \mathbf{r}+\mathbf{r} \cdot{grad}\left\{\frac{1}{r} f^{\prime}(r)\right\}=\frac{3}{r} f^{\prime}(r)+\mathbf{r} \cdot\left[\frac{d}{d r}\left\{\frac{1}{r} f^{\prime}(r)\right\}{grad} r\right]\)

⇒ \(\frac{3}{r} f^{\prime}(r)+\mathbf{r} \cdot\left[\left\{-\frac{1}{r^2} f^{\prime}(r)+\frac{1}{r} f^{\prime \prime}(r)\right\} \frac{1}{r} \mathbf{r}\right]=\frac{3}{r} f^{\prime}(r)+\left[\frac{1}{r}\left\{-\frac{1}{r^2} f^{\prime}(r)+\frac{1}{r} f^{\prime \prime \prime}(r)\right\}\right](\mathbf{r} \cdot \mathbf{r})\)

⇒ \(\frac{3}{r} f^{\prime}(r)+\left[\frac{1}{r}\left\{-\frac{1}{r^2} f^{\prime}(r)+\frac{1}{r} f^{\prime \prime}(r)\right\}\right] r^2=\frac{3}{r} f^{\prime}(r)-\frac{1}{r} f^{\prime}(r)+f^{\prime \prime}(r)=f^{\prime \prime}(r)+\frac{2}{r} f^{\prime}(r)\)

120. Show that ∇\(\left[r \nabla\left(\frac{1}{r^3}\right)\right]\)\(=\frac{3}{r^4}\)

Solution:

⇒ \(\nabla\left(\frac{i}{r^3}\right)={grad} r^{-3}=\frac{\partial}{\partial x}\left(r^{-3}\right) \mathbf{i}+\frac{\partial}{\partial y}\left(r^{-3}\right) \mathbf{j}+\frac{\partial}{\partial z}+\left(r^{-3}\right) \mathbf{k}\)

⇒ \(-3 r^{-4}\left[\frac{\partial r}{\partial x} \mathbf{i}+\frac{\partial r}{\partial y} \mathbf{j}+\frac{\partial r}{\partial z} \mathbf{k}\right]=-3 r^{-4}\left[\frac{x}{r} \mathbf{i}+\frac{y}{r} \mathbf{j}+\frac{z}{r} \mathbf{k}\right]=-3 r^{-5} \mathbf{r}\)

∴ \(r \nabla\left(\frac{1}{r^3}\right)=-3 r^{-4}(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})\)

∴ \(\nabla \cdot\left[r \nabla\left(\frac{1}{r^3}\right)\right]=\frac{\partial}{\partial x}\left(-3 r^{-4} x\right)+\frac{\partial}{\partial y}\left(-3 r^{-4} y\right)+\frac{\partial}{\partial z}\left(-3 r^{-4} z\right)\)

⇒ \(12 r^{-5} x \frac{\partial r}{\partial x}-3 r^{-4}+12 r^{-5} \frac{\partial r}{\partial y} y-3 r^{-4}+12 r^{-5} \frac{\partial r}{\partial y} z-3 r^{-4}\)

⇒ \(-9 r^{-4}+12 r^{-5}\left(\frac{x^2}{r}+\frac{y^2}{r}+\frac{z^2}{r}\right)=12 r^{-4}-9 r^{-4}=3 r^{-4}\)

121. If F=grad(x³+y³+z³-3xyz) then find div F, curl F.

Solution:

F = grad \(\left(x^3+y^3+z^3-3 x y z\right)=\mathbf{i}\left(3 x^2-3 y z\right)+\mathbf{j}\left(3 y^2-3 x z\right)+\mathbf{k}\left(3 z^2-3 x y\right)\)

div F = \(\nabla \cdot \mathbf{F}=\frac{\partial}{\partial x}\left(3 x^2-3 y z\right)+\frac{\partial}{\partial y}\left(3 y^2-3 x z\right)+\frac{\partial}{\partial z}\left(3 z^2-3 x y\right)\)

⇒ 6x+6y+6z = 6 (x+y+z)

⇒ \({curl} \mathbf{F}=\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
3 x^2-3 y z & 3 y^2-3 x z & 3 z^2-3 x y
\end{array}\right|\)

⇒ I (-3x+3x) -j (-3y + 3y )+k (-3z + 3z) = 0

122. If φ =2x³y²z⁴,find div(grad)

Solution:

⇒ \(\phi=2 x^3 y^2 z^4 \Rightarrow \frac{\partial \phi}{\partial x}=6 x^2 y^2 z^4, \frac{\partial \phi}{\partial y}=4 x^3 y z^4, \frac{\partial \phi}{\partial z}=8 x^3 y^2 z^3\)

⇒ \({grad} \phi=\nabla \phi=\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}=6 x^2 y^2 z^4
{i}+4 x^3 y z^4 \mathbf{j}+8 x^3 y^2 z^3 \mathbf{k}\)

⇒ \({div}{grad} \phi=\nabla \cdot(\nabla \phi)=\frac{\partial}{\partial x}\left(6 x^2 y^2 z^4\right)+\frac{\partial}{\partial y}\left(4 x^3 y z^4\right)+\frac{\partial}{\partial z}\left(8 x^3 y^2 z^3\right)\)

⇒ \(12 x y^2 z^4+4 x^3 z^4+24 x^3 y^2 z^2\)

123. If φ= 2x³y²z⁴ ,then show that

Solution:

⇒ \(\phi=2 x^3 y^2 z^4 \Rightarrow \frac{\partial \phi}{\partial x}=6 x^2 y^2 z^4, \frac{\partial \phi}{\partial y}=4 x^3 y z^4, \frac{\partial \phi}{\partial z}=8 x^3 y^2 z^3\)

⇒ \(\frac{\partial^2 \phi}{\partial x^2}=12 x y^2 z^4, \frac{\partial^2 \phi}{\partial y^2}=4 x^3 z^4, \frac{\partial^2 \phi}{\partial z^2}=24 x^3 y^2 z^2\)

⇒ \(\nabla \phi=\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}=6 x^2 y^2 z^4 \mathbf{i}+4 x^3 y z^4 \mathbf{j}+8 x^3 y^2 z^3 \mathbf{k}\)

⇒ \(\nabla \cdot(\nabla \phi)=\frac{\partial}{\partial x}\left(6 x^2 y^2 z^4\right)+\frac{\partial}{\partial y}\left(4 x^3 y z^4\right)+\frac{\partial}{\partial z}\left(8 x^3 y^2 z^3\right)=12 x y^2 z^4+4 x^3 z^4+24 x^3 y^2 z^2\)

⇒ \(\nabla^2 \phi=\frac{\partial^2 \phi}{\partial x^2}\left(6 x^2 y^2 z^4\right)+\frac{\partial^2 \phi}{\partial y^2}\left(4 x^3 y z^4\right)+\frac{\partial^2 \phi}{\partial z^2}\left(8 x^3 y^2 z^3\right)=12 x y^2 z^4+4 x^3 z^4+24 x^3 y^2 z^2\)

∴ \(\nabla \cdot(\nabla \phi)=\nabla^2 \phi\)

124.If u= x²+y²+z²,find curl ∇u

Solution:

⇒ \(\nabla u=\mathbf{i} \frac{\partial u}{\partial x}+\mathrm{j} \frac{\partial u}{\partial y}+\mathbf{k} \frac{\partial u}{\partial z}=2 x \mathbf{i}+2 y \mathrm{j}+2 z \mathbf{k}\)

⇒ \({curl} \nabla u=\left|\begin{array}{lll}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x & 2 y & 2 z
\end{array}\right|=\mathbf{i}(0-0)-\mathbf{j}(0-0)+\mathbf{k}(0-0)=0\)

125. If φ =x²yz then find curl (grad φ).

Solution:

⇒ \({grad} \varphi=\nabla \varphi=\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial v}+\mathbf{k} \frac{\partial \varphi}{\partial z}=2 x y \mathbf{i}+x^2 z \mathbf{j}+x^2 y \mathbf{k}\)

⇒ \(\text { curl grad } \varphi=\nabla \times(\nabla \varphi)=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x y z & x^2 z & x^2 y
\end{array}\right|=\mathbf{i}\left(x^2-x^2\right)-\mathbf{j}(2 x y-2 x y)+\mathbf{k}(2 x z-2 x z)\)

= 0

126. If A=2xz² i-yzj+3xz³k then find curl (curlA) at (1,1,1).

Solution:

⇒ \({curl} \mathbf{A}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x z^2 & -y z & 3 x z^3
\end{array}\right|=\mathbf{i}(0+y)-\mathbf{j}\left(3 z^3-4 x z\right)+\mathbf{k}(0-0)=y \mathbf{i}+\left(4 x z-3 z^3\right) \mathbf{j}\)

curl (curl A) = \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y & 4 x z-3 z^3 & 0
\end{array}\right|=\mathbf{i}\left(0-4 x+9 z^2\right)-\mathbf{j}(0-0)+\mathbf{k}(4 z-1)\)

⇒ \(\left(9 z^2-4 x\right) \mathbf{i}+(4 z-1) \mathbf{k}=5 \mathbf{i}+3 \mathbf{k} \text { at }(1,1,1)\)

127. If f= x²yi-2xzj+2yzk a the point (1,-1,1) then find div f, curl (curl f).

Solution:

Given f = \(x^2 y \mathbf{i}-2 x z \mathbf{j}+2 y z \mathbf{k}\)

div f = \(\frac{\partial}{\partial x}\left\{x^2 y\right\}+\frac{\partial}{\partial y}\{-2 x z\}+\frac{\partial}{\partial z}\{2 y z\}=2 x y+2 y\)

= 2(1)(-1)+2(-1) = -4 at (1,-1,1)

curl f = \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x^2 y & -2 x z & 2 y z
\end{array}\right|=\mathbf{i}(2 z+2 x)-\mathbf{j}(0-0)+\mathbf{k}\left(-2 z-x^2\right)=(2 x+2 z) \mathbf{i}-\left(x^2+2 z\right) \mathbf{k}\)

curl (curl f) = \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x+2 z & 0 & -x^2-2 z
\end{array}\right|=\mathbf{i}(0-0)-\mathbf{J}(-2 x-2)+\mathbf{k}(0-0)=(2 x+2) \mathbf{j}\)

= 4j at (1,-1,1)

128.If f=x²yz, g=xy-3z²,find div (grad f× grad g).

Solution:

grad f = \(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}=\mathbf{i} 2 x y z+\mathbf{j} x^2 z+\mathbf{k} x^2 y\)

grad g = \(\mathbf{i} \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial g}{\partial z}=\mathbf{i} y+\mathbf{j} x+\mathbf{k}(-6 z)\)

grad f × grad g = \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 x y z & x^2 z & x^2 y \\
y & x & -6 z
\end{array}\right|\)

⇒ \(\mathbf{i}\left(-6 x^2 z^2-x^3 y\right)-\mathbf{j}\left(-12 x y z^2-x^2 y^2\right)+\mathbf{k}\left(2 x^2 y z-x^2 y z\right)\)

⇒ \(\left(-6 x^2 z^2-x^3 y\right) \mathbf{i}+\left(12 x y z^2+x^2 y z\right) \mathbf{j}+\left(x^2 y z\right) \mathbf{k}\)

div (grad f × grad g) = \(\frac{\partial}{\partial x}\left(-6 x^2 z^2-x^3 y\right)+\frac{\partial}{\partial y}\left(12 x y z^2+x^2 y^2\right)+\frac{\partial}{\partial z}\left(x^2 y z\right)\)

⇒ \(-12 x z^2-3 x^2 y+12 x z^2+2 x^2 y+x^2 y=0\)

129.If φ  =x²yz and A=2xz²i-yz-j+3xz³ K then find ∇×( φA)

Solution:

⇒ \(\nabla \times(\varphi \mathbf{A})={curl}\left(2 x^3 y z^2 \mathbf{i}-x^2 y^2 z^2 \mathbf{j}+3 x^3 y z^4 \mathbf{k}\right)=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x^3 y z^2 & -x^2 y^2 z^2 & 3 x^3 y z^4
\end{array}\right|\)

⇒ \(\mathbf{i}\left(3 x^3 z^4+2 x^2 y^2 z\right)-\mathbf{j}\left(9 x^2 y z^4-4 x^3 y z\right)+\mathbf{k}\left(-2 x y^2 z^2-2 x^3 z^2\right)=5 \mathbf{i}-5 \mathbf{j}-4 \mathbf{k} \text { at }(1,1,1)\)

130. If F = 3xyzi+4xyj-xyzk  the find  ∇ ( ∇.F) and ∇× ( ∇×F) at (-1,2,1) .Also verify that  ∇ ( ∇.F)=∇× ( ∇×F)+ ∇²F.

Solution:

⇒ \(\nabla \cdot \mathbf{F}=\frac{\partial}{\partial x}\left\{3 x y z^3\right\}+\frac{\partial}{\partial y}\left\{4 x^3 y\right\}-\frac{\partial}{\partial z}\left\{x y^2 z\right\}=3 y z^3+4 x^3-x y^2\)

⇒ \(\nabla(\nabla \cdot \mathbf{F})=\mathbf{i}\left(12 x^2-y^2\right)+\mathbf{j}\left(3 z^3-2 x y\right)+\mathbf{k}\left(9 y z^2\right)\)

⇒ \(\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
3 x y z^3 & 4 x^3 y & -x y^2 z
\end{array}\right|=\mathbf{i}(-2 x y z-0)-\mathbf{j}\left(-y^2 z-9 x y z^2\right)+\mathbf{k}\left(12 x^2 y-3 x z^3\right)\)

⇒ \(\nabla \times(\nabla \times \mathbf{F})=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
-2 x y z & 9 x y z^2+y^2 z & 12 x^2 y-3 x z^3
\end{array}\right|\)

⇒ \(\mathbf{i}\left(12 x^2-18 x y z-y^2\right)-\mathbf{j}\left(24 x y-3 z^3+2 x y\right)+\mathbf{k}\left(9 y z^2+2 x z\right)\)

⇒ \(\mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=-\frac{3 x}{r^5} \mathbf{i} \times(\mathbf{a} \times \mathbf{r})+\frac{i}{r^3} \mathbf{i} \times(\mathbf{a} \times \mathbf{i})\)

⇒ \(-\frac{3 x}{r^5}[(\mathbf{i} \cdot \mathbf{r}) \mathbf{a}-(\mathbf{i} \cdot \mathbf{a}) \mathbf{r}]+\frac{1}{r^3}[(\mathbf{i} \cdot \mathbf{i}) \mathbf{a}-(\mathbf{i} \cdot \mathbf{a}) \mathbf{i}]\)

⇒ \(-\frac{3 x}{r^5} x a+\frac{3 x}{r^5} a_1 \mathbf{r}+\frac{1}{r^3} \mathbf{a}-\frac{1}{r^3} a_1 \mathbf{i}\left[\text { where } \mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}\right]\)

⇒ \(-\frac{3 x^2}{r^5} a+\frac{3}{r^5} a_1 x \mathbf{r}+\frac{1}{r^3} a-\frac{1}{r^3} a_1 \mathbf{i}\)

⇒ \(\Sigma\left\{\mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)\right\}=\left\{-\frac{3}{r^5} \Sigma x^2\right\} \mathbf{a}+\left\{\frac{3}{r^5} \Sigma a_1 x\right\} \mathbf{r}+\frac{3}{r^3} \mathbf{a}-\frac{1}{r^3} \Sigma a_1 \mathbf{i}\)

⇒ \(-\frac{3}{r^5} r^2 a+\frac{3}{r^5}(r \cdot a) r+\frac{3}{r^3} a-\frac{1}{r^3} a=-\frac{a}{r^3}+\frac{3}{r^5}(a \cdot r) r\)

131. If a is a constant vector  then show that curl\(\frac{a \times r}{r^3}\) =\(\frac{-a}{r^3}+3 \frac{r}{r^5}\)(a.r).

Solution:

curl \(\frac{\mathbf{a} \times \mathbf{r}}{r^3}=\nabla \times\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=\Sigma\left\{\mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)\right\}\)

Now \(\frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=-\frac{3}{r^4} \frac{\partial r}{\partial x}(\mathbf{a} \times \mathbf{r})+\frac{1}{r^3}\left(\mathbf{a} \times \frac{\partial \mathbf{r}}{\partial x}\right)+\frac{1}{r^3}\left(\frac{\partial \mathbf{a}}{\partial x} \times \mathbf{r}\right)\)

⇒ \(\frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=-\frac{3}{r^4} \frac{x}{r}(\mathbf{a} \times \mathbf{r})+\frac{1}{r^3}(\mathbf{a} \times \mathbf{i})=-\frac{3 x}{r^5}(\mathbf{a} \times \mathbf{r})+\frac{1}{r^3}(\mathbf{a} \times \mathbf{i})\)

⇒ \(\mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=-\frac{3 x}{r^5} \mathbf{i} \times(\mathbf{a} \times \mathbf{r})+\frac{1}{r^3} \mathbf{i} \times(\mathbf{a} \times \mathbf{i})\)

⇒ \(\mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=-\frac{3 x}{r^5} \mathbf{i} \times(\mathbf{a} \times \mathbf{r})+\frac{1}{r^3} \mathbf{i} \times(\mathbf{a} \times \mathbf{i})\)

⇒ \(-\frac{3 x}{r^5}[(\mathbf{i} \cdot \mathbf{r}) \mathbf{a}-(\mathbf{i} \cdot \mathbf{a}) \mathbf{r}]+\frac{1}{r^3}[(\mathbf{i} \cdot \mathbf{i}) \mathbf{a}-(\mathbf{i} \cdot \mathbf{a}) \mathbf{i}]\)

⇒ \(-\frac{3 x}{r^5} x \mathbf{a}+\frac{3 x}{r^5} a_1 \mathbf{r}+\frac{1}{r^3} \mathbf{a}-\frac{1}{r^3} a_1 \mathbf{i}\left[\text { where } \mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}\right]\)

⇒ \(-\frac{3 x^2}{r^5} a+\frac{3}{r^5} a_1 x \mathbf{r}+\frac{1}{r^3} a-\frac{1}{r^3} a_1 \mathbf{i}\)

⇒ \(\Sigma\left\{\mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)\right\}=\left\{-\frac{3}{r^5} \Sigma x^2\right\} \mathbf{a}+\left\{\frac{3}{r^5} \Sigma a_1 x\right\} \mathbf{r}+\frac{3}{r^3} \mathbf{a}-\frac{1}{r^3} \Sigma a_1 \mathbf{i}\)

⇒ \(-\frac{3}{r^5} r^2 a+\frac{3}{r^5}(r \cdot a) r+\frac{3}{r^3} a-\frac{1}{r^3} a=-\frac{a}{r^3}+\frac{3}{r^5}(a \cdot r)\)

132. If F is a vector point function, show that(F×∇).r=0.

Solution:

(F×∇).r \(=\left\{\boldsymbol{F} \times \Sigma \mathbf{i} \frac{\partial}{\partial x}\right\}\)

\(=\left\{\Sigma(\mathbf{F} \times \mathbf{i}) \cdot \frac{\partial \mathbf{r}}{\partial x}\right\}\)

=∑(F×i).i=0.

133. If f is a vector point function, prove that(f×∇)×r=-2f.

Solution:

⇒ \((\mathbf{f} \times \nabla) \times \mathbf{r}=(\mathbf{f} \times \mathbf{i}) \times \frac{\partial \mathbf{r}}{\partial x}+(\mathbf{f} \times \mathbf{j}) \times \frac{\partial \mathbf{r}}{\partial y}+(\mathbf{f} \times \mathbf{j}) \times \frac{\partial \mathbf{r}}{\partial z}\)

⇒ (f × i) × i + (f × j × j + (f × k) × k

= (i.f) i – (i.i)f + (j.f) j – (j.j) f + (k.f) k – (k.k)f

= (i.f)  i – (i.i) f + (j.f) j – (j.j)f + (k.f)k- (k.k) f

= (i.f)i+(j.f) j + (k.f) k – 3f = f – 3f = -2f

134. If F is a vector point function and a is a constant vector then show that

(1) ∇(a.F)=(a.∇)F+a×(curl f)

(2) ∇.(a×F)=-a.curlF

(3) ∇×(a×F) = a(div F)-(a.∇)F.

Solution:

⇒ \(\mathbf{a} \times({curl} \mathbf{F})=\mathbf{a} \times(\nabla \times \mathbf{F})=\mathbf{a} \times\left[\mathbf{i} \times \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{F}}{\partial z}\right]\)

⇒ \(=\mathbf{a} \times\left(\mathbf{i} \times \frac{\partial \mathbf{F}}{\partial x}\right)+\mathbf{a} \times\left(\mathbf{j} \times \frac{\partial \mathbf{F}}{\partial y}\right)+\mathbf{a} \times\left(\mathbf{k} \times \frac{\partial \mathbf{F}}{\partial z}\right)\)

⇒ \(\left(\mathbf{a} \cdot \frac{\partial \mathbf{F}}{\partial x}\right) \mathbf{i}-(\mathbf{a} \cdot \mathbf{i}) \frac{\partial \mathbf{F}}{\partial x}+\left(\mathbf{a} \cdot \frac{\partial \mathbf{F}}{\partial y}\right) \mathbf{j}-(\mathbf{a} \cdot \mathbf{j}) \frac{\partial \mathbf{F}}{\partial y}+\left(\mathbf{a} \cdot \frac{\partial \mathbf{F}}{\partial z}\right) \mathbf{k}-(\mathbf{a} \cdot \mathbf{k}) \frac{\partial \mathbf{F}}{\partial z}\)

⇒ \(i\left(\mathbf{a} \cdot \frac{\partial \mathbf{F}}{\partial x}\right)+\mathbf{j}\left(\mathbf{a} \cdot \frac{\partial \mathbf{F}}{\partial y}\right)+\mathbf{k}\left(\mathbf{a} \cdot \frac{\partial \mathbf{F}}{\partial z}\right)-\left[(\mathbf{a} \cdot \mathbf{i}) \frac{\partial \mathbf{F}}{\partial x}+(\mathbf{a} \cdot \mathbf{j}) \frac{\partial \mathbf{F}}{\partial y}+(\mathbf{a} \cdot \mathbf{k}) \frac{\partial \mathbf{F}}{\partial z}\right]\)

⇒ \(\mathbf{i} \frac{\partial}{\partial x}(\mathbf{a} \cdot \mathbf{F})+\mathbf{j} \frac{\partial}{\partial y}(\mathbf{a} \cdot \mathbf{F})+\mathbf{k} \frac{\partial}{\partial z}(\mathbf{a} \cdot \mathbf{F})-(\mathbf{a} \cdot \nabla) \mathbf{F}=\nabla(\mathbf{a} \cdot \mathbf{F})-(\mathbf{a} \cdot \nabla) \mathbf{F}\)

⇒ \(\nabla \cdot(\mathbf{a} \times \mathbf{F})=\mathbf{F} \cdot(\nabla \times \mathbf{a})-\mathbf{a} \cdot(\nabla \times \mathbf{F})=-\mathbf{a} \cdot({curl} \mathbf{F})\)

⇒ \(\nabla \times(\mathbf{a} \times \mathbf{F})=\mathbf{i} \times \frac{\partial}{\partial x}(\mathbf{a} \times \mathbf{F})+\mathbf{j} \times \frac{\partial}{\partial y}(\mathbf{a} \times \mathbf{F})+\mathbf{k} \times \frac{\partial}{\partial z}(\mathbf{a} \times \mathbf{F})\)

⇒ \(\mathbf{i} \times\left(\mathbf{a} \times \frac{\partial \mathbf{F}}{\partial x}\right)+\mathbf{j} \times\left(\mathbf{a} \times \frac{\partial \mathbf{F}}{\partial y}\right)+\mathbf{k} \times\left(\mathbf{a} \times \frac{\partial \mathbf{F}}{\partial z}\right)\)

⇒ \(\left.\left(\mathbf{i} \cdot \frac{\partial \mathbf{F}}{\partial x}\right) \mathbf{a}-(\mathbf{i} \cdot \mathbf{a}) \frac{\partial \mathbf{F}}{\partial x}+\left(\mathbf{j} \cdot \frac{\partial \mathbf{F}}{\partial y}\right) \mathbf{a}-\mathbf{j} \cdot \mathbf{a}\right) \frac{\partial \mathbf{F}}{\partial y}+\left(\mathbf{k} \cdot \frac{\partial \mathbf{F}}{\partial z}\right) \mathbf{a}-(\mathbf{k} \cdot \mathbf{a}) \frac{\partial \mathbf{F}}{\partial z}\)

⇒ \(\mathbf{a}\left[\mathbf{i} \cdot \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{F}}{\partial z}\right]-\left[(\mathbf{a} \cdot \mathbf{i}) \frac{\partial \mathbf{F}}{\partial x}+(\mathbf{a} \cdot \mathbf{j}) \frac{\partial \mathbf{F}}{\partial y}+(\mathbf{a} \cdot \mathbf{k}) \frac{\partial \mathbf{F}}{\partial z}\right]\)

⇒ \(\mathbf{a}(\nabla \cdot \mathbf{F})-(\mathbf{a} \cdot \nabla) \mathbf{F}=\mathbf{a}({div} \mathbf{F})-(\mathbf{a} \cdot \nabla) \mathbf{F}\)

135.Show that curl rⁿ (a×r)=(n+2)rⁿ a-nr^n-2 (a.r)r.

Solution: 

⇒ \({curl} r^n(\mathbf{a} \times \mathbf{r})=\nabla \times r^n(\mathbf{a} \times \mathbf{r})=\nabla r^n \times(\mathbf{a} \times \mathbf{r})+r^n \nabla \times(\mathbf{a} \times \mathbf{r})\)

⇒ \(n r^{n-2} \mathbf{r} \times(\mathbf{a} \times \mathbf{r})+r^n 2 \mathbf{a}=n r^{n-2}[(\mathbf{r} \cdot \mathbf{r}) \mathbf{a}-(\mathbf{r} \cdot \mathbf{a}) \mathbf{r}]+2 r^n \mathbf{a}\)

⇒ \(n r^n a+2 r^n a-n r^{n-2} r(\mathbf{r} \cdot \mathbf{a})=(n+2) r^n a-n r^{n-2} r(\mathbf{r} \cdot \mathbf{a})\)

136. If f and g are two scalar point functions then prove that div(f∇g)=f∇² g+∇ f.∇ g

Solution:

⇒ \({div}(f \nabla g)={i} \cdot(f \nabla g)={i} \cdot \frac{\partial}{\partial x}(f \nabla g)+{j}\frac{\partial}{\partial y}(f \nabla g)+\mathbf{k} \cdot \frac{\partial}{\partial z}(f \nabla g)\)

⇒ \(\mathbf{i} \cdot\left(\frac{\partial f}{\partial x} \nabla g+f \frac{\partial}{\partial x} \nabla g\right)+\mathbf{j} \cdot\left(\frac{\partial f}{\partial y} \nabla g+f \frac{\partial}{\partial y} \nabla g\right)+\mathbf{k} \cdot\left(\frac{\partial f}{\partial z} \nabla g+f \frac{\partial}{\partial z} \nabla g\right)\)

⇒ \(\mathbf{i} \frac{\partial f}{\partial x} \cdot \nabla g+f\left(\mathbf{i} \cdot \frac{\partial}{\partial x} \nabla g\right)+\mathbf{j} \frac{\partial f}{\partial y} \cdot \nabla g+f\left(\mathbf{j} \cdot \frac{\partial}{\partial y} \nabla g\right)+\mathbf{k} \frac{\partial f}{\partial z} \cdot \nabla g+f\left(\mathbf{k} \cdot \frac{\partial}{\partial z} \nabla g\right)\)

⇒ \(\left(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\right) \cdot \nabla g+f\left(\mathbf{i} \cdot \frac{\partial}{\partial x} \nabla \dot{g}+\mathbf{j} \cdot \frac{\partial}{\partial y} \nabla g+\mathbf{k} \cdot \frac{\partial}{\partial z} \nabla g\right)\)

⇒ \(\nabla f \cdot \nabla g+f \nabla \cdot \nabla g=\nabla f \cdot \nabla g+f \nabla^2 g\)

137. Show that div (f∇g)-div(g∇f)=f∇²g-g∇²f.

Solution:

⇒ \(f \nabla g=\mathbf{i} f \frac{\partial g}{\partial x}+\mathbf{j} f \frac{\partial g}{\partial y}+\mathbf{k} f \frac{\partial g}{\partial z}\)

⇒ \({div}(f \nabla g)=\nabla \cdot(f \nabla g)=\frac{\partial}{\partial x}\left(f \frac{\partial g}{\partial x}\right)+\frac{\partial}{\partial y}\left(f \frac{\partial g}{\partial y}\right)+\frac{\partial}{\partial z}\left(f \frac{\partial g}{\partial z}\right)\)

⇒ \(f\left(\frac{\partial^2 g}{\partial x^2}+\frac{\partial^2 g}{\partial y^2}+\frac{\partial^2 g}{\partial z^2}\right)+\frac{\partial f}{\partial x} \frac{\partial g}{\partial x}+\frac{\partial f}{\partial y} \frac{\partial g}{\partial y}+\frac{\partial f}{\partial z} \frac{\partial g}{\partial z}\)

⇒ \(f \nabla^2 g+\left(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\right) \cdot\left(\mathbf{i} \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial g}{\partial z}\right)=f \nabla^2 g+\nabla f \cdot \nabla g\)

Similarly div \((g \nabla f)=g \nabla^2 f+\nabla g \cdot \nabla f\)

∴ \({div}(f \nabla g)-{div}(g \nabla f)=f \nabla^2 g-g \nabla^2 f\)

Vector Differentiation- 3 Exercise 3 Solved Problems

1. Define a vector function.

Vector function: Let S⊆ R . If to each t ∈ S there corresponds a unique vector f(t) then the corresponding f is called a vector function with scalar variable t of domain S.

2. Define the limit of a vector function.

Limit: Let f be a vector function over the domain S and an as be a limit point of S. A vector L is said to be the limit of at an if to each given ε>0,∃δ>0∋t ∈ S, 0<|t-a| δ ⇒|f(t)-L|<ε It is denoted by\(\underset{t \rightarrow a}{L t}\)f(t) = L.

3. Define the continuity of a vector function.

1. Continuous at a point: let f be a vector function with domain S and a ∈ S. Then f is said to be continuous at if\(\underset{t \rightarrow a}{L t}\)f(t)=f(a)
2. Continuity on a set: let f be a vector function with domain S. Then f is said to be continuous on S if f is continuous at a for all a ∈ S.

4. Define the derivative of a vector function.

Derivative: Let f be a vector function with domain S and a ∈ Then f is said to be differentiable at an if \(\underset{t \rightarrow a}{L t}\)\(\frac{f(t)-f(a)}{t-a}\)exits. At the limit is called the derivative of f at a. It is denoted by f'(a) or \(\left(\frac{d f}{d t}\right)_{t=a}\)

5. If f, g are two differentiable vector functions at t  then prove that f + g is differentiable at t and\(\frac{d}{d t}\{\mathbf{f}+\mathbf{g}\}\)=\(=\frac{d f}{d t}+\frac{d g}{d t}\)

Solution:

⇒ \(\underset{h \rightarrow 0}{L t} \frac{(\mathbf{f}+\mathbf{g})(t+h)-(\mathbf{f}+\mathbf{g})(t)}{h}=\underset{h \rightarrow 0}{L t} \frac{\mathbf{f}(t+h)+\mathbf{g}(t+h)-\mathbf{f}(t)-\mathbf{g}(t)}{h}\)

= \(\underset{h \rightarrow 0}{L t} \frac{\mathbf{f}(t+h)-\mathbf{f}(t)}{h}+\underset{h \rightarrow 0}{L t} \frac{\mathbf{g}(t+h)-\mathbf{g}(t)}{h}=\frac{d \mathbf{f}}{d t}+\frac{d \mathbf{g}}{d t}\)

∴ \(\mathbf{f}+\mathbf{g}\)are differentiable at t and \(\frac{d}{d t}\{\mathbf{f}+\mathbf{g}\}=\frac{d \mathbf{f}}{d t}+\frac{d \mathbf{g}}{d t}\)

6. If f and g are two differentiable vector functions at t then prove that

1. f • g is differentiable at t and \(\frac{d}{d t}\{\mathbf{f} \cdot \mathbf{g}\}\)=\(\frac{d \mathbf{f}}{d t} \cdot \mathbf{g}+\mathbf{f} \cdot \frac{d \mathbf{g}}{d t}\)
2. f x g is differentiable at t and \(\frac{d}{d t}\{\mathbf{f} \times \mathbf{g}\}\)=\(\frac{d \mathbf{f}}{d t} \times \mathbf{g}+\mathbf{f} \times \frac{d \mathbf{g}}{d t}\)

Solution:

1. \(\underset{h \rightarrow 0}{L t} \frac{(\mathbf{f} \cdot \mathbf{g})(t+h)-(\mathbf{f} \cdot \mathbf{g})(t)}{h}=\underset{h \rightarrow 0}{L t} \frac{\mathbf{1}(t+h) \cdot \mathbf{g}(t+h)-\mathbf{f}(t) \cdot \mathbf{g}(t)}{h}\)

= \(\text{Lt}_{h \rightarrow 0} \frac{\mathbf{f}(t \cdot+h) \cdot \mathbf{g}(t+h)-\mathbf{f}(t+h) \cdot \mathbf{g}(t)+\mathbf{f}(t+h) \cdot \mathbf{g}(t)-\mathbf{f}(t) \cdot \mathbf{g}(t)}{h}\)

= \(\underset{h \rightarrow 0}{L t}\left\{\mathbf{f}(t+h) \cdot \frac{[\mathbf{g}(t+h)-\mathbf{g}(t)]}{h}+\frac{[\mathbf{f}(t+h)-\mathbf{f}(t)]}{h} \cdot \mathbf{g}(t)\right\}\)

= \(\left[{ }_{h \rightarrow 0}^{L t} \mathbf{f}(t+h)\right] \cdot\left[{ }_{h \rightarrow 0}^{L t} \frac{\mathbf{g}(t+h)-\mathbf{g}(t)}{h}\right]+\left[{ }_{h \rightarrow 0}^{L t} \frac{\mathbf{f}(t+h)-\mathbf{f}(t)}{h}\right] \cdot \mathbf{g}(t)\)

= \(\mathbf{f} \cdot \frac{d \mathbf{g}}{d t}+\frac{d \mathbf{f}}{d t} \cdot \mathbf{g}\)

∴ \(\mathbf{f} \cdot \mathbf{g}\) is differentiable at t and \(\frac{d}{d t}\{\mathbf{f} \cdot \mathbf{g}\}=\frac{d \mathbf{f}}{d t} \cdot \mathbf{g}+\mathbf{f} \cdot \frac{d \mathbf{g}}{d t}\)

2. \(\underset{h \rightarrow 0}{L t} \frac{(\mathbf{f} \times \mathbf{g})(t+h)-(\mathbf{f} \times \mathbf{g})(t)}{h}=\underset{h \rightarrow 0}{L t} \frac{\mathbf{f}(t+h) \times \mathbf{g}(t+h)-\mathbf{f}(t) \times \mathbf{g}(t)}{h}\)

= \(\text{Lt}_{h \rightarrow 0} \frac{\mathbf{f}(t+h) \times \mathbf{g}(t+h)-\mathbf{f}(t+h) \times \mathbf{g}(t)+\mathbf{f}(t+h) \times \mathbf{g}(t)-\mathbf{f}(t) \times \mathbf{g}(t)}{h .}\)

= \(\text{Lt}_{h \rightarrow 0}\left\{\mathbf{f}(t+h) \times \frac{[\mathbf{g}(t+h)-\mathbf{g}(t)]}{h}+\frac{[\mathbf{f}(t+h)-\mathbf{f}(t)]}{h} \times \mathbf{g}(t)\right\}\)

= \([\underset{h \rightarrow 0}{L t} \mathbf{f}(t+h)] \times\left[\underset{h \rightarrow 0}{L t} \frac{\mathbf{g}(t+h)-\mathbf{g}(t)}{h}\right]+\left[\underset{h \rightarrow 0}{L t} \frac{\mathbf{f}(t+h)-\mathbf{f}(t)}{h}\right] \times \mathbf{g}(t)\)

= \(\mathbf{f} \times \frac{d \mathbf{g}}{d t}+\frac{d \mathbf{f}}{d t} \times \mathbf{g}\) .

∴ \(\mathbf{f} \times \mathbf{g}\) is differentiable at t and \(\frac{d}{d t}\{\mathbf{f} \times \mathbf{g}\}=\frac{d \mathbf{f}}{d t} \times \mathbf{g}+\mathbf{f} \times \frac{d \mathbf{g}}{d t}\)

7. If f, g, h are three differentiable vector functions at t then prove that

1. f×(g×h) is differentiable at t and \(\frac{d}{d t}\{\mathbf{f} \times(\mathbf{g} \times \mathbf{h})\}\)=\(\frac{d \mathbf{f}}{d t} \times(\mathbf{g} \times \mathbf{h})+\mathbf{f} \times\left(\frac{d \boldsymbol{g}}{d t} \times \mathbf{h}\right)+\mathbf{f} \times\left(\mathbf{g} \times \frac{d \mathbf{h}}{d t}\right)\)

2. (f×g)×h is differentiable at t and \(\frac{d}{d t}\{(\mathbf{f} \times \mathbf{g}) \times \mathbf{h}\}\)=\(\left(\frac{d \mathbf{f}}{d t} \times \mathbf{g}\right) \times \mathbf{h}+\left(\mathbf{f} \times \frac{d \boldsymbol{g}}{d t}\right) \times \mathbf{h}+(\mathbf{f} \times \mathbf{g}) \times \frac{d \mathbf{h}}{d t}\)

Solution:

1. g,h is differentiable at  t ⇒ g × h is differentiable at t

f,g × h are differentiable at t ⇒  f×(g × h) is differentiable at t

⇒ \(\frac{d}{d t}\{\mathbf{f} \times(\mathbf{g} \times \mathbf{h})\}\)

= \(\frac{d \mathbf{f}}{d t} \times(\mathbf{g} \times \mathbf{h}), \mathbf{f} \times \frac{d}{d t}(\mathbf{g} \times \mathbf{h})\)

= \(\frac{d \mathbf{f}}{d t} \times(\mathbf{g} \times \mathbf{h})+\mathbf{f} \times\left(\frac{d \mathbf{g}}{d t} \times \mathbf{h}+\mathbf{g} \times \frac{d \mathbf{h}}{d t}\right)\)

= \(\frac{d \mathbf{f}}{d t} \times(\mathbf{g} \times \mathbf{h})+\mathbf{f} \times\left(\frac{d \mathbf{g}}{d t} \times \mathbf{h}\right)+\mathbf{f} \times\left(\mathbf{g} \times \frac{d \mathbf{h}}{d t}\right)\)

2. f, g is differentiable at t ⇒ (f×g)  is differentiable at t

f×g,h are differentiable at t ⇒ (f×g) ×h is differentiable at t

⇒ \(\frac{d}{d t}\{(\mathbf{f} \times \mathbf{g}) \times \mathbf{h}\}=\left(\frac{d \mathbf{f}}{d t} \times \mathbf{g}+\mathbf{f} \times \frac{d \mathbf{g}}{d t}\right) \times \mathbf{h}+(\mathbf{f} \times \mathbf{g}) \times \frac{d \mathbf{h}}{d t}\)

= \(\left(\frac{d \mathbf{f}}{d t} \times \mathbf{g}\right) \times \mathbf{h}+\left(\mathbf{f} \times \frac{d \mathbf{g}}{d t}\right) \times \mathbf{h}+(\mathbf{f} \times \mathbf{g}) \times \frac{d \mathbf{h}}{d t}\)

8. If f  is a differentiable vector function at t and φ is a differentiable scalar function at t then prove that φ f  is differentiable at t and\(\frac{d}{d t}\{\varphi \boldsymbol{f}\}\)=\(\varphi \frac{d f}{d t}+\frac{d \varphi}{d t} \mathbf{f}\)

Solution:

⇒ \(\text{Lt}_{h \rightarrow 0} \frac{(\varphi \mathbf{f})(t+h)-(\varphi t)(t)}{h}=\text{Lt}_{h \rightarrow 0} \frac{\varphi(t+h) \mathbf{f}(t+h)-\varphi(t) \mathbf{f}(t)}{h}\)

= \(\text{Lt}_{h \rightarrow 0} \frac{\varphi(t+h) \mathbf{f}(t+h)-\varphi(t+h) \mathbf{f}(t)+\varphi(t+h) \mathbf{f}(t)-\varphi(t) \mathbf{f}(t)}{h}\)

= \(\text{Lt}_{h \rightarrow 0}\left\{\varphi(t+h) \frac{[\mathbf{f}(t+h)-\mathbf{f}(t)]}{h}+\frac{[\varphi(t+h)-\varphi(t)]}{h}(t)\right\}\)

= \(\underset{h \rightarrow 0}{L t} \varphi(t+h) \underset{h \rightarrow 0}{L t} \frac{f(t+h)-\mathbf{f}(t)}{h}+\underset{h \rightarrow 0}{L t} \frac{\varphi(t+h)-\varphi(t)}{h} \mathbf{f}(t)=\varphi(t) \frac{d \mathbf{t}}{d t}+\frac{d \varphi}{d t} \mathbf{f}\)

∴ φ  is differentiable at t and \(\frac{d}{d t}\){φf}= φ \(\frac{d t}{dt}+\frac{d \varphi}{d t}\) f

9. Prove that a vector function f is constant if f \(\frac{d f}{d t}\)=0

Solution:

Suppose f is constant  then \(\frac{d \mathbf{f}}{d t}\)=0

Conversely suppose that \(\frac{d \mathbf{f}}{d t}\)=0 .Let f= f₁i+f₂j+f₃k

⇒ \(\frac{d f_1}{d t} \mathbf{i}+\frac{d f_2}{d t} \mathbf{j}+\frac{d f_3}{d t} \mathbf{k}\)=0

⇒ \(\frac{d f_1}{d t}\)=0, \(\frac{d f_2}{d t}\)=0,\(\frac{d f_3}{d t}\)=0

⇒ f₁,f₂,f₃  are constants ⇒ f is a  constant vector function.

10. Prove that a vector function f is o fconstant magnitude if f.\(\frac{d f}{d t}\)=0

Solution:

Suppose f is constant magnitude. Then f²=|f|²= a constant.

∴ \(\frac{d}{d t}\){f²}=0 ⇒ 2f. \(\frac{d \mathbf{f}}{d t}\) =0  ⇒ f. \(\frac{d \mathbf{f}}{d t}\)

Conversely suppose that f. \(\frac{d \mathbf{f}}{d t}\) =0⇒ 2f. \(\frac{d \mathbf{f}}{d t}\) =0  ⇒ \(\frac{d}{d t}\){f²}=0

∴ f² is constant  ⇒|f|²= constant  ⇒ f is of constant length.

11. Prove that a vector function I has constant direction if f  x \(\frac{d f}{d t}\)=0

Solution:

Let f (t)=f (t)F where f (t) =| f (t)| and F (t) is a vector function with unit magnitude, for every t in the domain of f.

∴  \(\frac{d \mathbf{f}}{d t}\) = \(\frac{d}{d t}\){fF} =f\(\frac{d \mathbf{F}}{d t}+\frac{d f}{d t}\) F

f× \(\frac{d \mathbf{f}}{d t}\)=f F ×(f\(\frac{d \mathbf{F}}{d t}+\frac{d f}{d t}\) F)= f²( F ×\(\frac{d \mathbf{F}}{d t}\)) +f\(\frac{d \mathbf{f}}{d t}\)

(F×F) = f(F×\(\frac{d \mathbf{F}}{d t}\))

Suppose f has a constant direction.

Then F is constant  ⇒ \(\frac{d \mathbf{F}}{d t}\))=0

⇒ F× \(\frac{d \mathbf{F}}{d t}\)=0  ⇒ f×\(\frac{d \mathbf{f}}{d t}\)= f²( F ×\(\frac{d \mathbf{F}}{d t}\)) =0

Conversely suppose that f×\(\frac{d \mathbf{f}}{d t}\) = 0. Then  f²( F ×\(\frac{d \mathbf{F}}{d t}\)) =0⇒ F ×\(\frac{d \mathbf{F}}{d t}\)=0

F is of unit length ⇒ F.\(\frac{d \mathbf{F}}{d t}\) = 0

F ×\(\frac{d \mathbf{F}}{d t}\)=0 ⇒  F.\(\frac{d \mathbf{F}}{d t}\) = 0 ⇒\(\frac{d \mathbf{F}}{d t}\) = 0 ⇒F is constant ⇒ f have constant direction.

12. If r = e^-ti  + log(t²+ l)J-tan t k then find \(\frac{d \mathbf{r}}{d t}\),\(\frac{d^2 \mathbf{r}}{d t^2}\), \(\left|\frac{d t}{d t}\right|\), and \(\left|\frac{d^2 \mathbf{r}}{d t^2}\right|\) at =0

Solution:

r = \(e^{-t} \mathbf{i}+\log \left(t^2+1\right) \mathbf{J}-\tan t \mathbf{k} \Rightarrow \frac{d \mathbf{r}}{d t}=-e^{-t} \mathbf{i}+\frac{2 t}{t^2+1} \mathbf{J}-\sec ^2 t \mathbf{k} \)

⇒ \(\frac{d^2 \mathbf{r}}{d t^2}=e^{-t} \mathbf{I}+\frac{\left(t^2+1\right) 2-2 t \cdot 2 t}{\left(t^2+1\right)^2} \mathbf{J}-2 \sec ^2 t \tan t \mathbf{k}\)

At t = 0, \(\frac{d \mathbf{r}}{d t}=-\mathbf{I}-\mathbf{k}, \frac{d^2 \mathbf{r}}{d t^2}=\mathbf{I}+2 \mathbf{J},\left|\frac{d \mathbf{r}}{d t}\right|=\sqrt{1+1}=\sqrt{2},\left|\frac{d^2 \mathbf{r}}{d t^2}\right|=\sqrt{1+4}=\sqrt{5}\)

13. Ifr = t² i- tj + (2t+ 1) k, find the values of \(\frac{d \mathbf{r}}{d t}\),\(\frac{d^2 \mathbf{r}}{d t^2}\), \(\left|\frac{d r}{d t}\right|\), and \(\left|\frac{d^2 \mathbf{r}}{d t^2}\right|\)  at =0

Solution:

r = \(t^2 \mathbf{I}-t \mathbf{j}+(2 t-1) \mathbf{k} \Rightarrow \frac{d \mathbf{r}}{d t}=2 t \mathbf{I}-\mathbf{J}+2 \mathbf{k}, \frac{d^2 \mathbf{r}}{d \mathbf{t}^2}=2 \mathbf{r} . \)

At t = 0, \(\frac{d \mathbf{r}}{d t}=-\mathbf{J}+2 \mathbf{k}, \frac{d^2 \mathbf{r}}{d t^2}=2 \mathbf{i},\left|\frac{d \mathbf{r}}{d t}\right|=\sqrt{1+4}=\sqrt{5},\left|\frac{d^2 \mathbf{r}}{d t^2}\right|=2\)

14. If r = e^-ti  + + log(t²+ l)J-tan t k find\(\left(\frac{d r}{d t} \times \frac{d^2 r}{d t^2}\right)\)  at  t=0

Solution:

r = \(e^{-t} \mathbf{i}+\log \left(t^2+1\right) \mathbf{j}-\tan t \mathbf{k} \Rightarrow \frac{d \mathbf{r}}{d t}=-e^{-t} \mathbf{i}+\frac{2 t}{t^2+1} \mathbf{j}-\sec ^2 t \mathbf{k}\)

and \(\frac{d^2 \mathbf{r}}{d t^2}=e^{-t} \mathbf{i}+\frac{\left(t^2+1\right) 2-2 t \cdot 2 t}{\left(t^2+1\right)^2} \mathbf{j}-2 \sec ^2 t \tan t \mathbf{k}\)

At t=0, \(\frac{d \mathbf{r}}{d t}=-\mathbf{i}-\mathbf{k}, \frac{d^2 \mathbf{r}}{d t^2}=\mathbf{i}+2 \mathbf{j}\)

⇒ \(\frac{d \mathbf{r}}{d t} \times \frac{d^2 \mathbf{r}}{d t^2}=\left|\begin{array}{rrr}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-1 & 0 & -1 \\
1 & 2 & 0
\end{array}\right|\)= \(\mathbf{i}(0+2)-\mathbf{j}(0+1)+\mathbf{k}(-2-0)=2 \mathbf{i}-\mathbf{j}-2 \mathbf{k}\)

15. If r=e^t ( c cos 2t +d sin 2t) where c and d are constant vectors then show that \(\frac{d^2 r}{d t^2}-2 \frac{d r}{d t}+5 r\)=0

Solution: 

Given that \(\mathbf{r}=e^t(\mathbf{c} \cos 2 t+\mathbf{d} \sin 2 t)\)

⇒ \(\frac{d \mathbf{r}}{d t}=e^t(-2 \mathbf{c} \sin 2 t+2 \mathbf{d} \cos 2 t)+e^t(\mathbf{c} \cos 2 t+\mathbf{d} \sin 2 t)\)

= \(e^t(-2 \mathbf{c} \sin 2 t+2 \mathbf{d} \cos 2 t+\mathbf{c} \cos 2 t+\mathbf{d} \sin 2 t)\)

⇒ \(\frac{d^2 \mathbf{r}}{d t^2}=e^t(-4 \mathbf{c} \cos 2 t-4 \mathbf{d} \sin 2 t-2 \mathbf{c} \sin 2 t+2 \mathbf{d} \cos 2 t)\)

+ \(e^t(-2 \mathbf{c} \sin 2 t+2 \mathbf{d} \cos 2 t+\mathbf{c} \cos 2 t+\mathbf{d} \sin 2 t)\)

= \(e^t(-3 c \cos 2 t-3 \mathbf{d} \sin 2 t-4 \mathbf{c} \sin 2 t+4 \mathbf{d} \cos 2 t)\)

⇒ \(\frac{d^2 \mathbf{r}}{d t^2}-2 \frac{d \mathbf{r}}{d t}+5 \mathbf{r}=e^t(-3 \mathbf{c} \cos 2 t-3 \mathbf{d} \sin 2 t-4 \mathbf{c} \sin 2 t+4 \mathbf{d} \cos 2 t)\)

-2 \(e^t(-2 \mathrm{e} \sin 2 t+2 \mathrm{~d} \cos 2 t+\mathrm{c} \cos 2 t+\mathrm{d} \sin 2 t)+5 e^t(\mathrm{c} \cos 2 t+\mathrm{d} \sin 2 t)\)

= \(e^t(-3 \mathbf{c} \cos 2 t-3 \mathbf{d} \sin 2 t-4 \mathbf{c} \sin 2 t+4 \mathbf{d} \cos 2 t+4 \mathbf{c} \sin 2t\)

-4 \(\mathbf{d} \cos 2 t-2 c \cos 2 t-2 \mathbf{d} \sin 2 t+5 \mathbf{c o s} 2 t+5 \mathbf{d} \sin 2 t)=\mathbf{0} \text {. }\)

16. If A= t i-t²j+(2t=1) k and B = (2t-3)i+j-tk, find

1. (A×B)’
2. (|A+B|)’ at t=1.

Solution:

1. \(\mathbf{A} \times \mathbf{B}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
t^2 & -t & 2 t+1 \\
2 t-3 & 1 & -t
\end{array}\right|\)

= \(\mathbf{i}\left(t^2-2 t-1\right)-\mathbf{j}\left(-t^3-4 t^2+4 t+3\right)+\mathbf{k}\left(t^2+2 t^2-3 t\right)\)

= \(\left(t^2-2 t-1\right) \mathbf{i}+\left(t^3+4 t^2-4 t-3\right) \mathbf{j}+\left(3 t^2-3 t\right) \mathbf{k}\)

∴ \(\frac{d}{d t}(\mathbf{A} \times \mathbf{B})=(2 t-2) \mathbf{i}+\left(3 t^2+8 t-4\right) \mathbf{j}+(6 t-3) \mathbf{k}=7 \mathbf{j}+3 \mathbf{k}\) at t=1 .

2. A+B = \(\left[t^2 \mathbf{i}-t \mathbf{j}+(2 t+1) \mathbf{k}\right]+[(2 t-3) \mathbf{i}+\mathbf{j}-t \mathbf{k}]\)

= \(\left(t^2+2 t-3\right) \mathbf{i}+(1-t) \mathbf{j}+(t+1) \mathbf{k}\)

⇒ \(|\mathbf{A}+\mathrm{B}|=\sqrt{\left(t^2+2 t-3\right)^2+(1-t)^2+(t+1)^2}\)

= \(\sqrt{t^4+4 t^2+9+4 t^3-6 t^2-12 t+2+2 t^2}\)

= \(\sqrt{t^4+4 t^3-12 t+11}\)

⇒ \(\frac{d}{d t}(|\mathbf{A}+\mathrm{B}|)=\frac{1}{2 \sqrt{t^4+4 t^3-12 t+11}} \times 4 t^3+12 t^2-12=\frac{4+12-12}{2 \sqrt{1+4-12+11}}=\frac{2}{2}\)

=1 at t=1 .

17. If A= 5t²i+tj-t k and B = sin ti- cost j then find

(1) \(\frac{d}{d t}\) (A.B)

(2) \(\frac{d}{d t}\) (AB)

(3) \(\frac{d}{d t}\) (A.A)

Solution:

(1)  A.B = 5t² sint-t cos t

∴ \(\frac{d}{d t}\) ( A.B) 5t² cos t +10 t sin t+ t sin t -cos t =5t² cos t + 11 t sin t -cos t

(2) \(\mathbf{A} \times \mathbf{B}\)

= \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
5 t^2 & t & -t^3 \\
\sin t & -\cos t & 0
\end{array}\right|\)

= \(\mathbf{i}\left(0-t^3 \cos t\right)-\mathbf{j}\left(0+t^3 \sin t\right)+\mathbf{k}\left(-5 t^2 \cos t-t \sin t\right)\)

∴ \(\frac{d}{d t}(\mathbf{A} \times \mathbf{B})=\left(t^3 \sin t-3 t^2 \cos t\right) \mathbf{i}-\left(t^3 \cos t+3 t^2 \sin t\right) \mathbf{J}\)

+ \(\left(5 t^2 \sin t-10 t \cos t-t \cos t-\sin t\right) \mathbf{k}\)

(3) \(\mathbf{A} \cdot \mathbf{A}=\left(5 t^2\right)^2+(t)^2+\left(-t^3\right)^2=25 t^4+t^2+t^6=t^6+25 t^4+t^2\)

∴ \(\frac{d}{d t}(\mathbf{A} \cdot \mathbf{A})=\frac{d}{d t}\left(t^6+25 t^4+t^2\right)=6 t^5+100 t^3+2 t \text {. }\)

18. If r= a cos t i+ a sin t j+ at tan θ k then find \(\left|\frac{d r}{d t} \times \frac{d^2}{d t^2}\right|\) and \(\left[\frac{d \mathbf{r}}{d t} \frac{d^2 \mathbf{r}}{d t^2} \frac{d^3 \mathbf{r}}{d t^3}\right]\)

Solution: r- a cos t i+ a sin t j+ at tan θ k

r = \(a \cos t \mathbf{i}+a \sin t \mathbf{j}+a t \tan \theta \mathbf{k}\)

∴ \(\frac{d \mathbf{r}}{d t}=-a \sin t \mathbf{i}+a \cos t \mathbf{j}+a \tan \theta \mathbf{k}\), \(\frac{d^2 \mathbf{r}}{d t^2}=-a \cos t \mathbf{i}-a \sin t \mathbf{j}\),  \(\frac{d^3 \mathbf{r}}{d t^3}=a \sin t \mathbf{i}-a \cos t \mathbf{j}\)

∴ \(\frac{d \mathbf{r}}{d t} \times \frac{d^2 \mathbf{r}}{d t^2}\)

= \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-a \sin t & a \cos t & a \tan \theta \\
-a \cos t & -a \sin t & 0
\end{array}\right|\)

= \(\mathbf{i}\left(0+a^2 \sin t \tan \theta\right)-\mathbf{j}\left(0+a^2 \cos t \tan \theta\right)+\mathbf{k}\left(a^2 \sin ^2 t+a^2 \cos ^2 t\right)\)

= \(a^2 \sin t \tan \theta \mathbf{i}-a^2 \cos t \tan \theta \mathbf{j}+a^2 \mathbf{k}\)

⇒ \(\left|\frac{d \mathbf{r}}{d t} \times \frac{d^2 \mathbf{r}}{d t^2}\right|\)

= \(\sqrt{a^4 \sin ^2 t \tan ^2 \theta+a^4 \cos ^2 t \tan ^2 \theta+a^4}=a^2 \sqrt{\tan ^2 \theta+1}=a^2 \sec \theta\)

⇒ \({\left[\begin{array}{lll}
\frac{d \mathbf{r}}{d t} & \frac{d^2 \mathbf{r}}{d t^2} & \frac{d^3 \mathbf{r}}{d t^3}
\end{array}\right]}\)

= \({\left|\begin{array}{ccc}
-a \sin t & a \cos t & a \tan \theta \\
-a \cos t & -a \sin t & 0 \\
a \sin t & -a \cos t & 0
\end{array}\right|}\)

= \(a \tan \theta\left(a^2 \cos ^2 t+a^2 \sin ^2 t\right)=a^3 \tan \theta\)

19.  If r= a cos t i+ a sin t j+ at tan θ k  , find \(\left(\frac{d \mathbf{r}}{d t} \times \frac{d^2 \mathbf{r}}{d t^2}\right)\) at t=0.

Solution: r= a cos t i+ a sin t j+ at tan θ k

r = \(a \cos t \mathbf{I}+a \sin t \mathbf{j}+a t \tan \theta \mathbf{k}\)

∴ \(\frac{d \mathbf{r}}{d t}=-a \sin t \mathbf{i}+a \cos t \mathbf{j}+a \tan \theta \mathbf{k}\), \(\frac{d^2 \mathbf{r}}{d t^2}=-a \cos t \mathbf{i}-a \sin t \mathbf{j}\), \(\frac{d^3 \mathbf{r}}{d t^3}=a \sin t \mathbf{i}-a \cos t \mathbf{j}\)

∴ \(\frac{d \mathbf{r}}{d t} \times \frac{d^2 \mathbf{r}}{d t^2}\)

= \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-a \sin t & a \cos t & a \tan \theta \\
-a \cos t & -a \sin t & 0
\end{array}\right|\)

= \(\mathbf{I}\left(0+a^2 \sin t \tan \theta\right)-\mathbf{J}\left(0+a^2 \cos t \tan \theta\right)+\mathbf{k}\left(a^2 \sin ^2 t+a^2 \cos ^2 t\right)\)

= \(a^2 \sin t \tan \theta \mathbf{i}-a^2 \cos t \tan \theta \mathbf{j}+a^2 \mathbf{k}\)

∴ \(\left(\frac{d \mathbf{r}}{d t} \times \frac{d^2 \mathbf{r}}{d t^2}\right)_{t=0}=-a^2 \tan \theta \mathbf{J}+a^2 \mathbf{k}\)

20.  If A= sint i+ cos t j+tk, B= cos t i-sin t j-3k and C 2i+3j-k then find \(\frac{d}{d t}[\mathbf{A} \times(\mathbf{B} \times \mathbf{C})]\) at t =0

Solution: B× C\(=\left|\begin{array}{ccc}1 & \mathbf{k} & \mathbf{k} \\
\cos t & -\sin t & -3 \\2 & 3 & -1\end{array}\right|\)=i(sin t+9)-j(-cost+6) + k (3 cos t+2 sin t)

A x B x C = \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\sin t & \cos t & t \\
\sin t+9 & \cos t-6 & 3 \cos t+2 \sin t
\end{array}\right|\)

= \(\mathbf{i}\left(3 \cos ^2 t+2 \cos t \sin t-t \cos t+6 t\right)-\mathbf{j}\left(3 \sin t \cos t+2 \sin ^2 t-t \sin t-9 t\right)\)

+ \(\mathbf{k}(\sin t \cos t-6 \sin t-\sin t \cos t-9 \cos t)\)

= \(\mathbf{i}\left(3 \cos ^2 t+2 \cos t \sin t-t \cos t+6 t\right)-\mathbf{j}\left(3 \sin t \cos t+2 \sin ^2-t \sin t-9 t\right)\)

+ \(\mathbf{k}(-6 \sin t-9 \cos t)\)

∴ \(\frac{d}{d t}[\mathbf{A} \times(\mathbf{B} \times \mathbf{C})]=\mathbf{i}\left(-6 \cos t \sin t+2 \cos ^2 t-2 \sin ^2 t+t \sin t-\cos t+6\right)\)

– \(\mathbf{j}\left(3 \cos ^2 t-3 \sin ^2 t+4 \sin t \cos t-t \cos t-\sin t-9\right)+\mathbf{k}(-6 \cos t+9 \sin t)\)

∴ \(\frac{d}{d t}[\mathbf{A} \times(\mathbf{B} \times \mathbf{C})]=7 \mathbf{t}=0 \mathbf{i}-6 \mathbf{k} \text {. }\)

21. If r is a vector function such that |r| = r then show that

(1) [r r’ r”]’=[r r’ r”’]

(2) [r×(r’×r”)]’= (r ×(r’×r”’)+ r×(r’×r”’).

Solution: (1) [r r’ r”]= [r’ r’ r”] + [r r” r”]+[r r’ r”’]=[r r’ r”’]

(2)  [r×(r’×r”)]’=[r×(r’×r”)]’+[r×(r”×r”)]+[r×(r’×r”’)]

=r’×(r’×r”)+r×(r’×r”’)

22. Define the partial derivative of a vector function.

Solution:

Partial derivative: Let f = f (p, q, t) be a vector function of scalar variables that exists, then the limit is called the ”partial derivative”  of f  with respect to t. It is denoted by \(\frac{\partial f}{\partial f}\) . Similarly we can define \(\frac{\partial f}{\partial p}\),\(\frac{\partial f}{\partial q}\).

23. If f=cos xyi+(3xy-2x²)j-(3x+2y) k then find \(\frac{\partial^2 \mathbf{f}}{\partial x^2}\),\(\frac{\partial^2 \mathbf{f}}{\partial x \partial y}\),\(\frac{\partial^2 \mathbf{f}}{\partial y^2}\).

Solution:

f = \(\cos x y \mathbf{i}+\left(3 x y-2 x^2\right) \mathbf{j}-(3 x+2 y) \mathbf{k}\)

⇒ \(\frac{\partial \mathbf{f}}{\partial x}=-y \sin x y \mathbf{i}+(3 y-4 x) \mathbf{j}-3 \mathbf{k}\)

⇒ \(\frac{\partial \mathbf{f}}{d y}=-x \sin x y \mathbf{i}+3 x \mathbf{j}-2 \mathbf{k}\)

⇒ \(\frac{\partial^2 \mathbf{f}}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial \mathbf{f}}{\partial x}\right)=-y^2 \cos x y \mathbf{i}-4 \mathbf{j}\)

⇒ \(\frac{\partial^2 \mathbf{f}}{\partial x \partial y}=\frac{\partial}{\partial x}\left(\frac{\partial \mathbf{f}}{\partial y}\right)\)

= \(-(x y \sin x y+\sin x y) \mathbf{i}+3 \mathbf{j}\)

∴ \(\frac{\partial^2 \mathbf{f}}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial \mathbf{f}}{\partial y}\right)=-x^2 \cos x y \mathbf{i}\)

24.If f=(2x²y-x⁴)i+(e^xy-y sin x)j+(x² cos y) k, find \(\frac{\partial^2 \mathbf{f}}{\partial x^2}\) and  \(\frac{\partial^2 \mathbf{f}}{\partial x \partial y}\)

Solution:

f = \(\left(2 x^2 y-x^4\right) \mathbf{i}+\left(e^{x y}-y \sin x\right) \mathbf{j}+\left(x^2 \cos y\right) \mathbf{k}\)

⇒ \(\frac{\partial f}{\partial x}=\left(4 x y-4 x^3\right) \mathbf{I}+\left(y e^{x y}-y \cos x\right) \mathbf{j}+(2 x \cos y) \mathbf{k}\)

and \(\frac{\partial f}{\partial y} \doteq 2 x^2 \mathbf{I}+\left(x e^{x y}-\sin x\right) \mathbf{j}-x^2 \sin y \mathbf{k}\)

⇒ \(\frac{\partial^2 f}{\partial x^2}=\left(4 y-12 x^2\right) \mathbf{I}+\left(y^2 e^{x y}+y \sin x\right) \mathbf{j}+2 \cos y \mathbf{k}\)

and \(\frac{\partial^2 f}{\partial x \partial y}=4 x \mathbf{I}+\left(x y e^{x y}+e^{x y}-\cos x\right) \mathbf{j}-2 x \sin y \mathbf{k} .\)

25. If f= a cos  nt+b sin nt then prove that \(\frac{\partial^2 r}{\partial t^2}\)+n² r =0 where a,b,n are constants .

Solution:

Given r = \(a \cos n t+b \sin n t\). Then \(\frac{\partial r}{\partial t}=-a n \sin n t+b n \cos n t\)

∴ \(\frac{\partial^2 r}{\partial t^2}=-a n^2 \cos n t-b n^2 \sin n t=-n^2(a \cos n t+b \sin n t)=-n^2 r \Rightarrow \frac{\partial^2 r}{\partial t^2}+n^2 r=0 \text {. }\)

26. If f=2x²i-3yzj+xz²k and φ=2z-x³y then find

(1) f.  \(\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right)\)

(2) \(\frac{\partial \mathbf{f}}{\partial x} \cdot\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{\jmath} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right)\)  and

(3) \(\frac{\partial \mathbf{f}}{\partial z} \times\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right)\) at (1,-1,1) .

Solution:

⇒ \(\frac{\partial \mathbf{f}}{\partial x}=4 x \mathbf{i}+z^2 \mathbf{k}, \frac{\partial \mathbf{f}}{\partial z}=-3 y \mathbf{j}+2 x z \mathbf{k} . \text { At }(1,-1,1), \frac{\partial \mathbf{f}}{\partial x}=4 \mathbf{i}+\mathbf{k}, \frac{\partial \mathbf{f}}{\partial z}=3 \mathbf{j}+2 \mathbf{k}\)

⇒ \(\frac{\partial \varphi}{\partial x}=-3 x^2 y, \frac{\partial \varphi}{\partial y}=-x^3, \frac{\partial \varphi}{\partial z}=2\)

At (1,-1,1), \(\mathbf{f}=2 x^2 \mathbf{i}-3 y z \mathbf{j}+x z^2 \mathbf{k}=2 \mathbf{i}+3 \mathbf{j}+\mathbf{k}\) and \(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}=\mathbf{i}\left(-3 x^2 y\right)+\mathbf{j}\left(-x^3\right)+\mathbf{k}(2)=3 \mathbf{i}-\mathbf{j}+2 \mathbf{k}\)

1. \(\mathbf{f} \cdot\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right)=(2 \mathbf{i}+3 \mathbf{j}+\mathbf{k}) \cdot(3 \mathbf{i}-\mathbf{j}+2 \mathbf{k})=6-3+2=5\)
2. \(\frac{\partial \mathbf{f}}{\partial x} \cdot\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right)=(4 \mathbf{i}+\mathbf{k}) \cdot(3 \mathbf{i}-\mathbf{j}+2 \mathbf{k})=12+2=14\)
3. \(\frac{\partial \mathbf{f}}{\partial z} \times\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right)=(3 \mathbf{j}+2 \mathbf{k}) \times(3 \mathbf{i}-\mathbf{j}+2 \mathbf{k})\)

= \(\left|\begin{array}{rrr}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 3 & \mathbf{2} \\ 3 & -1 & 2\end{array}\right|\)

= \(\mathbf{i}(6+2)-\mathbf{j}(0-6)+\mathbf{k}(0-9)=8 \mathbf{i}+6 \mathbf{j}-9 \mathbf{k}\).

27. If r=xi+yj+zk and  a is a constant vector prove that

(1) \(\frac{\partial}{\partial x}(\mathbf{a} \cdot \mathbf{r}) \mathbf{i}+\frac{\partial}{\partial y}(\mathbf{a} \cdot \mathbf{r}) \mathbf{j}+\frac{\partial}{\partial z}(\mathbf{a} \cdot \mathbf{r}) \mathbf{k}\)=a

(2) \(\frac{\partial}{\partial x}(\mathbf{a} \times \mathbf{r}) \times \mathbf{i}+\frac{\partial}{\partial y}(\mathbf{a} \times \mathbf{r}) \times \mathbf{I}+\frac{\partial}{\partial z}(\mathbf{a} \times \mathbf{r}) \times \mathbf{k}\)=-2a

Solution:

Let \(\mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}\)

Given \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

Now \(\frac{\partial \mathbf{r}}{\partial x}=\mathbf{i}, \frac{\partial \mathbf{r}}{\partial y}=\mathbf{j}, \frac{\partial \mathbf{r}}{\partial z}=\mathbf{k}\).

1. \(\frac{\partial}{\partial x}(\mathbf{a} \cdot \mathbf{r}) \mathbf{i}+\frac{\partial}{\partial y}(\mathbf{a} \cdot \mathbf{r}) \mathbf{j}+\frac{\partial}{\partial z}(\mathbf{a} \cdot \mathbf{r}) \mathbf{k}\)

= \(\left(\mathbf{a} \cdot \frac{\partial \mathbf{r}}{\partial x}\right) \mathbf{i}+\left(\mathbf{a} \cdot \frac{\partial \mathbf{r}}{\partial y}\right) \mathbf{j}+\left(\mathbf{a} \cdot \frac{\partial \mathbf{r}}{\partial z}\right) \mathbf{k}\)

= \((\mathbf{a} \cdot \mathbf{i}) \mathbf{i}+(\mathbf{a} \cdot \mathbf{j}) \mathbf{j}+(\mathbf{a} \cdot \mathbf{k}) \mathbf{k}=\mathbf{a}\)

2. \(\frac{\partial}{\partial x}(\mathbf{a} \times \mathbf{r}) \times \mathbf{i}+\frac{\partial}{\partial x}(\mathbf{a} \times \mathbf{r}) \times \mathbf{j}+\frac{\partial}{\partial z}(\mathbf{a} \times \mathbf{r}) \times \mathbf{k}\)

= \(\left(\mathbf{a} \times \frac{\partial \mathbf{r}}{\partial x}\right) \times \mathbf{i}+\left(\mathbf{a} \times \frac{\partial \mathbf{r}}{\partial y}\right) \times \mathbf{j}+\left(\mathbf{a} \times \frac{\partial \mathbf{r}}{\partial z}\right) \times \mathbf{k}\)

= \((\mathbf{a} \times \mathbf{i}) \times \mathbf{i}+(\mathbf{a} \times \mathbf{j}) \times \mathbf{j}+(\mathbf{a} \times \mathbf{k}) \times \mathbf{k}\)

=  \((\mathbf{i} \cdot \mathbf{a}) \mathbf{i}-(\mathbf{i} \cdot \mathbf{i}) \mathbf{a}+(\mathbf{j} \cdot \mathbf{a}) \mathbf{j}-(\mathbf{j} \cdot \mathbf{j}) \mathbf{a}+(\mathbf{k} \cdot \mathbf{a}) \mathbf{k}-(\mathbf{k} \cdot \mathbf{k}) \mathbf{a}\)

= \({[(\mathbf{i} \cdot \mathbf{a}) \mathbf{i}+(\mathbf{j} \cdot \mathbf{a}) \mathbf{j}+(\mathbf{k} \cdot \mathbf{a}) \mathbf{a}]-3 \mathbf{a}=\mathbf{a}-3 \mathbf{a}=-2 \mathbf{a} }\)

28. If f=yzi+zxj+xyk, prove that \(\mathbf{i} \times \frac{\partial f}{\partial x}+\mathbf{j} \times \frac{\partial f}{\partial y}+\mathbf{k} \times \frac{\partial f}{\partial z}\)=0.

Solution:

f = \(y z \mathbf{i}+z x \mathbf{j}+x y \mathbf{k} \Rightarrow \frac{\partial f}{\partial \dot{x}}=z \mathbf{j}+y \mathbf{k}, \frac{\partial f}{\partial y}=z \mathbf{i}+x \mathbf{k}, \frac{\partial f}{\partial z}=y \mathbf{i}+x \mathbf{j}\)

⇒ \(\mathbf{i} \times \frac{\partial f}{\partial x}+\mathbf{j} \times \frac{\partial f}{\partial y}+\mathbf{k} \times \frac{\partial f}{\partial z}=\mathbf{i} \times(z \mathbf{j}+y \mathbf{k})+\mathbf{j} \times(z \mathbf{i}+x \mathbf{k})+\mathbf{k} \times(y \mathbf{i}+x \mathbf{j})\)

= \(z \mathbf{i}-y \mathbf{j}-z \mathbf{k}+x \mathbf{i}+y \mathbf{j}-x \mathbf{i}=\mathbf{0}\)

29. If φ=2xz⁴-x²y then find the value of \(\left|\frac{\partial \phi}{\partial x} \mathbf{i}+\frac{\partial \phi}{\partial y} \mathbf{j}+\frac{\partial \phi}{\partial k} \mathbf{k}\right|\) at (2,-2,-1).

Solution:

⇒ \(\phi=2 x z^4-x^2 y \Rightarrow \frac{\partial \phi}{\partial x}=2 z^4-2 x y, \frac{\partial \phi}{\partial y}=-x^2, \frac{\partial \phi}{\partial z}=8 x z^3\)

⇒ \(\frac{\partial \phi}{\partial x} \mathbf{i}+\frac{\partial \phi}{\partial y} \mathbf{j}+\frac{\partial \phi}{\partial z} \mathbf{k}=\left(2 z^4-2 x y\right) \mathbf{i}-x^2 \mathbf{j}+8 x z^3 \mathbf{k}=(2+8) \mathbf{i}-4 \mathbf{j}-16 \mathbf{k}=10 \mathbf{i}-4 \mathbf{j}-16 \mathbf{k}\)

⇒ \(\left|\frac{\partial \phi}{\partial x}+\frac{\partial \phi}{\partial y} \mathbf{j}+\frac{\partial \phi}{\partial z} \mathbf{k}\right|=\sqrt{100+16+256}=\sqrt{372} .\)

30. If φ =x²yz+4xz² and A=2i-j-2k , find A \(\left[\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right]\) at(2,-2,-1).

Solution:

⇒ \(\phi=2 x z^4-x^2 y \Rightarrow \frac{\partial \phi}{\partial x}=2 z^4-2 x y, \frac{\partial \phi}{\partial y}=-x^2, \frac{\partial \phi}{\partial z}=8 x z^3\)

⇒ \(\frac{\partial \phi}{\partial x} \mathbf{i}+\frac{\partial \phi}{\partial y} \mathbf{j}+\frac{\partial \phi}{\partial z} \mathbf{k}=\left(2 z^4-2 x y\right) \mathbf{i}-x^2 \mathbf{j}+8 x z^3 \mathbf{k}=(2+8) \mathbf{i}-4 \mathbf{j}-16 \mathbf{k}=10 \mathbf{i}-4 \mathbf{j}-16 \mathbf{k}\)

⇒ \(\left|\frac{\partial \phi}{\partial x}+\frac{\partial \phi}{\partial y} \mathbf{j}+\frac{\partial \phi}{\partial z} \mathbf{k}\right|=\sqrt{100+16+256}=\sqrt{372}\) .

31. Define a scalar point function.

Solution:

Scalar point function: Let S be a domain in space. If to each point P ∈ S there corresponds a unique scalar (real number) φ  (P) then the correspondence cp is called a “scalar point function” over the domain S.

32. Define a vector point function.

Solution:

Vector point function: Let S be a domain in space. If to each point P e S there corresponds a unique vector F (P) then the correspondence F is called a “vector point function “ over the domain S.

33. Define distance function.

Solution:

Distance function: Let O be the origin in space. For each point P (x, y, z) in space if we define r (P) = OP = \(\sqrt{x^2+y^2+z^2}\) then r  is a scalar point function. It is called the “Distance function”.

34. Define the position vector point function

Solution:

Position vector point function: Let O be the origin in space. For each point P (x, y, z) in space if we define r (P)- \(\overrightarrow{O P}\) = xi+yj + zk, then r is a vector point function, r is called ”position vector point function”.

35. Define the directional derivative of a scalar point function.

Solution:

Directional derivative of scalar point function: Let φ be a scalar point function defined on a neighborhood D of a point P. Let L be a ray from P in the direction of the unit vector e. Let Q ∈ L ∩ D and Q≠ P. If \(\stackrel{L t}{Q \rightarrow P}\)\(\frac{\varphi(Q)-\varphi(P)}{Q P}\)exists then the limit is called the “Directional derivative“ of φ  at P in the direction of e. It is denoted by \(\frac{\partial \varphi}{\partial e}\) or  \(\frac{\partial \varphi}{\partial s}\) when s = QP.

36. Define the directional derivative of a vector point function.

Solution:

Directional derivative of a vector point function: Let F be a vector point function defined on a neighborhood D of a point P. Let L be a ray from P in the direction of the unit vector e. Let Q ∈ L∩D and Q≠P If \(\) exist then the limit is called the “Directional derivative” of F at P in the direction of e. It is denoted by\(\frac{\partial \varphi}{\partial e}\) when s = 0

37. If r is the position vector point function and e is a unit vector then prove that \(\frac{\partial r}{\partial e}\)=e

Solution:

Let P be a point and Q be a point in the ray from P in the direction of e such that Q≠P.

Let \(\mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}\)

Given \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

Now \(\frac{\partial \mathbf{r}}{\partial x}=\mathbf{i}, \frac{\partial \mathbf{r}}{\partial y}=\mathbf{j}, \frac{\partial \mathbf{r}}{\partial z}=\mathbf{k}\).

1. \(\frac{\partial}{\partial x}(\mathbf{a} \cdot \mathbf{r}) \mathbf{i}+\frac{\partial}{\partial y}(\mathbf{a} \cdot \mathbf{r}) \mathbf{j}+\frac{\partial}{\partial z}(\mathbf{a} \cdot \mathbf{r}) \mathbf{k}\)

= \(\left(\mathbf{a} \cdot \frac{\partial \mathbf{r}}{\partial x}\right) \mathbf{i}+\left(\mathbf{a} \cdot \frac{\partial \mathbf{r}}{\partial y}\right) \mathbf{j}+\left(\mathbf{a} \cdot \frac{\partial \mathbf{r}}{\partial z}\right) \mathbf{k}\)

= \((\mathbf{a} \cdot \mathbf{i}) \mathbf{i}+(\mathbf{a} \cdot \mathbf{j}) \mathbf{j}+(\mathbf{a} \cdot \mathbf{k}) \mathbf{k}=\mathbf{a}\)

2. \(\frac{\partial}{\partial x}(\mathbf{a} \times \mathbf{r}) \times \mathbf{i}+\frac{\partial}{\partial x}(\mathbf{a} \times \mathbf{r}) \times \mathbf{j}+\frac{\partial}{\partial z}(\mathbf{a} \times \mathbf{r}) \times \mathbf{k}\)

= \(\left(\mathbf{a} \times \frac{\partial \mathbf{r}}{\partial x}\right) \times \mathbf{i}+\left(\mathbf{a} \times \frac{\partial \mathbf{r}}{\partial y}\right) \times \mathbf{j}+\left(\mathbf{a} \times \frac{\partial \mathbf{r}}{\partial z}\right) \times \mathbf{k}=(\mathbf{a} \times \mathbf{i}) \times \mathbf{i}+(\mathbf{a} \times \mathbf{j}) \times \mathbf{j}+(\mathbf{a} \times \mathbf{k}) \times \mathbf{k}\)

= \((\mathbf{i} \cdot \mathbf{a}) \mathbf{i}-(\mathbf{i} \cdot \mathbf{i}) \mathbf{a}+(\mathbf{j} \cdot \mathbf{a}) \mathbf{j}-(\mathbf{j} \cdot \mathbf{j}) \mathbf{a}+(\mathbf{k} \cdot \mathbf{a}) \mathbf{k}-(\mathbf{k} \cdot \mathbf{k}) \mathbf{a}\)

= \({[(\mathbf{i} \cdot \mathbf{a}) \mathbf{i}+(\mathbf{j} \cdot \mathbf{a}) \mathbf{j}+(\mathbf{k} \cdot \mathbf{a}) \mathbf{a}]-3 \mathbf{a}=\mathbf{a}-3 \mathbf{a}=-2 \mathbf{a} }\)

38. If r is the distance point function and e is a unit vector then prove that\(\frac{\partial \mathbf{r}}{\partial e}\) = r.e/ r.

Solution:

r= \(|\boldsymbol{r}| \Rightarrow r^2=\mathbf{r}^2 \Rightarrow \frac{\partial}{\partial \bullet}\left(r^2\right)=\frac{\partial}{\partial \bullet}\left(\mathbf{r}^2\right) \Rightarrow 2 r \frac{\partial r}{\partial \bullet}\)

= \(2 \mathbf{r} \cdot \frac{\partial \mathbf{r}}{\partial \bullet} \Rightarrow r \frac{\partial r}{\partial \bullet}=\mathbf{r} \cdot \bullet\)

⇒ \(\frac{\partial r}{\partial \bullet}=\frac{\mathbf{r} \cdot \bullet}{r}\)

39. Define the gradient of a scalar point function.

Solution:

Gradient s If φ is a scalar point function having directional derivatives in the i,j,k  then \(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)is called gradient of φ. It is denoted by grad φ or ∇φ.

40. If f and g are two scalar point functions then prove that

(1) grad (f±g)=grad f ±grad g

(2) grad (fg) =(grad f)g+ f (grad g)

(3) grad \(\left(\frac{f}{g}\right)\)=\(\frac{1}{g^2}\)[g (grad f)-f(grad g)](grad g)

Solution: 

(1) \(\text{grad}(f+g)=\mathbf{i} \frac{\partial}{\partial x}(f+g)+\mathbf{j} \frac{\partial}{\partial y}(f+g)+\mathbf{k} \frac{\partial}{\partial z}(f+g)\)

= \(\mathbf{i}\left(\frac{\partial f}{\partial x}+\frac{\partial g}{\partial x}\right)+\mathbf{j}\left(\frac{\partial f}{\partial y}+\frac{\partial g}{\partial y}\right)+\mathbf{k}\left(\frac{\partial f}{\partial z}+\frac{\partial g}{\partial z}\right)\)

= \(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{i} \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{j} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}+\mathbf{k} \frac{\partial g}{\partial z}\)

= \(\left(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\right)+\left(\mathbf{i} \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial g}{\partial z}\right)\)=\(\text{grad} f+\text{grad} g.\)

Similarly we can prove that \(\text{grad}(f-g)=\text{grad} f-\text{grad} g\).

(2) grad(f g) = \(\mathbf{i} \frac{\partial}{\partial x}(f g)+\mathbf{j} \frac{\partial}{\partial y}(f g)+\mathbf{k} \frac{\partial}{\partial z}(f g)\)

= \(\mathbf{i}\left(\frac{\partial f}{\partial x} g+f \frac{\partial g}{\partial x}\right)+\mathbf{j}\left(\frac{\partial f}{\partial y} g+f \frac{\partial g}{\partial y}\right)+\mathbf{k}\left(\frac{\partial f}{\partial z} g+f \frac{\partial g}{\partial z}\right)\)

= \(\mathbf{i} \frac{\partial f}{\partial x} g+\mathbf{i} f \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y} g+\mathbf{j} f \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z} \mathbf{g}+\mathbf{k} f \frac{\partial g}{\partial z}\)

= \(\left(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\right) g+f\left(\mathbf{i} \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial g}{\partial z}\right)=(g r a d f) g+f(\text{grad} g)\)

(3) \(\text{grad}\left(\frac{f}{g}\right)=\mathbf{i} \frac{\partial}{\partial x}\left(\frac{f}{g}\right)+\mathbf{j} \frac{\partial}{\partial y}\left(\frac{f}{g}\right)+\mathbf{k} \frac{\partial}{\partial z}\left(\frac{f}{g}\right)\)

= \(\mathbf{I} \frac{\left(g \frac{\partial f}{\partial x}-f \frac{\partial g}{\partial x}\right)}{g^2}+\mathbf{j} \frac{\left(g \frac{\partial f}{\partial y}-f \frac{\partial g}{\partial y}\right)}{g^2}+\mathbf{k} \frac{\left(\frac{\partial f}{\partial z} g+f \frac{\partial g}{\partial z}\right)}{g^2}\)

= \(\frac{1}{g^2}\left[g\left(\mathbf{I} \frac{\partial f}{\partial x}+\mathbf{J} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\right)-f\left(\mathbf{i} \frac{\partial g}{\partial x}+\mathbf{J} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial g}{\partial z}\right)\right]\)

= \(\frac{1}{g^2}[g(g r a d f)-f(g r a d g)]\)

41. If φ  is a scalar point function and c is a scalar then prove that grad (c φ) = c (grad φ).

Solution:

grad\((c \varphi)=\mathbf{I} \frac{\partial}{\partial x}(c \varphi)+\mathbf{J} \frac{\partial}{\partial y}(c \varphi)+\mathbf{k} \frac{\partial}{\partial z}(c \varphi)\)

= \(\mathbf{I} c \frac{\partial \varphi}{\partial x}+\mathbf{J} c \frac{\partial \varphi}{\partial y}+\mathbf{k} c \frac{\partial \varphi}{\partial z}\)

= \(c\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{J} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right)=c(\text{grad} \varphi)\)

42. Prove that a scalar point function φ is constant if grad φ = 0.

Solution:

Suppose \(\varphi\) is constant. Then \(\frac{\partial \varphi}{\partial x}=0, \frac{\partial \varphi}{\partial y}=0, \frac{\partial \varphi}{\partial z}=0\)

∴ \(\text{grad} \varphi=1 \frac{\partial \varphi}{\partial x}+\mathrm{J} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}=0\)

Conversely, suppose that grad \(\varphi=0\).

Then \(I \frac{\partial \varphi}{\partial x}+J \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}=0\)

⇒ \(\frac{\partial \varphi}{\partial x}=0, \frac{\partial \varphi}{\partial y}=0, \frac{\partial \varphi}{\partial z}=0 \Rightarrow \varphi\) is constant.

43. If φ + x³-y³+x² z then find grad φ at (1, 1,-2).

Solution:

⇒ \(\frac{\partial \varphi}{\partial x}=3 x^2+2 x z, \frac{\partial \varphi}{\partial y}=-3 y^2, \frac{\partial \varphi}{\partial z}=x^2\)

grad \(\varphi=\mathbf{1} \frac{\partial \varphi}{\partial x}+\mathbf{J} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}=\left(3 x^2+2 x z\right) \mathbf{1}-3 y^2 \mathbf{J}+x^2 \mathbf{k}\)

At (1,1,-2), \(\text{grad} \varphi=-\mathbf{I}-3 \mathbf{J}+\mathbf{k}\) .

44. Find grad f  at the point (1, 1,- 2) where f= x³+y³+3xyz.

Solution:

f = \(x^3+y^3+3 x y z \Rightarrow \frac{\partial f}{\partial x}=3 x^2+3 y z, \frac{\partial f}{\partial y}=3 y^2+3 x z, \frac{\partial f}{\partial z}=3 x y\)

∴ grad f = \(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}=\left(3 x^2+3 y z\right) \mathbf{i}+\left(3 y^2+3 x z\right) \mathbf{j}+3 x y \mathbf{k}\)

At (1,1,-2), \(\text{grad} f=(3-6) \mathbf{i}+(3-6) \mathbf{j}+3 \mathbf{k}=-3 \mathbf{i}-3 \mathbf{j}+3 \mathbf{k}\).

45. If φ= x³ +y³ + z³+ 3xyz then find grad φ at  (1, 2, 3).

Solution:

⇒ \(\phi=x^3+y^3+z^3+3 x y z \Rightarrow \frac{\partial \phi}{\partial x}=3 x^2+3 y z, \frac{\partial \phi}{\partial y}=3 y^2+3 x z, \frac{\partial \phi}{\partial z}=3 z^2+3 x y\)

∴ grad \(\phi\)

= \(\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}=\left(3 x^2+3 y z\right) \mathbf{i}+\left(3 y^2+3 x z\right) \mathbf{j}+\left(3 z^2+3 x y\right) \mathbf{k}\)

At(1,2,3), \(\text{grad} \phi=(3+18) \mathbf{i}+(12+9) \mathbf{j}+(27+6) \mathbf{k}=21 \mathbf{i}+21 \mathbf{j}+33 \mathbf{k}\) .

r = \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

Then  r= \(|\mathbf{r}|=\sqrt{x^2+y^2+z^2} \Rightarrow r^2=x^2+y^2+z^2\)

2 r \(\frac{\partial r}{\partial x}=2 x, 2 r \frac{\partial r}{\partial y}=2 y, 2 r \frac{\partial r}{\partial z}=2 z \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}, \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r}\)

⇒ \(\nabla r=\mathbf{i} \frac{\partial r}{\partial x}+\mathbf{j} \frac{\partial r}{\partial y}+\mathbf{k} \frac{\partial r}{\partial z}=\mathbf{i} \frac{x}{r}+\mathbf{j} \frac{y}{r}+\mathbf{k} \frac{z}{r}=\frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{r}=\frac{\mathbf{r}}{r}\)

46. Show that ∇r\(=\frac{r}{r}\)

Solution:

r = \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

Then r = \(|\mathbf{r}|=\sqrt{x^2+y^2+z^2} \Rightarrow r^2=x^2+y^2+z^2\)

2 r \(\frac{\partial r}{\partial x}=2 x, 2 r \frac{\partial r}{\partial y}=2 y, 2 r \frac{\partial r}{\partial z}\)=2 z

⇒ \(\frac{\partial r}{\partial x}=\frac{x}{r}, \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r}\)

∴ \(\nabla r=\mathbf{i} \frac{\partial r}{\partial x}+\mathbf{j} \frac{\partial r}{\partial y}+\mathbf{k} \frac{\partial r}{\partial z}=\mathbf{i} \frac{x}{r}+\mathbf{j} \frac{y}{r}+\mathbf{k} \frac{z}{r}=\frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{r}=\frac{\mathbf{r}}{r}\)

47. Show that ∇\(\left(\frac{1}{r}\right)\)=\(\frac{-r}{r^3}\)

Solution:

⇒ \(\nabla\left(\frac{1}{r}\right)=\mathbf{i} \frac{\partial}{\partial x}\left\{\frac{1}{r}\right\}+\mathbf{j} \frac{\partial}{\partial y}\left\{\frac{1}{r}\right\}+\mathbf{k} \frac{\partial}{\partial z}\left\{\frac{1}{r}\right\}\)

= \(\mathbf{i}\left\{-\frac{1}{r^2}\right\} \frac{\partial r}{\partial x}+\mathbf{j}\left\{-\frac{1}{r^2}\right\} \frac{\partial r}{\partial y}+\mathbf{k}\left\{-\frac{1}{r^2}\right\} \frac{\partial r}{\partial z}\)

= \(-\frac{1}{r^2}\left[\mathbf{i} \frac{\partial r}{\partial x}+\mathbf{j} \frac{\partial r}{\partial y}+\mathbf{k} \frac{\partial r}{\partial z}\right]=-\frac{1}{r^2}\left[\mathbf{i} \frac{x}{r}+\mathbf{j} \frac{y}{r}+\mathbf{k} \frac{z}{r}\right]\)

= \(-\frac{1}{r^3}[x \mathbf{i}+y \mathbf{j}+z \mathbf{k}]=-\frac{\mathbf{r}}{r^3}\)

48. show that ∇f(r) =f’ (r) \(=\frac{r}{r}\).

Solution:

⇒ \(\nabla f(r)=\mathbf{i} \frac{\partial}{\partial x}\{f(r)\}+\mathbf{j} \frac{\partial}{\partial y}\{f(r)\}+\mathbf{k} \frac{\partial}{\partial z}\{f(r)\}=\mathbf{i} f^{\prime}(r) \frac{\partial r}{\partial x}+\mathbf{j} f^{\prime}(r) \frac{\partial r}{\partial y}+\mathbf{k} f^{\prime}(r) \frac{\partial r}{\partial z}\)

= \(f^{\prime}(r)\left[\mathbf{i} \frac{x}{r}+\mathbf{j} \frac{y}{r}+\mathbf{k} \frac{z}{r}\right]=f^{\prime}(r) \frac{\mathbf{r}}{r}\)

49. Show that ∇(log|r|)\(=\frac{r}{r^2}\)

Solution:

⇒ \(\nabla \log |r|=\mathbf{I} \frac{\partial}{\partial x}\{\log |r|\}+\mathbf{J} \frac{\partial}{\partial y}\{\log |r|\}+\mathbf{k} \frac{\partial}{\partial z}\{\log |r|\}\)

= \(\mathbf{I} \frac{1}{r} \frac{\partial r}{\partial x}+\mathbf{J} \frac{1}{r} \frac{\partial r}{\partial y}+\mathbf{k} \frac{1}{r} \frac{\partial r}{\partial z}=\mathbf{I} \frac{1}{r} \frac{x}{r}+\mathbf{J} \frac{1}{r} \frac{y}{r}+\mathbf{k} \frac{1}{r} \frac{z}{r}=\frac{\mathbf{r}}{r^2}\)

50. If x+y+z,b=x²+y²+z², c= xy+yz+zx then show that =0

Solution:

⇒ \(\nabla a=\left(\mathbf{1} \frac{\partial}{\partial x}+\mathbf{1} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right)(x+y+z)\)

= \(\mathbf{I} \frac{\partial}{\partial x}(x+y+z)+\mathbf{J} \frac{\partial}{\partial y}(x+y+z)+\mathbf{k} \frac{\partial}{\partial z}(x+y+z)=\mathbf{1}+\mathbf{j}+\mathbf{k}\)

⇒ \(\nabla b=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{J} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right)\left(x^2+y^2+z^2\right)=2 x \mathbf{i}+2 y \mathbf{J}+2 z \mathbf{k}\)

⇒ \(\nabla c=\left(\mathbf{1} \frac{\partial}{\partial x}+\mathbf{J} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right)(x y+y z+z x)=(y+z) \mathbf{I}+(x+z) \mathbf{\jmath}+(y+x) \mathbf{k}\)

51. Show that gradr^m=m ^m-2r.

Solution:

⇒ \(\frac{\partial}{\partial x}\left(r^m\right)=m r^{m-1} \frac{\partial r}{\partial x}=m r^{m-1} \frac{x}{r}=m r^{m-2} x\)

⇒ \(\frac{\partial}{\partial y}\left(r^m\right)=m r^{m-1} \frac{\partial r}{\partial y}=m r^{m-1} \frac{y}{r}=m r^{m-2} y\)

⇒ \(\frac{\partial}{\partial z}\left(r^m\right)=m r^{m-1} \frac{\partial r}{\partial z}=m r^{m-1} \frac{z}{r}=m r^{m-2} z\)

grad \(r^m=\mathbf{i} \frac{\partial}{\partial x}\left(r^m\right)+\mathbf{j} \frac{\partial}{\partial y}\left(r^m\right)+\mathbf{k} \frac{\partial}{\partial z}\left(r^m\right)=m r^{m-2} x \mathbf{i}+m r^{m-\dot{2}} y \mathbf{j}+m r^{m-2} z \mathbf{k}\)

= \(m r^{m-2}(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})=m r^{m-2} \mathbf{r} \text {. }\)

52. Show that grad (r.a)=a

Solution:

Let a=a₁i+a₂j+a₃k= , r= xi+yj=zk, r.a=a₁x+a₂y+a₃z

grad(r.a) =∇(r.a)= \(=\mathbf{i} \frac{\partial}{\partial x}(\mathbf{r} \cdot \mathbf{a})+\mathbf{j} \frac{\partial}{\partial y}(\mathbf{r} \cdot \mathbf{a})+\mathbf{k} \frac{\partial}{\partial z}(\mathbf{r} \cdot \mathbf{a})\)

= i a₁+j a₂+k a₃=a.

53. Define the level surface of a scalar point function.

Leval surface: Let φ  be a scalar point function defined over the domain S and P ∈  S. The set of all points Q  ∈ S such thatφ  (Q)= φ   (P) is called a “Level surface” of (p through P. If c is a constant then the set of all points Q (x,y, z) ∈ S such that φ (Q)- c is called a “Level surface” at the level c. It is denoted by φ(x,y,z) = c.

54. Prove that the directional derivative of a scalar point function φ at a point P in the direction of the unit e vector is (grad φ). e.

Solution:

⇒ \(\frac{\partial \varphi}{\partial e}\)=\(\frac{\partial \varphi}{\partial s}\) =\(\frac{\partial \varphi}{\partial x}\)

⇒ \(\frac{\partial \varphi}{\partial s}\)

+ \(\frac{\partial \varphi}{\partial y}\)\(\frac{\partial \varphi}{\partial s}\)

+ \(\frac{\partial \varphi}{\partial z}\)\(\frac{\partial \varphi}{\partial s}\)

= \(\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right) \cdot\left(\mathbf{i} \frac{\partial x}{\partial s}+\mathbf{j} \frac{\partial y}{\partial s}+\mathbf{k} \frac{\partial z}{\partial s}\right)\)

= (grad φ).e.

55. Find the directional derivative of φ =x²yz + 4x²z at the point (1, -2,- 1) in the direction of 2i − j −2k

Solution:

If e is the unit vector in the direction of 2i-j-2k then

e= \(=\frac{2 \mathbf{i}-\mathbf{j}-2 \mathbf{k}}{|2 \mathbf{i}-\mathbf{j}-2 \mathbf{k}|}=\frac{2 \mathbf{i}-\mathbf{j}-2 \mathbf{k}}{3}\)

=\(\frac{\partial \varphi}{\partial x}\)=2xyz+4z²,  \(\frac{\partial \varphi}{\partial y}\)=x²z,\(\frac{\partial \varphi}{\partial z}\)=x²y+8xz

grad φ =\(i \frac{\partial \varphi}{\partial x}+j \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)(2xyz+4z²)i+(x²z)j+(x²y+8xz)k

The directional derivative of in the direction of e is (grad φ) e

⇒ \(=\frac{2\left(2 x y z+4 z^2\right)-x^2 z-2\left(x^2 y+8 x z\right)}{3}\)

At (1,-2,-1) , directional derivative \(=\frac{2(4+4)-(-1)-2(-2-8)}{3}\)=\(\frac{37}{3}\)

56. Find the directional derivative of φ=xyz at (1, 1, 1) in the direction of the vector i+j+kSolution:

grad φ \(=1 \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+k \frac{\partial \varphi}{\partial z}\) =yzi+xzj+xyz

It is the vector in the direction of I+J+K, then e \(=\frac{1+j+k}{\sqrt{1+1+1}}\)=\(\frac{1+j+k}{\sqrt{3}}\)

Directional derivative =(grad φ).  e (yzi+xzj+xyk).(i+j+k)/\(\sqrt{3}\)

(yz+zx+xy)/\(\sqrt{3}\)

Directional derivative at (1,1,1) is (1+1+1)/\(\sqrt{3}\)=\(\sqrt{3}\).

57. Find the directional derivative of φ= xy + yz + zx at the point (1, 2, 0) in the direction i+2j+2k.

Solution:

If e is the unit vector in the direction of i+2j+2k, then

e \(=\frac{\mathbf{i}+2 \mathbf{j}+2 \mathbf{k}}{\sqrt{1+4+4}}\)=\(\frac{\mathbf{i}+2 \mathbf{j}+2 \mathbf{k}}{3}\)

⇒ \(\frac{\partial \varphi}{\partial x}\)=y+z,\(\frac{\partial \varphi}{\partial x}\)=x+z,\(\frac{\partial \varphi}{\partial x}\)=y+x.

grad φ =\(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)=(y+z)i+(z+x)j+(x+y)k

The directional derivative of  in the direction of e is

(grad φ) . e\(=\frac{(y+z)+2(x+z)+2(y+x)}{3}\)

At directional (1,2,0) derivative \(=\frac{(2+0)+2(1+0)+2(2+1)}{3}\)=\(\frac{10}{3}\).

58. Find the directional derivative of φ =xy+yz+zx at A in the direction of \(\overrightarrow{A B}\) where A=(1,2,-1) , B=(-1,2,3).

Solution:

f = xy+yz+zx

grad f = \(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}=(y+z) \mathbf{i}+(z+x) \mathbf{j}+(x+y) \mathbf{k}\)

A B =(-1-1) \(\mathbf{i}+(2-2) \mathbf{j}+(3+1) \mathbf{k}=-2 \mathbf{i}+4 \mathbf{k}\)

If is the unit vector in the direction of -2 i+4 k then \(\frac{-2 i+4 k}{\sqrt{4+16}}=\frac{1}{\sqrt{5}}(-1+2 k)\)

∴ The directional derivative =\(\mathbf{e} \cdot \text{grad} f\)

= \((1 / \sqrt{5})(-\mathbf{i}+2 \mathbf{k}) \cdot[(y+z) \mathbf{i}+(z+x) \mathbf{j}+(x+y) \mathbf{k}]=(1 / \sqrt{5})[-y-z+2 x+2 y]\)

= \((1 / \sqrt{5})[2 x+y-z]=1 / \sqrt{5}(2+2+1)\) at (1,2,-1)=\((5 / \sqrt{5})=\sqrt{5}\) at (1,2,-1)

59. Find the directional derivative of f=x²-y² + 2z² at the point P(1, 2, 3) in the direction of the line \(\overrightarrow{P Q}\)where Q = (5, 0, 4).

Solution:

Given that \(\overrightarrow{OP}\) =i+2j+3k,\(\overrightarrow{OQ}\) = 5i+4k.

∴  \(\overrightarrow{PQ}\)=\(\overrightarrow{OQ}\)–\(\overrightarrow{OP}\) = 4i-2j+k.

Unit vector in the direction of  \(\overrightarrow{PQ}\) is

e = \(\frac{\overrightarrow{P Q}}{|\overrightarrow{P Q}|}\) = \(\frac{4 i-2 j+k}{\sqrt{16+4+1}}\) = \(\frac{1}{\sqrt{21}}(4 i-2 j+k)\)

∇ f \(=i \frac{\partial f}{\partial x}+i \frac{\partial f}{\partial y}+k \frac{\partial f}{\partial z}\)=12x+j(-2y)+k(4z)

Directional derivative =e.∇f

= \(\frac{1}{\sqrt{21}}(4 i-2 j+k) \cdot(2 x i-2 y j+4 z k)\)

= \(\frac{1}{\sqrt{21}}(8 x+4 y+4 z)\)

60. Find the directional derivative of  φ = xy +yz2 +x2 along the tangent to the curve x = t,y = t²,z = t³ at (1, 1, 1)

Solution:

The position vector of any point on the given curve is r=xi+yj+zk

⇒ r= ti+t²j+t³k

⇒ \(\frac{d r}{d t}\)= i+2tj+3t²k

Unit vector along the tangent is e \(=\frac{\mathbf{i}+2 t \mathbf{j}+3 t^2 \mathbf{k}}{\sqrt{1+4 t^2+9 t^4}}\)=\(\frac{\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}}{\sqrt{14}}\) at (1,1,1)

Directional derivative along e is ∇φ \(\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right)\).e

=[i(y²+2x)+j(2xy+z²) +k (2yz)].e

=(3i+3j+2k). \(\frac{(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k})}{\sqrt{14}}\)

=\(=\frac{3+6+6}{\sqrt{14}}\)

=\(\frac{15}{\sqrt{14}}\) at(1,1,1).

61. Find the directional derivative of the function xy²+yz²+ zx² along the tangent to the curve x =t,y  = t², z = t³at the point (1, 1, 1).

Solution:

The position vector of any point on the given curve is r=xi+yj+zk

⇒ r=ti+t²j+t³k

⇒ \(\frac{d r}{d t}\) = i+2tj+3t²k

Unit vector along the tangent is e \(=\frac{\mathbf{i}+2 t \mathbf{j}+3 t^2 \mathbf{k}}{\sqrt{1+4 t^2+9 t^4}}\)=\(\frac{\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}}{\sqrt{14}}\) at(1,1,1)

Let φ =xy²+yz²+zx²

The directional derivative of φ  along e is  ∇ φ .e \(=\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right)\).e

=[i(y²+2zx)+j(2xy+z²)+k (2yz+x²)].e

= (3i+3j+3k).\(\frac{(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k})}{\sqrt{14}}\)

= \(\frac{3+6+9}{\sqrt{14}}\)

∴ \(\frac{18}{\sqrt{14}}\) at (1,1,1).

62. Prove that g grad φ is a normal vector to the level surface φ (x,y, z) = c where c is constant.

Solution:

Let p(x,y,z) be a point on the level surface and T be the unit tangent vector at P The position vector of P is r= xi+yj+zk

∴\(\frac{\partial r}{\partial s}=\mathbf{i} \frac{\partial x}{\partial s}+\mathbf{j} \frac{\partial y}{\partial s}+\mathbf{k} \frac{\partial z}{\partial s}\)

φ (x,y,z) = c ⇒ \(\frac{\partial \varphi}{\partial s}\)=0

⇒ \(\frac{\partial \varphi}{\partial x} \frac{\partial x}{\partial s}+\frac{\partial \varphi}{\partial y} \frac{\partial y}{\partial s}+\frac{\partial \varphi}{\partial z} \frac{\partial z}{\partial z}\)=0

\(\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right) \cdot\left(\mathbf{i} \frac{\partial x}{\partial s}+\mathbf{j} \frac{\partial y}{\partial s}+\mathbf{k} \frac{\partial z}{\partial s}\right)\)=0

⇒ (grad φ). \(\frac{\partial \varphi}{\partial s}\)=0

⇒ (grad φ). T =0 ⇒ grad φ is perpendicular to T

⇒ grad φ is a normal vector to the level surface φ(x,y,z)=c.

63. If φ is a scalar point function then prove that\(\frac{\partial \varphi}{\partial s}\) direction of grad φ.

Solution:

The directional derivative of φ at a point  P in the direction of a unit vector e is
\(\frac{\partial \varphi}{\partial s}\) =(grad φ ) .e =\(\frac{\partial \varphi}{\partial N}\)N.e where N is the normal vector to the surface. at P.

∴  \(\frac{\partial \varphi}{\partial s}\)=\(\frac{\partial \varphi}{\partial N}\) |N| |e| cos (N,e) = \(\frac{\partial \varphi}{\partial N}\) cos (N,e)

\(\frac{\partial \varphi}{\partial s}\) has maximum ⇔ cos (N,e) = 1 ⇔  N=e.

∴ The directional derivative has maximum value along the normal to the surface.

∴ \(\frac{\partial \varphi}{\partial s}\)  has maximum value in the direction of grad   φ

Maximum value of the directional derivative=\(\frac{\partial \varphi}{\partial \mathbf{N}}\)=|grad φ|

64 Find the maximum value of the directional derivative of φ = 2x²-y-z⁴ at (2,1,-1)

Solution:

⇒ \(\frac{\partial \varphi}{\partial x}=4 x, \frac{\partial \varphi}{\partial y}=-1, \frac{\partial \varphi}{\partial z}=-4 z^3\)

grad \(\varphi=\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}=4 x \mathbf{i}-\mathbf{j}-4 z^3 \mathbf{k}\)

At (2,-1,1), \(\text{grad} \varphi=8 \mathbf{i}-\mathbf{j}-4 \mathbf{k}\)

Maximum value of the directional derivative of \(\varphi\) at (2,-1,1) is \(|\text{grad} \varphi|=\sqrt{64+1+16}=\sqrt{81}=9\)

65. Find the greatest value of the directional derivative of the function f=£y£ at (2, 1,-1).

Solution:

grad f=2xyzi+xzj+3xyzk= -4i-4j+12k at (2,-1,1)

∴ Greatest value directional derivative of f = |∇f| =\(\sqrt{16+16+44}\)=\( \sqrt{11}\).

66. Find the maximum value of the directional derivative and the direction of the directional derivative when it is maximum, of φ =xy+ 2yz + 3xz at the point (1, 1, 1).

Solution:

⇒ \(\frac{\partial \varphi}{\partial x}=y+3 z, \frac{\partial \varphi}{\partial y}=x+2 z, \frac{\partial \varphi}{\partial z}=2 y+3 x\)

grad \(\varphi=\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}=(y+3 z) \mathbf{i}+(x+2 z) \mathbf{j}+(3 x+2 y) \mathbf{k}\)

At(1,1,1), \(\text{grad} \varphi=4 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k}\)

Maximum value of the directional derivative of \(\varphi\) at (1,1,1) is \(|\text{grad} \varphi|\)

= \(\sqrt{16+9+25}=5 \sqrt{2}\)

A directional derivative is maximum in the direction of the unit normal vector N.

N = \(\frac{\text{grad} \varphi}{|\text{grad} \varphi|}=\frac{4 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k}}{5 \sqrt{2}}\)

67. Define the angle between two surfaces.

Solution:

The angle between surfaces: Let P be a point of intersection (common point) to the level surfaces f(x, y, z) = 0,g (x, y, z) = 0. The angle between the normals to the surfaces f(x,y, z) = 0, g (x,y, z) = 0 at P is called the “Angle between the surfaces” at P.

68. Find the angle between the surfaces of the spheres x²+y² + z² = 29, x² +y² + z² + 4x- 6y- 8z- 47 = 0 at the point (4,- 3, 2).

Solution: Let f=x²+y² + z² − 29,  g=x² +y² + z² + 4x- 6y- 8z- 47

grad f = \(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}=2 x \mathbf{i}+2 y \mathbf{j}+2 z \mathbf{k}\)

grad g = \(\mathbf{i} \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial g}{\partial z}=(2 x+4) \mathbf{i}+(2 y-6) \mathbf{j}+(23-8) \mathbf{k}\)

At (4,-3,2), \(\text{gradf}=8 \mathbf{i}-6 \mathbf{j}+4 \mathbf{k}=\mathbf{a}\) (say), \(\text{grad} g=12 \mathbf{i}-12 \mathbf{j}-4 \mathbf{k}=\mathbf{b}\)  (say).

Now \(\mathbf{a}, \mathbf{b}\) are normal vectors to the surfaces at (4,-3,2)

∴ Angle between the surfaces at (4,-3,2) is equal to \((\mathbf{a}, \mathbf{b})\).

⇒ \(\cos (\mathbf{a}, \dot{b})=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{(8 \mathbf{i}-6 \mathbf{j}+4 \mathbf{k}) \cdot(12 \mathbf{I}-12 \mathbf{j}-4 \mathbf{k})}{\sqrt{64+36+16} \sqrt{144+144+16}}\)

= \(\frac{96+72-16}{\sqrt{166} \sqrt{304}}=\frac{152}{\sqrt{116 \times 304}}=\sqrt{\frac{19}{29}}\)

∴ \((\mathbf{a}, \mathbf{b})=\text{Cos}^{-1}\left(\sqrt{\frac{19}{29}}\right)\)

69.   Find the angle between the surfaces x²+y²+z²=9 and =x²+y²−z=3

Solution:

Let f=x²+y²+z²-9 and g=x²+y²+z²-3

∇f = \(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\)=2xi+2yj+2zk. At( 2,-1,2), ∇f=4I-2J+4K

The normal to the surface x+y+z= 9 at (2,-1,2), is 4i-2j+4k

= \(2xi+2yj-k At (2,-1,2), g=4i-2j-kx\)

The normal to the surface z= x²+y²-z at (2,-1,2) is 4i-2j+4k

If  θ  is the angle between the surfaces at (2,-1,2) then

cos θ \(=\frac{(4 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}) \cdot(4 \mathbf{i}-2 \mathbf{j}-\mathbf{k})}{|4 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}||4 \mathbf{i}-2 \mathbf{j}-\mathbf{k}|}\)

= \(\frac{16+4-4}{\sqrt{16+4+16} \sqrt{16+4+1}}\)

= \(\frac{16}{6 \sqrt{21}}\)=\(\frac{8}{3 \sqrt{21}}\)

∴ Angle between the surfaces = Cos^-1\((8 / 3 \sqrt{21)}\)

 

Multiple Integrals – 1 Exercise 1 Solved Problems

1. Evaluate \(\int_0^2 \int_0^3 x y d x d y\)

Solution: \(\int_0^2 \int_0^3 x y d x d y=\int_0^2 x d x \int_0^3 y d y\)

= \(\left[\frac{x^2}{2}\right]\left[\frac{y^2}{2}\right]=(2-0)\left(\frac{9}{2}-0\right)=9\)

2. Evaluate \(\int_0^3 \int_{-1}^1 x^2 y^2 d x d y\)

Solution: \(\left.\int_0^3 \int_{-1}^1 x^2 y^2 d x d y=\int_0^3 \int_{-1}^1 \int_{-1}^2 y^2 d y\right] d x\)

= \(\int_0^3 2\left[\int_0^1 x^2 y^2 d y\right] d x=2 \int_0^3\left[\frac{x^2 y^3}{3}\right] d x=2 \int_0^3 \frac{x^2}{3} d x\)

= \(2\left[\frac{x^3}{9}\right]_b^3=6\)

3. Evaluate \(\int_0^3 \int_1^2 x y(x+y) d y d x\)

Solution: \(\int_0^3 \int_1^2 x y(x+y) d y d x=\int_0^3\left[\int_1^2\left(x^2 y+x y^2\right) d x\right] d y\)

= \(\int_0^3\left[\frac{x^3 y}{3}+\frac{x^2 y^2}{2}\right]_0^2 d y\)

= \(\int_0^3\left(\frac{7 y}{3}+\frac{3 y^2}{2}\right) d y\)

= \(\left[\frac{7 y^2}{6}+\frac{y^3}{2}\right]=\frac{63}{6}+\frac{27}{2}=\frac{48}{2}=24\)

4. Evaluate \(\int_0^1 \int_0^1 \frac{d x d y}{\sqrt{1-x^2} \sqrt{1-y^2}}\)

Solution: \(I=\int_0^1 \int_0^1 \frac{d x d y}{\sqrt{1-x^2} \sqrt{1-y^2}}\)

= \(\int_0^1 \frac{d x}{\sqrt{1-x^2}} \int_0^1 \frac{d y}{\sqrt{1-y^2}}=\left[\ Sin ^-1 x \right]\left[\ Sin ^-1 y\right]\)

= \(\left(\ Sin^-1 1-\ Sin^-1 0\right)\left(\ Sin^-1 1-\ Sin^-1 0\right)=\frac{\pi}{2} \frac{\pi}{2}=\frac{\pi^2}{4}\)

5. Evaluate \(\iint x y\left(x^2+y^2\right) d x d y \text { over }[0, a ; 0, b]\)

Solution: \(\iint_R x y\left(x^2+y^2\right) d x d y=\int_0^a \int_0^b x y\left(x^2+y^2\right) d x d y\)

= \(\int_0^a\left[\int_0^b x y\left(x^2+y^2\right) d y\right] d x\)

= \(\int_0^a\left[\int_0^b\left(x^3 y+x y^3\right) d y\right] d x=\int_0^a\left[\frac{x^3 y^3}{2}+\frac{x y^4}{4}\right]_0^b d x\)

= \(\int_0^a\left[\frac{b^2 x^3}{2}+\frac{b^4 x}{4}\right] d x=\left[\frac{b^2 x^4}{8}+\frac{b^4 x^2}{8}\right]_0^a\)

= \(\frac{a^4 b^2+a^2 b^4}{8}=\frac{1}{8} a^2 b^2\left(a^2+b^2\right)\)

6. Evaluate \(\iint_R \frac{d x d y}{\left(1+x^2\right)\left(1+y^2\right)} \text { over }[0,1 ; 0,1]\)

Solution: \(\iint_R \frac{d x d y}{\left(1+x^2\right)\left(1+y^2\right)}\)

= \(\int_0^1 \int_0^1 \frac{d x d y}{\left(1+x^2\right)\left(1+y^2\right)}=\int_0^1 \frac{1}{1+x^2} d x \int_0^1 \frac{1}{1+y^2} d y\)

= \(\left[\text{Tan}^{-1} x\right]_{x=0}^{x=1}\left[\text{Sin}^{-1} y\right]_{y=0}^{y=1}=\frac{\pi}{4} \frac{\pi}{4}=\frac{\pi^2}{16}\)

7. Evaluate \(\iint y e^{x y} d x d y \text { over }[0, a ; 0, b]\)

Solution: \(\iint y e^{x y} d x d y=\int_0^a \int_0^b y e^{x y} d x d y=\int_0^b\left[\int_0^a y e^{x y} d x\right] d y\)

= \(\int_0^b\left[y \frac{e^{x y}}{y}\right]_{x=0}^{x=a} d y=\int_0^b\left(e^{a y}-1\right) d y\)

= \(\left[\frac{e^{a y}}{a}-y\right]_0^b=\frac{e^{a b}-1}{a}-b\)

8. Evaluate \(\iint \frac{x-y}{x+y} d x d y \text { over }[0,1 ; 0,1]\)

Solution: \(\iint \frac{x-y}{x+y} d x d y=\int_0^1 \int_0^1 \frac{x-y}{x+y} d x d y\)

= \(\int_0^1\left[\int_0^1 \frac{2 x-(x+y)}{x+y} d y\right] d x=\int_0^1\left[\int_0^1\left(\frac{2 x}{x+y}-1\right) d y\right] d x\)

= \(\int_0^1[2 x \log (x+y)-y] \int_{y=0}^{y=1} d x=\int_0^1[2 x \log (x+1)-1-2 x \log x] d x\)

= \(\left[\log (x+1) x^2\right]-\int_0^1 \frac{x^2}{x+1} d x-[x]-\left[\log \left(x^2\right)\right]_0^1+\int_0^1 \frac{x^2}{x} d x\)

= \(\log 2-\int_0^1\left(x-1+\frac{1}{x+1}\right) d x-1+\left[\frac{x^2}{2}\right]_0^1\)

= \(\log 2-\left[\frac{x^2}{2}-x+\log (x+1)\right]-1+\frac{1}{2}\)

= \(\log 2-\frac{1}{2}+1-\log 2-\frac{1}{2}=0\)

9. Evaluate \(\int_0^2 \int_0^x y d x d y\)

Solution: \(\int_0^2 \int_0^x y d x d y=\int_0^2\left[\int_0^x y d y\right] d x\)

= \(\int_0^2\left[\frac{y^2}{2}\right]=\int_0^2 \frac{x^2}{2} d x=\left[\frac{x^3}{6}\right]=\frac{8}{6}=\frac{4}{3} .\)

10. Evaluate \(\int_1^2 \int_1^x x y^2 d x d y\)

Solution: \(\int_1^2 \int_1^x x y^2 d x d y=\int_1^2\left[\int_1^x x y^2 d y\right] d x\)

= \(\int_1^2\left[x \frac{y^3}{3}\right]_1^x d x=\int_1^2\left[\frac{x^4}{3}-\frac{x}{3}\right] d x\)

= \(\left[\frac{x^5}{15}-\frac{x^2}{6}\right]^2=\frac{32}{15}-\frac{4}{6}-\frac{1}{15}+\frac{1}{6}\)

= \(\frac{31}{15}-\frac{3}{6}=\frac{62-15}{30}=\frac{47}{30}\)

11. Evaluate \(\int_0^2 \int_{x^2}^{2 x}(2 x+3 y) d x d y\)

Solution: \(\left.\int_0^2 \int_{x^2}^{2 x}(2 x+3 y) d x d y=\int_0^2 \int_{x^2}^{2 x}(2 x+3 y) d y\right] d x\)

= \(\int_0^2\left[2 x y+\frac{3 y^2}{2}\right]_{x^2}^{2 x} d x\)

= \(\int_0^2\left(4 x^2+6 x^2-2 x^3-\frac{3 x^4}{2}\right) d x\)

= \(\int_0^2\left[10 x^2-2 x^3-\frac{3 x^4}{2}\right] d x=\left[\frac{10 x^3}{3}-\frac{x^4}{2}-\frac{3 x^5}{10}\right]_0^2\)

= \(\frac{80}{3}-8-\frac{48}{5}=\frac{400-120-144}{15}=\frac{136}{15}\)

12. Evaluate \(\int_0^4 \int_0^{x^2} e^{y / x} d x d y\)

Solution: \(\int_0^4 \int_0^{x^2} e^{y / x} d x d y\)

= \(\int_0^4\left[\int_0^{x^2} e^{y / x} d y\right]\)dx

= \(\int_0^4\left[x e^{y / x}\right]_0^{x^2} d x\)

= \(\int_0^4\left(x e^x-x\right) d x\)

= \(\left[x e^x-e^x-\frac{x^2}{2}\right]_0^4\)

= \(\left(4 e^4-e^4-8\right)-(0-1-0)=3 e^4-7\)

13. Evaluate \(\int_0^1 \int_x^{\sqrt{x}}\left(x^2+y^2\right) d x d y\)

Solution: \(\int_0^1 \int_x^{\sqrt{x}}\left(x^2+y^2\right) d x d y\)

= \(\int_0^1\left[\int_x^{\sqrt{x}}\left(x^2+y^2\right) d y\right] d x=\int_0^1\left[x^2 y+\frac{y^3}{3}\right]_x^{\sqrt{x}} d x\)

= \(\int_0^1\left[x^{5 / 2}+\frac{x^{3 / 2}}{3}-x^3-\frac{x^3}{3}\right] d x\)

= \(\int_0^1\left[x^{5 / 2}+\frac{x^{3 / 2}}{3}-\frac{4 x^3}{3}\right] d x\)

= \(\left[\frac{x^{7 / 2}}{7 / 2}+\frac{x^{5 / 2}}{3(5 / 2)}-\frac{x^4}{3}\right]=\frac{2}{7}+\frac{2}{15}-\frac{1}{3}\)

= \(\frac{30+14-35}{105}=\frac{9}{105}=\frac{3}{35}\)

14. Evaluate \(\int_0^5 \int_0^{x^2} x\left(x^2+y^2\right) d x d y\).

Solution: \(\int_0^5 \int_0^{x^2} x\left(x^2+y^2\right) d x d y=\int_0^5\left[x^3 y+\frac{x y^3}{3}\right]_0^{x^2} d x\)

= \(\int_0^5\left[x^5+\frac{x^7}{3}\right] d x=\left[\frac{x^6}{6}+\frac{x^8}{24}\right]_0^5\)

= \(5^6\left[\frac{1}{6}+\frac{25}{24}\right]=\frac{29 \times 5^6}{24}\)

15. Evaluate \(\int_0^1 \int_x^{\sqrt{x}} x^2 y^2(x+y) d y d x\)

Solution: \(\int_0^1 \int_x^{\sqrt{x}} x^2 y^2(x+y) d x d y\)

= \(\int_0^1\left[\int_x^{\sqrt{x}}\left(x^3 y^2+x^2 y^3\right) d y\right] d x\)

= \(\int_0^1\left[\frac{x^3 y^3}{3}+\frac{x^2 y^4}{4}\right]_x^{\sqrt{x}} d x\)

= \(\int_0^1\left(\frac{x^{9 / 2}}{3}+\frac{x^4}{4}-\frac{x^6}{3}-\frac{x^6}{4}\right) d x\)

= \(\left[\frac{x^{11 / 2}}{3(11 / 2)}+\frac{x^5}{20}-\frac{x^7}{21}-\frac{x^7}{28}\right]_0^1\)

= \(\frac{2}{23}+\frac{1}{20}-\frac{1}{21}-\frac{1}{28}\)

= \(\frac{73}{660}-\frac{7}{84}=\frac{73}{660}-\frac{1}{12}=\frac{18}{660}=\frac{3}{110}\)

16. Evaluate \(\int_0^a \int_0^{\sqrt{a^2-x^2}} \sqrt{a^2-x^2-y^2} d x d y\).

Solution: \(\int_0^a \int_0^{\sqrt{a^2-x^2}} \sqrt{a^2-x^2-y^2} d x d y\)

= \(\int_0^a\left[\int_0^{\sqrt{a^2-x^2}} \cdot \sqrt{\left(a^2-x^2\right)-y^2} d y\right] d x\)

= \(\int_0^a\left[\frac{y}{2} \sqrt{a^2-x^2-y^2}+\frac{a^2-x^2}{2} \sin ^{-1} \frac{y}{\sqrt{a^2-x^2}}\right]_0^{\sqrt{a^2-x^2}} d x\)

= \(\int_0^a \frac{a^2-x^2}{2} \cdot \frac{\pi}{2} d x=\frac{\pi}{4} \int_0^a\left(a^2-x^2\right) d x\)

= \(\frac{\pi}{4}\left[a^2 x-\frac{x^3}{3}\right]_0^a=\frac{\pi}{4}\left[a^3-\frac{a^3}{3}\right]=\frac{\pi a^3}{6} .\)

17. Evaluate \(\int_0^1 \int_{\sqrt{y}}^{2-y} x^2 d x d y\).

Solution: \(\int_0^1 \int_{\sqrt{y}}^{2-y} x^2 d x d y=\int_0^1\left[\int_{\sqrt{y}}^{2-y} x^2 d x\right] d y\)

= \(\int_0^1\left[\frac{x^3}{3}\right]_{\sqrt{y}}^{2-y} d y=\int_0^1 \frac{(2-y)^3-y \sqrt{y}}{3} d y\)

= \(\left[\frac{-(2-y)^4}{12}-\frac{y^{5 / 2}}{3(5 / 2)}\right]_0^1\)

= \(-\frac{1}{12}-\frac{2}{15}+\frac{16}{12}=\frac{5}{4}-\frac{2}{15}=\frac{67}{60} .\)

18. Evaluate \(\int_1^{\log 8} \int_0^{\log y} e^{x+y} d x d y\).

Solution: \(\int_1^{\log 8} \int_0^{\log y} e^{x+y} d x d y\)

= \(\int_1^{\log 8}\left[\int_0^{\log y} e^{x+y} d x\right] d y=\int_1^{\log 8}\left[e^{x+y}\right]_0^{\log y} d y\)

= \(\int_1^{\log 8}\left(y e^y-e^y\right) d y\)

= \(\int_0^{\log 8}(y-1) e^y d y=\left[(y-1) e^y\right]_1^{\log 8}-\int_1^{\log 8} e^y d y=8(\log 8-1)-\left[e^y\right]_1^{\log 8}\)

= \(8 \log 8-8-8+e=8 \log 8-16+e\)

19. Prove that \(\int_0^1\left\{\int_0^1 \frac{x-y}{(x+y)^3} d y\right\} d x=\frac{1}{2} \neq-\frac{1}{2}=\int_0^1\left\{\int_0^1 \frac{x-y}{(x+y)^3} d x\right\} d y\).

Solution: \(\int_0^1\left\{\int_0^1 \frac{x-y}{(x+y)^3} d y\right\} d x\)

= \(\int_0^1\left\{\int_0^1\left[\frac{2 x}{(x+y)^3}-\frac{1}{(x+y)^2}\right] d y\right\} d x\)

= \(\int_0^1\left[\frac{1}{x+y}-\frac{x}{(x+y)^2}\right] d x\)

= \(\int_0^1\left[\frac{y}{(x+y)^2}\right]_0^1 d x=\int_0^1 \frac{1}{(1+x)^2} d x\)

= \(\left[-\frac{1}{1+x}\right]=-\frac{1}{2}+1=\frac{1}{2}\)

= \(\int_0^1\left[\int_0^1 \frac{x-y}{(x+y)^3} d x\right] d y\)

= \(\int_0^1\left[\int_0^1\left\{\frac{1}{(x+y)^2}-\frac{2 y}{(x+y)^3}\right\} d x\right] d y\)

= \(\int_0^1\left[-\frac{1}{x+y}+\frac{y}{(x+y)^2}\right]_0^1 d y\)

= \(\int_0^1\left[-\frac{x}{(x+y)^2}\right]_0^1 d y=-\int_0^1 \frac{1}{(1+y)^2} d y=\left[\frac{1}{1+y}\right]=\frac{1}{2}-1=-\frac{1}{2} .\)

20. Evaluate \(\iint(5-2 x-y) d x d y \text { where } \mathrm{R} \text { is given by } y=0, x+2 y=3, x=y^2 \text {. }\).

Solution: The region R bounded by \(y=0, x+2 y=3, x=y^2\) is shown in the figure.

It is convenient to consider a horizontal strip, with one end moving on x=y^2 and another end moving on x=3-2y.

To cover the region, this strip has to be slid from y=0 to y=1.

 

 

Hence \(\iint_R(5-2 x-y) d x d y=\int_0^1 \int_{y^2}^{3-2 y}(5-2 x-y) d y d x\)

= \(\int_0^1\left[\int_{y^2}^{3-2 y}(5-2 x-y) d x\right] d y=\int_0^1\left[5 x-x^2-x y\right]_2^{3-2 y} d y\)

= \(\int_0^1\left[5(3-2 y)-(3-2 y)^2-(3-2 y) y-5 y^2+y^4+y^3\right] d y\)

= \(\int_0^1\left(y^4+y^3-7 y^2-y+6\right) d y\)

= \(\left[\frac{y^5}{5}+\frac{y^4}{4}-\frac{7 y^3}{3}-\frac{y^2}{2}+6 y\right]_0^1\)

= \(\frac{1}{5}+\frac{1}{4}-\frac{7}{3}-\frac{1}{2}+6=\frac{217}{60} .\)

21. Evaluate \(\iint x y(x+y) d x d y \text { over the area between } y=x^2 \text { and } y=x\).

Solution:The curves y=x^2 and y=x intersect at (0,0),(1,1).

The region of integration is 0≤x≤1,x^2≤y≤x.

∴ \(\iint x y(x+y) d x d y=\int_0^1\left[\int_x^2[x y(x+y) d y]\right] d x\)

= \(\int_0^1\left[\int_{x^2}^x\left(x^2 y+x y^2\right) d y\right] d x=\int_0^1\left[\frac{x^2 y^2}{2}+\frac{x y^3}{3}\right]_{x^2}^x d x\)

= \(\int_0^1\left[\frac{x^4}{2}+\frac{x^4}{3}-\frac{x^6}{2}-\frac{x^7}{3}\right] d x=\int_0^1\left(\frac{5 x^4}{6}-\frac{x^6}{2}-\frac{x^7}{3}\right) d x\)

= \(\left[\frac{x^5}{6}-\frac{x^7}{14}-\frac{x^8}{24}\right]_0^1=\frac{1}{6}-\frac{1}{14}-\frac{1}{24}=\frac{9}{162}=\frac{3}{56}\)

22. Evaluate \(\iint e^{2 x+3 y} d x d y\) over the triangle bounded by x=0,y=0, and x+y=1.

Solution: The region of integration R is the shaded area shown in the figure.

The integral can be evaluated over R by considering either a horizontal strip or a vertical strip.

Imagine a vertical strip such that one end is moving on y=0 and the other end is moving on y=1-x.

By sliding this strip from x=0 to x=1 the region can be covered.

Hence \(\iint_R e^{2 x+3 y} d x d y=\int_0^1 \int_0^{1-x} e^{2 x+3 y} d x d y=\int_0^1\left[\int_0^{1-x} e^{2 x+3 y} d y\right] d x\)

= \(\int_0^1\left[\frac{e^{2 x+3 y}}{3}\right]_0^{1-x} d x\)

= \(\int_0^1 \frac{1}{3}\left(e^{3-x}-e^{2 x}\right) d x=\frac{1}{3}\left[-e^{3-x}-\frac{e^{2 x}}{2}\right]_0^1\)

= \(\frac{1}{3}\left[-e^2-\frac{e^2}{2}+e^3+\frac{1}{2}\right]=\frac{1}{3}\left[\frac{2 e^3-3 e^2+1}{2}\right]\)

= \(\frac{1}{6}(e-1)^2(2 e+1) .\)

23. Evaluate \(\iint_R x^2 d x d y\) where R is the region in the first quadrant bounded by the hyperbola xy=16 and the lines y=x,y=0, and x=8.

Solution:

The region is as a shaded area. It may be noted that the entire region can’t be covered with a single horizontal strip or a vertical strip.

Let us divide the region into two regions R₁ and R₂ and evaluate. Hence

∴ \(\iint_{\boldsymbol{R}} x^2 d x d y\) =\(\iint_{R_1} x^2 d x d y+\iint_{R_2} x^2 d x d y\)

Imagine a vertical strip in R₁ ∋ one end moves on the x-axis and another end moves on y = x. Thus the limits of y are 0 to x

And the strip must be slid from x = 0 to x = 4 to cover R₁

 

 

Imagine a vertical strip in R₂  ∋ one end moves on the x-axis and another end moves on = 16/x. Thus the limits of y are 0, 16/x.

This strip must be moved from x = 4 tox = 8 to cover R₂

Hence \(\iint_R x^2 d x d y=\iint_{R_1} x^2 d x d y+\iint_{R_2} x^2 d x d y\)

= \(\int_0^4 \int_0^x x^2 d x d y+\int_4^8 \int_{10}^{16 / x} x^2 d x d y \cdot\)

= \(\int_0^4\left[\int_0^x x^2 d y\right] d x+\int_4^8\left[\int_{10}^{16 / x} x^2 d y\right] d x\)

= \(\int_0^4\left[x^2 \cdot y\right]_0^x d x+\int_4^8\left[x^2 y\right]_0^{16 / x} d x=\int_0^4 x^3 d x+\int_4^8 16 x d x\)

= \(\left[\frac{x^4}{4}\right]_0^4+\left[\frac{16 x^2}{2}\right]_4^8=\frac{256}{4}+\frac{16 \cdot 64}{2}-\frac{16 \cdot 16}{2}=64+512-128=448\).

24. Evaluate \(\iint\left(x^2+y^2\right) d x d y\) over the area bounded by x+y≤1 in the first quadrant.

Solution: 

⇒ \(\iint_R\left(x^2+y^2\right) d x d y=\int_{x=0}^1 \int_{y=0}^{1-x}\left(x^2+y^2\right) d x d y\)

= \(\int_{x=0}^1\left[\int_{y=0}^{1-x}\left(x^2+y^2\right) d y\right] d x\)

= \(\int_0^1\left[x^2 y+\frac{y^3}{3}\right]_0^{1-x} d x d y=\int_0^1\left[x^2(1-x)+\frac{1}{3}(1-x)^3\right] d x\)

= \(\left[\frac{x^3}{3}-\frac{x^4}{4}-\frac{1}{12}(1-x)^4\right]_0^1=\left[\frac{1}{3}-\frac{1}{4}-0\right]-\left[0-0-\frac{1}{12}\right]\)

= \(\frac{4-3+1}{12}=\frac{1}{6} .\)

25. Evaluate \(\iint\left(x^2+y^2\right) d x d y\) over the region bounded by y=x,y=2x,x=1 in the first quadrant .

Solution:

⇒ \(\iint_R\left(x^2+y^2\right) d x d y=\int_{x=0}^1 \int_{y=x}^{y=2 x}\left(x^2+y^2\right) d x d y\)

= \(\int_{x=0}^1\left[\int_{y=x}^{y=2 x}\left(x^2+y^2\right) d y\right] d x\)

= \(\int_0^1\left[x^2 y+\frac{y^3}{3}\right]_{y=x}^{2 x} d x=\int_0^1\left[2 x^3+\frac{8 x^3}{3}-x^3-\frac{x^3}{3}\right] d x\)

= \(\int_0^1 \frac{10}{3} x^3 d x=\frac{10}{3}\left[\frac{x^4}{4}\right]_0^1=\frac{5}{6}\)

26. Evaluate \(\iint x y d x d y\) taken over the positive quadrant of the circle \(x^2+y^2=a^2\).

Solution: The region of integration is the positive quadrant of the circle x²+y²=a²

⇒ y various from 0 to \(\sqrt{a^2-x^2}\) and x varies from 0 to a.

∴ \(\iint_0 x y d x d y=\int_0^a \int_0^{\sqrt{a^2-x^2}} x y d x d y=\int_0^a\left[\int_0^{\sqrt{a^2-x^2}} x y d y\right] d x\)

= \(\int_0^a\left[x \frac{y^2}{2}\right] d x=\int_0^{\sqrt{a^2-x^2}} \frac{x\left(a^2-x^2\right)}{2} d x=\frac{a^4}{8 .}\)

27. Evaluate \(\iint(x+y)^2 d x d y\) over the area bounded by the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).

Solution: 

Given \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow \frac{y^2}{b^2}=\frac{a^2-x^2}{a^2} \Rightarrow y= \pm \frac{b}{a} \sqrt{a^2-x^2}\)

The region of integration is \(-a \leq x \leq a\),

– \(\frac{b}{a} \sqrt{a^2-x^2} \leq y \leq \frac{b}{a} \sqrt{a^2-x^2} \text {. }\)

⇒ \(\iint_R(x+y)^2 d x d y=\iint_R\left(x^2+y^2+2 x y\right) d x d y\)

= \(\int_{x=-a}^a \int_{-\frac{b}{a} \sqrt{a^2-x^2}}^{\frac{b}{a} \sqrt{a^2-x^2}}\left(x^2+y^2+2 x y\right) d x d y\)

= \(\int_{-a}^a\left[\int_{-\frac{b}{a} \sqrt{a^2-x^2}}^{\frac{b}{a} \sqrt{a^2-x^2}}\left(x^2+y^2+2 x y\right) d y\right] d x\)

= \(\int_{-a}^a\left[x^2 y+\frac{y^3}{3}+x y^2\right]_{-\frac{b}{a}}^{\frac{b}{a} \sqrt{a^2-x^2}-x^2} d x \int_{-a}^a\left[x^2\left(\frac{2 b}{a} \sqrt{a^2-x^2}\right)+\frac{2}{3} \frac{b^3}{a^3}\left(a^2-x^2\right)^{3 / 2}+0\right] d x\)

= \(2 \int_0^a\left[x^2\left(\frac{2 b}{a} \sqrt{a^2-x^2}\right)+\frac{2}{3} \frac{b^3}{a^3}\left(a^2-x^2\right)^{3 / 2}\right] d x\)

Put x = \(a \sin \theta\) so that

dx = \(a \cos \theta d \theta\)

= \(4 \int_0^{\pi / 2}\left[\frac{b}{a} a^2 \sin ^2 \theta a \cos \theta+\frac{b^3}{3 a^3} a^3 \cos ^3 \theta\right] a \cos \theta d \theta\)

= \(4 b a^3 \int_0^{\pi / 2} \sin ^2 \theta \cos ^2 \theta d \theta+\frac{4 a b^3}{3} \int_0^{\pi / 2} \cos ^4 \theta d \theta\)

= \(4 a^3 b \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}+4 \frac{a b^3}{3} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\)

= \(\left(a^3 b+a b^3\right) \frac{\pi}{4}=\frac{a b\left(a^2+b^2\right) \pi}{4}\)

28. Evaluate \(\int_0^{\pi / 4} \int_0^{\sqrt{\cos 2 \theta}} \frac{r}{\left(1+r^2\right)^2} d \theta d r\)

Solution:

⇒ \(\int_0^{\pi / 4} \int_0^{\sqrt{\cos 2 \theta}} \frac{r}{\left(1+r^2\right)^2} d \theta d r\)

= \(\int_0^{\pi / 4}\left[\int_0^{\sqrt{\cos 2 \theta}} \frac{r}{\left(1+r^2\right)^2} d r\right] d \theta\)

= \(\int_0^{\pi / 4} \frac{1}{2}\left[-\frac{1}{1+r^2}\right]_0^{\sqrt{\cos 2 \theta}} d \theta\)

= \(\frac{1}{2} \int_0^{\pi / 4}\left(\frac{-1}{1+\cos 2 \theta}+1\right) d \theta=\frac{1}{2} \int_0^{\pi / 4}\left(1-\frac{1}{2} \sec ^2 \theta\right) d \theta\)

= \(\frac{1}{2}\left[\theta-\frac{1}{2} \tan \theta\right]_0^{\pi / 4}=\frac{1}{2}\left[\frac{\pi}{4}-\frac{1}{2}\right]\)

= \(\frac{\pi-2}{8}\)

29. Evaluate \(\int_0^{\pi / 4} \int_0^{a \sin \theta} \frac{r d \theta d r}{\sqrt{a^2-r^2}}\).

Solution:  \(\int_0^{\pi / 4} \int_0^{a \sin \theta}\left[\frac{r d r}{\sqrt{a^2-r^2}}\right] d \theta\)

= \(\int_0^{\pi / 4}\left[-\left(a^2-r^2\right)^{1 / 2}\right]_0^{a \sin \theta} d \theta=\int_0^{\pi / 4}[-a \cos \theta+a] d \theta\)

= \([-a \sin \theta+a \theta]{ }_0^{\pi / 4}=a\left[\frac{\pi}{4}-\sin \frac{\pi}{4}\right]=a\left(\frac{\pi}{4}-\frac{1}{\sqrt{2}}\right)\)

30. Evaluate \(\int_0^\pi \int_0^{a \sin \theta} r d \theta d r\)

Solution: \(\int_0^\pi \int_0^{a \sin \theta} r d r d \theta=\int_0^\pi\left[\int_0^{a \sin \theta} r d r\right] d \theta\)

= \(\int_0^\pi\left[\frac{r^2}{2}\right]_0^{a \sin \theta} d \theta=\int_0^\pi \frac{a^2}{2} \sin ^2 \theta d \theta=a^2 \int_0^{\pi / 2} \sin ^2 \theta d \theta\)

= \(a^2\left(\frac{1}{2}\right)\left(\frac{\pi}{2}\right)=\frac{\pi a^2}{4}\)

31. Evaluate \(\int_0^{\pi / 2} \int_0^{2 a \cos \theta} r^2 \sin \theta d \theta d r\)

Solution: \(\int_0^{\pi / 2} \int_0^{2 a \cos \theta} r^2 \sin \theta d r d \theta\)

= \(\int_0^{\pi / 2}\left[\int_0^{2 a \cos \theta} r^2 \sin \theta d r\right] d \theta\)

= \(\int_0^{\pi / 2}\left[\frac{r^3}{3} \sin \theta\right]_0^{2 a \cos \theta} d \theta\)

⇒ \(\int_0^{\pi / 2}\left[\frac{8 a^3}{3} \cos ^3 \theta \sin \theta\right] d \theta\)

= \(\frac{8 a^3}{3}\left[\frac{-\cos ^4 \theta}{4}\right]_0^{\pi / 2}\)=\(\frac{2 a^3}{3}\)

32. Evaluate \(\int_0^{\pi / 2} \int_{a(1-\cos \theta)}^a r^2 d \theta d r\)

Solution:  \(\int_0^{\pi / 2} \int_{a(1-\cos \theta)}^a r^2 d \theta d r=\int_0^{\pi / 2}\left[\int_{a(1-\cos \theta)}^a r^2 d r\right] d \theta\)

= \(\int_0^{\pi / 2} d \theta[\frac{r^3}{3}]=\frac{1}{a} \int_0^{\pi / 2}[a^3-a^3(1-\cos \theta)\)

= \(\frac{a^3}{3} \int_0^{\pi / 2}\left[1-(1-\cos \theta)^3\right] d \theta\)

= \(\frac{a^3}{3} \int_0^{\pi / 2}\left[1-\left(1-3 \cos \theta+3 \cos ^2 \theta-\cos ^3 \theta\right)\right] d \theta\)

= \(\frac{a^3}{3} \int_0^{\pi / 2}\left[3 \cos \theta-3 \cos ^2 \theta+\cos ^3 \theta\right] d \theta\)

= \(\frac{a^3}{3}\left[3(\sin \theta)-3 \int_0^{\pi / 2} \cos ^2 \theta d \theta+\int_0^{\pi / 2} \cos ^3 \theta d \theta\right]\)

= \(\frac{a^3}{3}\left[3-3 \cdot \frac{1}{2} \cdot \frac{\pi}{2}+\frac{2}{3} \cdot 1\right]\)

= \(\frac{a^3}{3}\left[3-\frac{3 \pi}{4}+\frac{2}{3}\right]=\frac{a^3}{36}(44-9 \pi) .\)

33. Evaluate \(\iint r \sin \theta d \theta d r\) over the cardioid r=a(1-cos⁡θ) above the initial line.

Solution:  The cardioid is symmetric about the initial line and through the pole O. Above the initial line the region of integration is r=0, r=a(1-a cos⁡θ) and  θ=0, θ=π

⇒ \(\iint_R r \sin \theta d r d \theta=\int_{r=0}^{a(1-\cos \theta)} \int_{\theta=0}^\pi r \sin \theta d r d \theta\)

= \(\int_{\theta=0}^\pi \sin \theta\left[\frac{r^2}{2}\right]_0^{a(1-\cos \theta)} d \theta\)

= \(\int_0^\pi \frac{a^2(1-\cos \theta)^2}{2} \sin \theta d \theta\)

= \(\frac{a^2}{2}\left[\frac{(1-\cos \theta)^3}{3}\right]_0^\pi\)

= \(\frac{a^2}{6}\left[(1-\cos \pi)^3-(1-\cos 0)^3\right]=\frac{a^2}{6}[8-0]=\frac{4 a^2}{3}\)

34. Evaluate \(\iint r \sqrt{a^2-r^2} d \theta d r\)over the upper half of the circle r=a cos⁡θ.

Solution:

The required integral = \(\int_0^{\pi / 2} \int_{r=0}^{r=a \cos \theta} r \sqrt{a^2-r^2} d r d \theta\)

= \(\int_0^{\pi / 2}\left[-\frac{1}{3}\left(a^2-r^2\right)^{3 / 2}\right]_0^{a \cos \theta} d \theta\)

= \(-\frac{1}{3} \int_0^{\pi / 2}\left\{\left(a^2-a^2 \cos ^2 \theta\right)^{3 / 2}-\left(a^2\right)^{3 / 2}\right\} d \theta\)

= \(\frac{a^3}{3} \int_0^{\pi / 2}\left(1-\sin ^3 \theta\right) d \theta=\frac{a^3(3 \pi-4)}{18} .\)

35. Evaluate \(\iint \frac{\sqrt{\left(a^2 b^2-b^2 x^2-a^2 y^2\right)}}{\sqrt{\left(a^2 b^2+b^2 x^2+a^2 y^2\right)}} d x d y\) the field of integration being the positive quadrant of the ellipse \(x^2 / a^2+y^2 / b^2=1\).

Solution: Changing the variables x,y to , where x=aX,y=bY

We see that, since ∂ (x,y)/ ∂ (x,y)= ab, the integral = \(a b\iint \sqrt{\left(\frac{1-X^2-Y^2}{1+X^2+Y^2}\right)}\)dX dY,

The new field of integration is the positive quadrant of the circle X²+Y²=1.

Changing X,Y to r, where X= r cosθ , Y= =r sinθ , so that(X,Y)/(r,θ )=r,

We see that the integral = ab\(=a b \iint \frac{\sqrt{\left(1-r^2\right)}}{\sqrt{\left(1+r^2\right)}}\)r dr dθ.

It is easily seen that the positive quadrant of the circle X²+Y²= 1, will be described if 0 varies from 0 to π/2 and corresponds to each value of θ between 0 and  π/2, r varies from 0 to 1.

This new domain of integration, therefore, is the rectangle [ 0, 1:0, 1/2π].

Thus the integral = \(a b \int_0^{\pi / 2} d \theta \int_0^1 \frac{\sqrt{\left(1-r^2\right)}}{\sqrt{\left(1+r^2\right)}} r d r\)

= \(\frac{\pi}{2} a b \int_0^1 \frac{\sqrt{\left(1-r^2\right)}}{\sqrt{\left(1+r^2\right)}} r d r=\frac{1}{2} \pi(\pi-2) a b\)

where the integral has been evaluated by putting \(r^2=\cos t\).

36. Integrate the function \(\frac{1}{x y}\) over the area bounded by the four circles \(x^2+y^2=a x, a^{\prime} x, b y, b^{\prime} y \text { where } a, a^{\prime}, b, b^{\prime}\) are all positive.

Solution:  The integration is to be carried over the shaded area. We have supposed a’> a and b’ > b.

The region of integration is defined by ax≤ x+y ≤ a’ x; by≤x+y≤ b’y. We change the variables to u, v, where

u = \(\frac{\left(x^2+y^2\right)}{x, v}=\frac{\left(x^2+y^2\right)}{y} \Rightarrow x=\frac{u v^2}{u^2+v^2} ; y=\frac{u^2 v}{u^2+v^2}\)

It is easy to see that \(\frac{\partial(u, v)}{\partial(x, y)}=-\frac{\left(x^2+y^2\right)^2}{x^2 y^2}\)

∴ \(\frac{\partial(x, y)}{\partial(u, v)}=-\frac{x^2 y^2}{\left(x^2+y^2\right)^2}\).

Since the Jacobian is negative, the transformation is inverse. This fact may also be directly verified. The new field of integration is determined by the boundaries u=a, u=a’, v=b,v=b’, and is, therefore, the rectangle [a, a’; b, b’].

Thus we see that the integral = \(\iint \frac{1}{x y}|J| d u d v=\iint \frac{x y}{\left(x^2+y^2\right)^2} d u d v\)‘

= \(\iint \frac{1}{u v} d u d v=\int_a^{a^{\prime}} \frac{1}{u} d u \int_b^{b^{\prime}} \frac{1}{v} d v=\log \frac{a^{\prime}}{a} \cdot \log \frac{b^{\prime}}{b} .\)

37. By substituting x+y=u,x=uv, prove that the value of \(\iint \sqrt{[x y(1-x-y)]} d x d y\) taken over the interior of the triangle bounded by the lines x=0,y=0,x+y-1=0 is 2π/105.

Solution: 

Now x=uv,y=u (1-v)⇒ ∂(x,y)/∂(u,v)=-u, so that  the jacobian≤ 0.

The jacobian vanishes when u=0, i.e when x=y=0, but not otherwise.

It is easy to see that to the origin of the xy-plane corresponds to the whole line u=0 of the uv- plane so that the correspondence ceases to be one-to-one.

In order to exclude x=0, y=0, we look upon the given integral, which certainly exists, as the limit, when h→0, of the integral over the region bounded by x+y=1, x=0, y=h. (h>0)

The transformed region is, then, bounded by the lines u=1,v=0,u(1-v)=h, which correspond to the three boundaries of the region in the xy-plane. When h→0, this new region of the UV-plane tends, as the limit, to the square bounded by the lines u=1, v=1,u=0,v=0.

Thus the Integral

= \(\iint \sqrt{[u \nu u(1-v)(1-u)} u d u d v=\int_0^1 u^2 \sqrt{1-u} d u \int_0^1 \sqrt{[v(1-v)]} d v\)

Putting \(u=\sin ^2 \theta\) and \(v=\sin ^2 \psi\), we see that

⇒ \(\int_0^1 u^2 \sqrt{(1-u)} d u=2 \int_0^{\pi / 2} \sin ^5 \theta \cos ^2 \theta d \theta\)

= \(\frac{2 \cdot 4 \cdot 2 \cdot 1}{\not \cdot 5 \cdot 3 \cdot 1}=\frac{16}{105}\),

∴ \( \int_0^1 \sqrt{v(1-v)} d v=2 \int_0^{\pi / 2} \sin ^3 \psi \cos ^2 \psi d x=\frac{2 \cdot 1 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2}=\frac{\pi}{8}\).

Note:  We have u=x+y, v=x/(x+y) and x= uv, y= u(1-v).

38. Using the transformation x+y=u,x-y=ν, evaluate \(\iint e^{\frac{x-y}{x+y}} d x d y\) over the region bounded by x=0,y=0,x+y=1.

Solution:

Solving x+y=u, x-y=v, we have x=\(\frac{u+v}{2}, y=\frac{u-v}{2}\)

We transform the function, curves in to u, v plane with this substitution. xy plane

Curves :

x=0

y=0

x+y=1

u v plane

u=-v

u=v

u=1

Now, \(f(x, y)=e^{\frac{x-y}{x+y}}\) is . \(f(u, v)=e^{\frac{v}{u}}\) . \(J\left(\begin{array}{ll}x & y \\ u & v\end{array}\right)\)

= \(\left|\begin{array}{ll}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{array}\right|\)

= \(\left|\begin{array}{cc}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{array}\right|=-\frac{1}{2}\).

Hence dx dy = \(\left|J\left(\begin{array}{ll}
x & y \\
u & v
\end{array}\right)\right| d u d v=\frac{1}{2} d u d v\)

⇒ \(\int_0^1 \int_{-u}^u e^{v / u} \cdot \frac{1}{2} d v d u\)

= \(\frac{1}{2} \int_0^1\left[u e^{v / u}\right]_{-u}^u d u=\frac{1}{2} \int_0^1\left(u \cdot e-u \cdot e^{-1}\right) d u\)

= \(\frac{1}{2} \int_0^1\left(u e-\frac{u}{e}\right) d u\)

= \(\frac{1}{2}\left(e-\frac{1}{e}\right)\left[\frac{u^2}{2}\right]_0^1=\frac{1}{4}\left(e-\frac{1}{e}\right)\).

In the new region of the u-v plane imagine a vertical strip with one end on v=-u and the other end on v=u. This has, to be slid from u=0 to u=1, to cover the region. With the transformation, the integral becomes

39. By changing into polar coordinates evaluate the integral \(\int_0^{2 a} \int_0^\sqrt{2 a x-x^2}\left(x^2+y^2\right) d x d y\).

Solution:  The region of integration is the semi-circle x+y=2ax above the x-axis. Changing into polar, the region becomes r=0 to r=2a cos θ from  θ=0 to θ= π/2. Hence the required integral

= \(\int_0^{\pi / 2} \int_0^{2 a \cos \theta}\left(r^2 \cos ^2 \theta+r^2 \sin ^2 \theta\right) r d r d \theta\)

=\(\int_0^{\pi / 2} \cdot \int_0^{2 a \cos \theta} r^2 d r d \theta\)

= \(\int_0^{\pi / 2}\left[\frac{r^4}{4}\right]_0^{2 a \cos \theta} d \theta=4 a^4 \int_0^{\pi / 2} \cos ^4 \theta d \theta=\frac{3 \pi a^4}{4} .\)

40. Change into polar coordinates and evaluate \(\int_0^a \int_0^{\sqrt{a^2-x^2}} e^{-\left(x^2+y^2\right)} d x d y\).

Solution: 

Here f(x,y) =e^-(x2+y2) and the region is bounded by y=0, y=\(\sqrt{a^2-x^2}\), x=0,x=a.

The region is shown in the figure. Substitute x=r cosθ,y=r sinθ. f(x,y)= e^{-(r2cos2θ+r sin2θ)} = e^-r2.

The curve y= \(\sqrt{a^2-x^2}\) is a circle whose equation is x²+y²=a².

Substituting x= r cosθ , y= r sinθ, this becomes r=a

Imagine a wedge with one end at the pole O and the other end moving on r=a.To cover the region the wedge has to be moved from θ=0 to θ= π/2

Thus \(\int_0^a \int_0^{\sqrt{a^2-x^2}} e^{-\left(x^2+y^2\right)} d x d y=\int_0^{\pi / 2} \int_0^a e^{-r^2} r d r d \theta\)

= \(\int_0^{\pi / 2}\left[-\frac{e^{-r^2}}{2}\right]_0^a d \theta\)

= \(\int_0^{\pi / 2}\left[-\frac{e^{-a^2}}{2}+\frac{1}{2}\right] d \theta\)

= \(\frac{1-e^{-a^2}}{2}[\theta]_0^{\pi / 2}=\frac{\pi}{4}\left(1-e^{-a^2}\right)\)

41. Change into polar coordinates and evaluate \(\int_0^{\infty} \int_0^{\infty} \frac{1}{\left(a^2+x^2+y^2\right)^{3 / 2}} d x d y\).

Solution:

The region of integration is y-=0, y = ∞ , x = 0, x = ∞ , and hence the entire region lies in the first quadrant. Put x = r cos θ, y = r sin θ.

Now \(\frac{1}{\left(a^2+x^2+y^2\right)^{3 / 2}}\)

= \(\frac{1}{\left(a^2+r^2 \cos ^2 \theta+r^2 \sin ^2 \theta\right)^3 / 2}\)\(=\frac{1}{\left(a^2+r^2\right)^{3 / 2}}\)

Imagine a wedge of angular thickness δθ with one end at pole r=0 and the other end on r=∞. To cover the first quadrant this wedge has to be moved from θ =0 to θ =π/2.

Hence \(\int_0^{\infty} \int_0^{\infty} \frac{1}{\left(a^2+x^2+y^2\right)^{3 / 2}} d x d y\)

= \(\int_0^{\pi / 2} \int_0^{\infty} \frac{1}{\left(a^2+r^2\right)^{3 / 2}} r d r d \theta\)

= \(\int_0^{\pi / 2}-\left[\frac{1}{\left(a^2+r^2\right)^{1 / 2}}\right]_0^{\infty} d \theta\)

= \(\int_0^{\pi / 2} \frac{1}{a} d \theta=\frac{1}{a} \cdot \frac{\pi}{2}=\frac{\pi}{2 a}\)

42. Change into polar coordinates and evaluate \(\int_0^a \int_{\sqrt{a x-x^2}}^{\sqrt{a^2-x^2}} \frac{1}{\sqrt{a^2-x^2-y^2}} d x d y\).

Solution:

The region of integration is surrounded by y = \(\sqrt{a x-x^2}, y=\sqrt{a^2-x^2}, x=0, x=a\).

y = \(\sqrt{a x-x^2}\) i.e., \(y^2=a x-x^2\) i.e., \(\left(x-\frac{a}{2}\right)^2+y^2=\frac{a^2}{4}\) represents a circle with center (a/2,0), radius a/2.

y = \(\sqrt{a^2-x^2}\) represents a circle with center (0,0) and radius a.

The region of integration is shown in the figure

Put x = \(r \cos \theta, y=r \sin \theta\) so that \(\frac{1}{\sqrt{a^2-x^2-y^2}}=\frac{1}{\sqrt{a^2-r^2}}, x^2+y^2=a^2\) is r=a \(\left(x-\frac{a}{2}\right)^2+y^2=\frac{a^2}{4}\) is \(r=a \cos \theta\)

Imagine a wedge of angular thickness δθ with one end on r = a cos θ   and another end on r = a. To cover the region this wedge has to be moved from θ = 0 to 0 = π/2. Thus the given integral in polar coordinates is

⇒ \(\int_0^{\pi / 2} \int_{a \cos \theta}^a \frac{1}{\sqrt{\left.a^2-r^2\right)}} r d r d \theta\)

= \(\int_0^{\pi / 2}\left[-\sqrt{a^2-r^2}\right]_{a \cos \theta}^a d \theta\)

= \(\int_0^{\pi / 2} a \sin \theta d \theta\)\(=a[-\cos \theta]_0^{\pi / 2}\)=a

43. Evaluate the integral \(\int_0^{4 a} \int_{y^2 / 4 a}^y \frac{x^2-y^2}{x^2+y^2} d x d y\) by changing to polar coordinates.

Solution:  The region of integration is y= 0,y=4a,x=y/4a, x=y

∴ The regions is bounded by the line y=x and the parabola y= 4ax.

Put x = \(r \cos \theta, y=r \sin \theta\).

Then dx dy = r dr dθ

⇒ \(y^2=4 a x \Rightarrow r^2 \sin ^2 \theta=4 a \cdot r \cos \theta\)

⇒ r = \(\frac{4 a \cos \theta}{\sin ^2 \theta}\)

The limits of integration are r=0 to r = \(\frac{4 a \cos \theta}{\sin ^2 \theta}\) and for the line x=y, slope, \(\tan \theta=1 \Rightarrow \theta=\pi / 4\).

∴ limits for θ are θ = \(\pi / 4\) to \(\theta=\pi / 2\)

Also \(x^2+y^2=r^2, x^2-y^2=r^2\left(\cos ^2 \theta-\sin ^2 \theta\right)\).

∴ \(\int_0^{4 a} \int_{y^2 / 4 a}^y \frac{x^2-y^2}{x^2+y^2} d x d y\)

= \(\int_{\theta=\pi / 4}^{\pi / 2} \int_{r=0}^{4 a \cos \theta / \sin ^2 \theta}\left(\cos ^2 \theta-\sin ^2 \theta\right) r d r d \theta\)

= \(\int_{\pi / 4}^{\pi / 2}\left(\cos ^2 \theta-\sin ^2 \theta\right)\left[\frac{r^2}{2}\right]_0^{4 a \cos \theta / \sin ^2 \theta} d \theta\)

= \(8 a^2 \int_{\pi / 4}^{\pi / 2}\left(\cos ^2 \theta-\sin ^2 \theta\right) \frac{\cos ^2 \theta}{\sin ^4 \theta} d \theta\)

= \(8 a^2 \int_{\pi / 4}^{\pi / 2}\left(\cot ^4 \theta-\cot ^2 \theta\right) d \theta=8 a^2\left[-\frac{\cot ^3 \theta}{3}+2 \cot \theta+2 \theta\right]_{/ 4}^{\pi / 2}\)

= \(8 a^2\left[\pi+\frac{1}{3}-2-\frac{\pi}{2}\right]=\frac{4 a^2}{3}[3 \pi-10]\)

44. By changing into polar coordinates evaluate \(\iint \frac{x^2 y^2}{x^2+y^2} d x d y\) over the angular region between the circles.

Solution: 

Put x=r cos θ,y= r sin θ.

Then dx dy=r dr dθ and x+y=r.

For the circle x+y=a ⇒ r=a

⇒ r=a and x+y=b ⇒  r=b ⇒ r=b.

∴ \(\iint \frac{x^2 y^2}{x^2+y^2}\) dx dy \(=\int_{\theta=0}^{2 \pi} \int_{r=a}^b \frac{r^2 \cos ^2 \theta r^2 \sin ^2 \theta}{r^2} r d r d \theta\)

= \(\int_{\theta=0}^{2 \pi} \int_{r=a}^b \sin ^2 \theta \cos ^2 \theta \cdot r^3 d r d \theta\)

= \(\int_{\theta=0}^{2 \pi}\left[\frac{r^4}{4}\right]_a^b \sin ^2 \theta \cos ^2 \theta d \theta\)

= \(\frac{b^4-a^4}{4} \int_0^{2 \pi} \sin ^2 \theta \cos ^2 \theta d \theta\)

= \(\frac{b^4-a^4}{4} \cdot 2 \int_0^\pi \sin ^2 \theta \cos ^2 \theta d \theta\)

= \(\frac{b^4-a^4}{2} \times 2 \int_0^{\pi / 2} \sin ^2 \theta \cos ^2 \theta d \theta\)

= \(\left(b^4-a^4\right) \cdot \frac{1}{4} \times \frac{1}{2} \times \frac{\pi}{2}=\frac{\left(b^4-a^4\right) \pi}{16}\)

45. By changing the variables evaluate \(\iint_R(x+y)^2 d x d y\) where R is the region bounded by the parallelogram x+y=0,x+y=1,2x-y=0,2x-y=4.

Solution:  Put x+y =u,2x-y=v.Then the given parallelogram reduces to a rectangle.

⇒ \(\frac{\partial(u, v)}{\partial(x, y)}\)

= \(\left|\begin{array}{ll}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{array}\right|\)

= \(\left|\begin{array}{rr}
1 & 1 \\
2 & -1
\end{array}\right|\)=-3 .

∴ Jacobian J = \(\frac{\partial(x, y)}{\partial(u, v)}=-\frac{1}{3}\) .

u.varies from 0 to 1 and v varies from 0 to 4.

∴ \(\iint_R(x+y)^2 d x d y=\int_{u=0}^1 \int_{v=0}^4 u^2\left(\frac{1}{3}\right) d u d v\)

= \(\frac{1}{3} \int_{v=0}^4 \frac{u^3}{3} d v=\frac{1}{9} \int_0^4 d v=\frac{4}{9} \text {. }\)

46. Change the order of integration in the double integral \(\int_0^{2 a} \int_{\sqrt{2 a x-x^2}}^{\sqrt{(2 a x)}} f(x, y) d x d y\).

Solution:  The given domain of integration is described by a line that starts from x=0 and moving parallel to itself, goes over to x=2a; the extremities of the moving line on the parts of the parabola y= 2ax and the circle xy=2ax in the first quadrant.

We now regard the same region as described by a line moving parallel to the x-axis instead of the y-axis. In this way, the domain of integration is subdivided into three sub-regions to each of which corresponds a double integral. Thus we have

 

⇒ \(\int_0^{2 a} \frac{\sqrt{2 a x}}{\sqrt{2 a x-x^2}} f d x d y\)

= \(\int_0^a \int_{y^2 / 2 a}^{a-\sqrt{\left(a^2-y^2\right)}} f d y d x\)

+\(\int_0^{2 a} \int_{a+\sqrt{a^2-y^2}}^{2 a} f d y d x+\int_a^{2 a} \int_{y^2 / 2 a}^{2 a} f d y d x\)

47. Change the order of integration in \(\int_0^{\rho \cos a} \int_{\tan a}^{\sqrt{\left(a^2-x^2\right)}} f(x, y) d x d y\) and verify result when f(x,y)=1.

Solution: The limits of integration are given by y = x tan α (a straight line), y= \(\sqrt{\left(a^2-x^2\right)}\), i.e., x²+y²= a² (a circle); x = 0 (y-axis) and x = a cos α  (a straight line). Clearly, the area of integration is OMNO.

The strips parallel to x- x-axis change their character at point M. Draw a straight line AM parallel to the x-axis and this divides the area OMNO into two portions OMA and AMN.

 

 

For the area OMA, the limits of x are from 0 to y cot α, and the limits of y are from 0 to y sin α. For the area AMN, the limits of x are from 0 to \(\sqrt{\left(a^2-y^2\right)}\), and the limits for are from a sin α  to a.

Hence changing the order of integration, we have \(\int_0^{\cos \alpha} \int_{x \tan \alpha}^{\sqrt{\left(a^2-x^2\right)}} f(x, y) d x d y\)

= \(\int_0^{\sin \alpha} \int_0^{y \cot \alpha} f(x, y) \cdot d x d y+\int_{a \sin \alpha}^\rho \int_0^{\sqrt{\left(a^2-y^2\right)}} f(x, y) d x d y\)

To verify result when f(x, y)=1,

L.H.S. =\(\int_0^{\cos \alpha} \int_{x \tan \alpha}^{\sqrt{\left(a^2-x^2\right)}} d x d y=\int_0^{\cos \alpha}\left\{\sqrt{\left(a^2-x^2\right)}-x \tan \alpha\right\} d x\)

= \(\frac{a}{2} \cos \alpha \cdot a \sin \alpha+\frac{a^2}{2} \text{Sin}^{-1}(\cos \alpha)-\frac{a^2 \cos ^2 \alpha}{2} \tan \alpha\)

= \(\frac{1}{2} a^2 \text{Sin}^{-1}\left\{\sin \left(\frac{\pi}{2}-\alpha\right)\right\}=\frac{1}{2} a^2\left(\frac{\pi}{2}-\alpha\right)\)

R.H.S. = \(\int_0^{\sin \alpha} \int_0^{\cot \alpha} d y d x+\int_{a \sin \alpha}^{\infty} \int_0^{\left.\sqrt{\left(a^2-y^2\right.}\right)} d y d x\)

= \(\int_0^{\sin \alpha} y \cot \alpha d y+\int_{a \sin \alpha}^0 \sqrt{\left(a^2-y^2\right)} d x\)

= \(\frac{1}{2} a^2 \sin \alpha \cot \alpha+\frac{a^2}{2} \cdot \frac{\pi}{2}-\left\{\frac{a}{2} \sin \alpha \cdot a \cos \alpha+\frac{a^2}{2} \cdot \alpha\right\}\)

= \(\frac{a^2}{2}\left(\frac{\pi}{2}-\alpha\right)\)

48. Change the order of integration in the double integral \(\int_0^{\infty} \int_x^{\infty} \frac{e^{-y}}{y} d x d y\) and hence find the value.

Solution: The limits of integration are given by the straight line y = x, y = ∞ x=0, and x= ∞ i.e., the region of integration is bounded by x = 0,y = x, and an ‘infinite boundary. Hence taking the strips parallel to the x-axis, the limits for y are from 0 to ∞.

Hence changing the order of integration, we have \(\int_0^{\infty} \int_x^{\infty} \frac{e^{-y}}{y} d x d y\)

= \(\int_0^{\infty} \int_0^y \frac{e^{-y}}{y} d y d x\) \(=\int_0^{\infty} \frac{e^{-y}}{y}[x]_0^y d y\)

= \(\int_0^{\infty} e^{-y} d y\)=1

49. Change the order of integration in double integral \(\int_0^p \int_0^x \frac{\phi^{\prime}(y) d x d y}{\sqrt{(a-x)(x-y)}}\) and hence find its value.

Solution: The region of integration is ONM. The limit for x are from y to a and limits for y are from 0 to a.

Hence \(\int_0^e \int_0^x \frac{\phi^{\prime}(y) d x d y}{\sqrt{(a-x)(x-y)}}\) \(=\int_0^a \int_y^g \frac{\phi^{\prime}(y) d y d x}{\sqrt{(a-x)(x-y)}}\)

To find the value, let x=a sin²θ +y cos²θ

Also (a-x)=(a-y) cos²θ, x-y=(a-y) sin²θ.

For the limits of  θ, when x=y, we have  y=a sin²θ+y cos²θ =(y-a) sin² θ=0

⇒θ=0 and when x=a,we have a=a sin ²θ+y cos² θ

⇒ (a-y) cos² θ =0 ⇒ θ =π/2

Thus the limits of θ bare from 0 to π/2

 

∴ Given integral = \(\int_0^2 \int_y^{\infty} \frac{\phi^{\prime}(y) d y d x}{\sqrt{(a-x)(x-y)}}\)

= \(\int_0^\pi \int_0^{\pi / 2} \frac{\phi^{\prime}(y) \cdot 2(a-y)(\sin \theta \cos \theta) d y d \theta}{(a-y)(\sin \theta \cos \theta)}\)

= \(2 \int_0^e \int_0^{\pi / 2} \phi^{\prime}(y) d y d \theta=2 \int_0^{\infty} \phi^{\prime}(y) \cdot \frac{\pi}{2} d y\)

= \(\pi[\phi(y)]_0^a=\pi[\phi(a)-\phi(0)]\)

50. Change the order of integration in \(\int_0^{2 a} \int_0^{\sqrt{2a x-x}} \frac{\phi^{\prime}(y)\left(x^2+y^2\right) x d x d y}{\sqrt{4 a^2 x^2-\left(x^2+y^2\right)}}\) and hence evaluate it.

Solution:  Clearly, the area of integration is OMNO.

Solving x²-2ax+y²=0 for x, we get x=a ± \(\sqrt{\left(a^2-y^2\right)}\)

hence limits of x are from a-\(\sqrt{\left(a^2-y^2\right)}\) to a+ \(\sqrt{\left(a^2-y^2\right)}\)

The limits of y are from 0 t0 a.

Hence, given integral becomes \(\int_0^e \int_{a-\sqrt{\left(a^2-y^2\right)}}^{a+\sqrt{\left(a^2-y\right)}} \frac{\phi^{\prime}(y)\left(x^2+y^2\right) x d y d x}{\sqrt{4 a^2 x^2-\left(x^2+y^2\right)^2}}\)

To evaluate the integral, put \(x^2+y^2=t\).

2 x d x=d t.

Then = \(\int_0^2 \int_{a-\sqrt{2}+\sqrt{\left(a^2-y^2\right)}}^{\left.a^2-y^2\right)} \frac{\phi^{\prime}(y)\left(x^2+y^2\right) x d y d x}{\sqrt{4 a^2 x^2-\left(x^2+y^2\right)^2}} \)

= \(\frac{1}{2} \int_0^1 \int_{t_1}^{t^2} \frac{\phi^{\prime}(y) t d y d x}{\sqrt{4 a^2\left(t-y^2\right)-t^2}}\)

= \(-\frac{1}{4} \int_0^2 \int_{t_1}^2 \frac{\phi^{\prime}(y)\left(4 a^2-2 t\right) d y d t}{\sqrt{\left(4 a^2 t-t^2-4 a^2 y^2\right)}}+\int_0^2 \int_{t_1}^2 \frac{\phi^{\prime}(y) \cdot a^2 d y d t}{\sqrt{4 a^2\left(a^2-y^2\right)-\left(2 a^2+t^2\right)}}\)

= \(-\frac{1}{2} \int_0^\rho \phi^{\prime}(y)\left[\sqrt{4 a^2 t-t^2-4 a^2 y^2}+a^2 \text{Sin}^{-1}\left\{\frac{-2 a^2+t}{\sqrt{4 a^2\left(a^2-y^2\right)}}\right\}\right]_{t_1}^{t_2} d y\)

51. Show how the change in the order of integration leads to the evaluation of \(\int_0^{\infty} \frac{\sin r x}{x} d x \text { from } \int_0^{\infty} \int_0^{\infty} e^{-x y} \sin r x d x d y\)

Solution: 

Let \(I=\int_0^{\infty} \int_0^{\infty} e^{-x y} \sin r x d x d y\)

= \(\int_0^{\infty} \sin r\left[\frac{e^{-x y}}{-x}\right] d x=\int_0^{\infty} \frac{\sin r x}{x} d x=x\)…..(1)

Again, \(I=\int_0^{\infty} \int_0^{\infty} e^{-x y} \sin r x d x d y=\int_0^{\infty}\left[\int_0^{\infty} e^{-x y} \sin r x d x\right] d y\)

= \(\int_0^{\infty} \frac{r d y}{\left(r^2+y^2\right)}=\left[\text{Tan}^{-1}\frac{y}{r}\right]_0^{\infty}=\pi / 2\)……(2)

Hence from (1) and (2) we have \(\int_0^{\infty} \frac{\sin r x}{x} d x=\pi / 2\).

52. Change the order of integration in \(\int_0^a \int_0^{\sqrt{a^2-y^2}}\left(x^2+y^2\right) d y d x\).

Solution:  The limits on the inner integrals are functions of y and these are x limits. These limits are obtained by considering horizontal strips. This region of integration is surrounded by x=0, x=\(\sqrt{a^2-y^2}\), y=0, y=a, and this region is shown as a shaded area in the figure.

Now we have to consider vertical strips for changing the order of integration. Imagine a vertical strip with one end on the axis (y=0) and the other end on the curve x²+y²=a² (y=\(\sqrt{a^2-y^2}\)).  To cover the region, this strip has to be moved from x=0 to x=a.

Hence the new equivalent double integral is \(\int_0^a \int_0^{\sqrt{a^2-x^2}}\left(x^2+y^2\right) d y d x\)

= \(\int_0^a\left[x^2 y+\frac{y^3}{3}\right]_0^{\sqrt{a^2-x^2}} d x\)

= \(\int_0^a x^2 \sqrt{a^2-x^2} d x+\int_0^a \frac{\left(a^2-x^2\right)^{3 / 2}}{3} d x\)

⇒ \(\int_0^a x^2 \sqrt{a^2-x^2} d x=\int_0^{\pi / 2} a^2 \sin ^2 \theta \cdot a \cos \theta \cdot a \cos \theta d \theta\)(where x=\(a \sin \theta\))

= \(a^4 \int_0^{\pi / 2} \sin ^2 \theta \cos ^2 \theta d \theta=a^4 \cdot \frac{1}{4} \cdot \frac{1}{2} \frac{\pi}{2}=\frac{\pi a^4}{16}\)

⇒ \(\int_0^a \frac{\left(a^2-x^2\right)^{3 / 2}}{3} d x\)

= \(\int_0^{\pi / 2} \frac{a^3 \cos ^3 \theta a \cos \theta}{3} d \theta=\frac{a^{4^{\pi / 2}}}{3} \int_0 \cos ^4 \theta d \theta\)

= \(\frac{a^4}{3} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}=\frac{\pi a^4}{16}\)

Hence \(\int_0^a \int_0^{\sqrt{a^2-x^2}}\left(x^2+y^2\right) d y d x=\frac{\pi a^4}{16}+\frac{\pi a^4}{16}=\frac{\pi a^4}{8}\).

53. Change the order of integration and evaluate \(\int_0^4 a \int_{x^2 / 4 a}^{2 \sqrt{a x}} d x d y\).

Solution: The limits on the inner integral are functions of x= and these are y limits.

These limits are obtained by considering vertical strips. The region is bounded by  y\(=\frac{x^2}{4 a}\),y=\(\sqrt{a x}\),x=0,x=4a and the

To change the order of integration we consider a horizontal strip with one end on the y²=4ax i..e. x=\(=\frac{y^2}{4 a}\) and the other end on x²=4ay  i.e, x=2\(\sqrt{a y}\)

This strip has to be slid be from y=0 to y= 4a to cover the region.

Hence the double integral with new limits is \(\int_0^{4 a} \int_{y^2 / 4 a}^{2 \sqrt{a y}} d y d x=\int_0^{4 a}\left[\int_{y^2 / 4 a}^{2 \sqrt{a y}} d x\right] d y\)

= \(\int_0^{4 a}[x]_{y^2 / 4 a}^{2 \sqrt{a y}} d y=\int_0^{4 a} 2 \sqrt{a y}-\frac{y^2}{4 a} d y\)

= \(2 \sqrt{a} \cdot\left[\frac{y^{3 / 2}}{3 / 2}\right]_0^{4 a}-\frac{1}{4 a}\left[\frac{y^3}{3}\right]_0^{4 a}\)

= \(\frac{4 \sqrt{a}}{3}\left[2^3 a^{3 / 2}\right]-\frac{1}{4 a}\left[\frac{64 a^3}{3}\right]\)

= \(\frac{32 a^2}{3}-\frac{16 a^2}{3}=\frac{16 a^2}{3}\).

54. Change the order of integration and evaluate \(\int_0^a \int_2^{2 a-x} x y d y d x\).

Solution: The limits on the inner integral are functions of x and these are limits of y. These are obtained by considering vertical strips. The region is bounded by

y\(=\frac{x^2}{ a}\) , y=2a-x, x=0, x=a and this is

To change the order of integration we consider horizontal strips. Since two different curves surround the region let us divide the region into two regions R₁, and R_2, and in each region, we consider a horizontal strip.

Integral over \(R_1=\int_0^a \int_0^{\sqrt{a y}} x y d x d y\).

Integral over \(R_2=\int_a^{2 a} \int_0^{2 a-y} x y d x d y\)

Thus given the integral of changing the order of integration, it becomes \(\int_0^a \int_0^{\sqrt{a y}} x y d x d y+\int_a^{2 a} \int_0^{2 a-y} x y d x d y\)

= \(\int_0^a\left[\frac{x^2 y}{2}\right]_0^{\sqrt{a y}} d y+\int_a^{2 a}\left[\frac{x^2 y}{2}\right]_0^{2 a-y} d y\)

= \(\int_0^a \frac{a y^2}{2} d y+\int_a^{2 a} \frac{y(2 a-y)^2}{2} d y\)

= \(\frac{a}{2} \int_0^a y^2 d y+\int_a^{2 a} \frac{y\left(4 a^2+y^2-4 a y\right)}{2} d y\)

= \(\frac{a}{2}\left[\frac{y^3}{3}\right]_0^a+\frac{1}{2}\left[\frac{4 a^2 y^2}{2}+\frac{y^4}{4}-\frac{4 a y^3}{3}\right]_a^{2 a}\)

= \(\frac{a^4}{6}+\frac{1}{2}\left[\frac{4 a^2 4 a^2}{2}+\frac{4 a^2 \cdot 4 a^2}{4}-\frac{32 a^4}{3}-\frac{4 a^4}{2}-\frac{a^4}{4}+\frac{4 a^4}{3}\right]\)

= \(\frac{a^4}{6}+\frac{5 a^4}{24}=\frac{9 a^4}{24}=\frac{3}{8} a^4\).

55. Change the order of integration and evaluate \(\int_0^{2 a} \int_0^{\sqrt{a^2-(x-a)^2}} d x d y\)

Solution:  The region of integrals is surrounded by

y=0,y=\(\sqrt{a^2-(x-a)^2}\) , x= 0 and x=2a.

y=\(\sqrt{a^2-(x-a)^2}\) ⇒y²=a²-(x-a)²

⇒ (x-a)²+ y²=a².

This represents a circle with center (a,0) and radius a.

The region is in the given integral the limits are obtained by considering the vertical strip.

Now imagine a horizontal strip where both ends slide on two parts of the circle in the first quadrant.

(x-a)²+y²=a² ⇒ (x-a)²= a²-y² ⇒ x-a =± \(\sqrt{a^2-y^2}\) ⇒ x=a±\(\sqrt{a^2-y^2}\)

Thus the end L of the horizontal strip moves on a-\(\sqrt{a^2-y^2}\) whereas the end M of the horizontal strip moves on a + \(\sqrt{a^2-y^2}\).

To cover the region this strip slides from y = 0 to y = a.

Thus the given double integral with the change of order of integration becomes

⇒ \(\int_0^a \int_{a-\sqrt{a^2-y^2}}^{a+\sqrt{a^2-y^2}} d x d y=\int_0^a[x] \frac{a-\sqrt{a^2-y^2-y^2}}{[x} d y\)

= \(\int_0^a a+\sqrt{a^2-y^2}-a+\sqrt{a^2-y^2} d y\)

= \(2 \int_0^a \sqrt{a^2-y^2} d y=2 \cdot\left[\frac{y \sqrt{a^2-y^2}}{2}+\frac{a^2}{2} \text{Sin}^{-1} \frac{y}{a}\right]_0^a\)

= \( 2\left[\frac{a^2}{2} \cdot \frac{\pi}{2}\right]=\frac{\pi a^2}{2}\) .

56. Evaluate the following integral by changing the order \(\int_0^3 \int_1^{\sqrt{4-y}}(x+y) d x d y\).

Solution:  I=\(\int_0^3 \int_1^{\sqrt{4-y}}(x+y) d x d y\)

The integration first w.r.t. ‘x’ then w.r.t. ‘x’  The integration w.r.t. ‘x’ corresponds to along one edge of the horizontal strip and the integration w.r.t. ‘y’ corresponds to sliding the strip from y = 0 to y = 3. Horizontal strip ends lie on x = 1, x =\(\sqrt{4-y}\) i.e., y = 4- x². The region of integration is ABC.

Change of order of integration: On change of order of integration, we integrate first w.r.t. y from y=0 to y=4-x² then w.r.t x from x=1 to x=2

I = \(\int_1^2 \int_0^{4-x^2}(x+y) d y d x=\int_1^2\left[x y+\frac{y^2}{2}\right]_0^{4-x^2} d x\)

= \(\int_1^1\left[x\left(4-\dot{x}^2\right)+\frac{\left(4-x^2\right)^2}{2}\right] d x\)

= \(\int_1^2\left(4 x-x^3+\frac{x^4+16-8 x^2}{2}\right) d x\)

= \(\left[2 x^2-\frac{x^4}{4}+\frac{x^5}{10}+8 x-\frac{4 x^3}{3}\right]_1^2\)

= \(\left(8-4+\frac{16}{5}+16-\frac{32}{3}\right)-\left(\frac{1}{10}+2-\frac{1}{4}+8-\frac{4}{3}\right)=\frac{241}{60}\)

57. By changing the order of integration, evaluate \(\int_0^a \int_y^a \frac{x}{x^2+y^2} d x d y\)

Solution: The limits on the inner integral are functions of y and these are limits of x. These are obtained by considering horizontal strips. The region is a triangle bounded by y = 0, x = a, and y=x.

To change the order of integration, we consider the vertical strip.

So the limits of integration are y = 0 to y=x and x = 0 to x = a. On changing the order of integration, given integral becomes

⇒ \(\int_0^a \int_0^a \frac{x}{x^2+y^2} d y d x\)

= \(\int_0^a\left[\frac{1}{x} \text{Tan}^{-1} \frac{y}{x}\right]_0^x \cdot x d x\)

= \(\int_0^a \frac{1}{2} \text{Tan}^{-1}(1) \cdot x d x=\int_0^a \frac{\pi}{4} d x=\frac{\pi a}{4}\) .

58. Change the order of integration and evaluate the double integral \(\int_0^1 \int_1^{e^x} d y d x\).

Solution: The limits on the inner integral are functions of x and these are limits of y. These are obtained by considering the vertical strip. The region of integration is bounded by y=1, x=1 y =e^x To change the order of integration we consider the horizontal strip. So the limits of integration are x = 1 to x = log y and y = 1 to y = e.

 

By changing the order of integration the given integral becomes \(\int_1^e \int_{\log y}^1 d x d y=\int_1^e[x]_{\log y}^1 d y=\int_1^e[1-\log y] d y=[y-y \log y+y]_1^e=e-2\).

 

 

 

 

Multiple Integrals 2 Exercise 2 Solved Problems

 

1. Evaluate \(\begin{equation}\int_0^1 \int_0^2 \int_1^2 x^2 y z d x d y d z\end{equation}\).

Solution: \(\int_0^1 \int_0^2 \int_1^2 x^2 y z d x d y d z\)

= \(\int_0^1\left[\int_0^2\left\{\int_0^2 x^2 y z d z\right\} d y\right] d x=\int_0^1\left\{\int\left[x^2 y \frac{z^2}{2}\right] d y\right\} d x\)

= \(\int_0^1\left[\int_0^2 x^2 y\left(2-\frac{1}{2}\right) d y\right] d x\)

= \(\int_0^1 \frac{3}{2} x^2\left[\frac{y^2}{2}\right] d x=\int_0^1 \frac{3 x^2}{2}(2) d x=\int_0^1 3 x^2 d x=\left[x^3\right]_0^1=1\)

2. Evaluate\(\int_0^{2 a} \int_0^x \int_y^x x y z\) dx dy dz.

Solution: \(\int_0^{2 a} \int_0^x \int_y^x x y z d x d y d z\)

= \(\int_0^{2 a}\left[\int_0^x\left[\int_y^x x y z d z\right] d y\right] d x=\int_0^{2 a x} \int x y\left[\frac{z^2}{2}\right]_0^x d y d x\)

= \(\int_0^{2 a} \int_0^x \frac{x y}{2}\left(x^2-y^2\right) d y d x\)

= \(\int_0^{2 a}\left[\frac{x^3 y^2}{4}-\frac{x y^4}{8}\right]_0^x d x=\int_0^{2 a}\left[\frac{x^5}{4}-\frac{x^5}{8}\right] d x\)

= \(\int_0^{2 a} \frac{x^5}{8} d x=\left[\frac{x^6}{48}\right]_0^{2 a}=\frac{64 a^6}{48}=\frac{4 a^6}{3}\)

3. Evaluate \(\int_0^1 \int_y^1 \int_0^{1-x} x\)dz dx dy.

Solution: \(\int_0^1 \int_y^1 \int_0^{1-x} x d z d x d y\)

= \(\int_{z=0}^{z=1}\left\{\int_{x=y}^{x=1}\left[\int_{y=0}^{y=1-x} x d y\right] d x\right\} d z\)

= \(\int_{z=0}^{z=1}\left\{\int_{x=y}^{x=1}[x y]{ }_{y=0}^{y=1-x} d x\right\} d z\)

= \(\int_{z=0}^{z=1}\left\{\int_{x=y}^{x=1} x(1-x) d x\right\} d z=\int_{z=0}^{z=1}\left[\int_{x=y}^{x=1}\left(x-x^2\right) d x\right] d z\)

= \(\int_{z=0}^{x=1}\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{x=y}^1 d z\)

= \(\int_{z=0}^{z=1}\left(\frac{1}{2}-\frac{1}{3}-\frac{y^2}{2}+\frac{y^3}{3}\right) d z=\int_0^1\left(\frac{1}{6}-\frac{y^2}{2}+\frac{y^3}{3}\right) d y\)

= \(\left[\frac{y}{6}-\frac{y^3}{6}+\frac{y^4}{12}\right]=\frac{1}{6}-\frac{1}{6}+\frac{1}{12}=\frac{1}{12}\)

4. Evaluate\(\int_{-1}^1 \int_0^z \int_{x-z}^{x+z}(x+y+z)\)dz dx dy.

Solution: \(\int_{-1}^1 \int_0^z \int_{x-z}^{x+z}(x+y+z) d z d x d y\)

= \(\int_{z=-1}^{z=1}\left[\int_{x=0}^{x=z}\left\{\int_{y=x-z}^{y=x+z}(x+y+z) d y\right\} d x\right] d z\)

= \(\int_{z=-1}^{z=1}\left[\int_{x=0}^{x=z}\left(x y+\frac{y^2}{2}+z y\right)_{y=x-z}^{y=x+z} d x\right] d z\)

= \(\int_{z=-1}^{z=1}\left[\int_{x=0}^{x=z}\left\{x(x+z)+\frac{(x+z)^2}{2}+z(x+z)-x(x-z)-\frac{(x-z)^2}{2}-z(x-z)\right\} d x\right] d z\)

= \(\int_{z=-1}^{z=1}\left[\int_{x=0}^{x=z}\left(2 x z+2 x z+2 z^2\right) d x\right] dz=\int_{z=-1}^{z=1}[x^2 z+x^2 z+2 x z_{x=0}^{x=z} d x\)

= \(\int_{z=-1}^{z=1}\left(z^2+z^3+2 z^3\right) d z\)

= \(\int_{-1}^1 4 z^3 d z=0\)

5. Evaluate \(\int_1^e \int_1^{\log y} \int_1^{e^x} \log z\) dy dx dz.

Solution: \(\int_1^e \int_1^{\log y} \int_1^{e^x} \log z d y d x d z\)

= \(\int_1^e\left\{\int_1^{\log y}\left[\int_1^{e^x} \log z d z\right] d x\right\} d y\)

= \(\int_1^e\left\{\int_1^{\log y}[z \log z-z]_1^{e^x} d x\right\} d y\)

= \(\int_1^e\left[\int_1^{\log y}\left(x e^x-e^x+1\right) d x\right] d y=\int_1^e\left[x e^x-e^x-e^x+x\right] d y\)

= \(\int_1^e(y \log y+\log y-2 y+e-1) d y=\left[\left(\frac{y^2}{2}+y\right) \log y-\left(\frac{y^2}{4}+y\right)-y^2+(e-1) y\right]\)

= \(\frac{e^2}{4}-2 e+\frac{13}{4}=\frac{1}{4}\left(e^2-8 e+13\right)\).

6. Evaluate\(\int_0^a \int_0^x \int_0^{x+y} e^{x+y+z}\) dx dy dz.

Solution: \(\int_0^a \int_0^x \int_0^{x+y} e^{x+y+z} d x d y d z\)

= \(\int_0^a\left[\int_0^x\left[\int_a^{x+y+z} e^{x+y+z} d z\right\} d y\right] d x\)

= \(\int_0^a\left[\int_0^x\left(e^{x+y+z}\right)_0^{x+y} d y\right] d x\)

= \(\int_0^a\left[\int_0^x\left(e^{2 x+2 y}-e^{x+y}\right) d y\right] d x\)

= \(\int_0^a\left[\frac{e^{2 x+2 y}}{2}-e^{x+y}\right]_0^x d x=\int_0^{4 x}\left(\frac{e^{4 x}}{2}-e^{2 x}-\frac{e^{2 x}}{2}+e^x\right) d x\)

= \(\left[\frac{e^{4 x}}{8}-\frac{e^{2 x}}{2}-\frac{e^{2 x}}{4}+e^x\right]_0^a\)

= \(\frac{e^{4 a}}{8}-\frac{e^{2 a}}{2}-\frac{e^{2 a}}{4}+e^a=\frac{e^{4 a}}{8}-\frac{3}{4} e^{2 a}+e^a\) .

7. Evaluate \(\int_0^{\log 2} \int_0^x \int_0^{x+\log y} e^{x+y+z}\)dx dy dz

Solution: \(\int_0^{\log 2} \int_0^x \int_0^{x+\log y} e^{x+y+z} d x d y d z\)

= \(\int_0^{\log 2}\left(\int_0^x\left[\int_0^{x+\log y} e^{x+y+z} d z\right] d y\right) d x\)

= \(\int_0^{\log 2}\left(\int_0^x\left[e^{x+y+z}\right] d y\right) d x\)

= \(\int_0^{\log 2}\left(\int_0^x\left(e^{2 x+y+\log y}-e^{x+y}\right) d y\right) d x\)

= \(\int_0^{\log 2}\left(\int_0^x\left[y e^{2 x+y}-e^{x+y}\right] d y\right) d x\)

= \(\int_0^{\log 2}\left[y e^{2 x+y}-e^{2 x+y}-e^{x+y}\right]_{y=0}^{y=x} d x\)

= \(\int_0^{\log 2}\left(x e^{3 x}-e^{3 x}-e^{2 x}+e^{2 x}+e^x\right) d x\)

= \(\int_0^{\log 2}\left[(x-1) e^{3 x}+e^x\right] d x=\left[(x-1) \frac{e^{3 x}}{3}-\frac{e^{3 x}}{9}+e^x\right]_0^{\log 2}\)

= \(\frac{8}{3}(\log 2-1)-\frac{8}{9}+2+\frac{1}{3}+\frac{1}{9}-1=\frac{8}{3}(\log 2-1)+\frac{5}{9}=\frac{8}{3} \log 2-\frac{19}{9}\)

8. Evaluate \(\int_0^1 \int_0^{1-x} \int_0^{1-x-y} \frac{d x d y d z}{(x+y+z+1)^3}\)

Solution:

I = \(\int_0^1 \int_0^{1-x 1} \int_0^{1-x-y} \frac{d x d y d z}{(x+y+z+1)^3}\)

= \(\int_0^1 \int_0^{1-x}\left[-\frac{1}{2} \frac{1}{(x+y+z+1)^2}\right]_0^{1-x-y} d x d y\)

= \(-\frac{1}{2} \int_0^1 \int_0^{1-x}\left[\frac{1}{4}-\frac{1}{(x+y+1)^2}\right] d x d y\)

= \(-\frac{1}{2} \int_0^1\left[\frac{1}{4} y+\frac{1}{x+y+1}\right]_0^{1-x} d x\)

= \(-\frac{1}{2} \int_0^1\left[\frac{1}{4}(1-x)+\frac{1}{2}-\frac{1}{x+1}\right] d x\)

= \(\frac{1}{2}\left[\frac{3}{4} x-\frac{1}{8} x^2-\log (x+1)\right]=\frac{1}{2}\left(\log 2-\frac{5}{8}\right)\)

9. Evaluate \(\int_0^a \int_0^{\sqrt{a}-x^2} \int_0^{\sqrt{a^2-x}-y} \frac{d x d y d z}{\sqrt{a^2-x^2-y^2-z^2}}\)

Solution: \(\int_0^a \int_0^{\sqrt{a^2-x^2}} \int_0^{\sqrt{a}^2-x^2-y^2} \frac{d x d y d z}{\sqrt{a^2-x^2-y^2-z^2}}\)

= \(\int_0^a\left\{\int_0^{\sqrt{a^2-x^2}}\left[\int_0^{\sqrt{a^2-x^2-y^2}} \frac{1}{\sqrt{a^2-x^2-y^2-z^2}} d z\right] d y\right\} d x\)

= \(\int_0^a\left\{\int_0^{\sqrt{a^2-x^2}}\left[\text{Sin}^{-1} \frac{z}{\sqrt{a^2-x^2-y^2}}\right] d y\right\} d x\)

= \(\int_0^a\left\{\int_0^{\sqrt{a^2-x^2}}\left[\text{Sin}^{-1} 1-\text{Sin}^{-1} 0\right] d y\right\} d x\)

= \(\int_0^a\left\{\int_0^{\sqrt{a^2-x^2}} \frac{\pi}{2} d y\right\} d x\)

= \(\int_0^a\left[\frac{\pi}{2} y\right]_0^{\sqrt{a^2-x^2}} d x=\frac{\pi}{2} \int_0^a \sqrt{a^2-x^2} d x\)

= \(\frac{\pi}{2}\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \text{Sin}^{-1} \frac{x}{a}\right]_0^a\)

= \(\frac{\pi}{2} \frac{a^2}{2} \frac{\pi}{2}=\frac{\pi^2 a^2}{8}\)

10. Evaluate \(\int_0^4 \int_0^{2 \sqrt{2}} \int_0^{\sqrt{4 z-x^2}} d z d x d y\) dx dy dz

Solution: \(\int_0^4 \int_0^{2 \sqrt{z}} \int_0^{\sqrt{4 z-x^2}} d z d x d y\)

= \(\int_0^4\left[\int_0^{2 \sqrt{z}}\left\{\int_0^{\sqrt{4 z-x^2}} d y\right\} d x\right] d z\)

= \(\int_0^4\left[\int_0^{2 \sqrt{z}}[y]_0^{\sqrt{4 z-x^2}} d x\right] d z=\int_0^4\left[\int_0^{2 \sqrt{z}} \sqrt{4 z-x^2} d x\right] d z\)

= \(\int_0^4\left[\frac{x}{2} \sqrt{4 z-x^2}+2 z \text{Sin}^{-1} \frac{x^{2 \sqrt{2}}}{2 \sqrt{z}}\right]_{x=0}^4 d z\)

= \(\int_0^4\left(2 z \frac{\pi}{2}\right) d z=\frac{\pi}{2}\left[z^2\right]=8 \pi\).

11. Evaluate \(\iint_D∫\) dx dy dz over the region D taken through the positive octant of the sphere x² +y² + z² = a².

Solution: To cover the region of positive octant of the sphere x² + y² + z² = a², z varies from \(\sqrt{a^2-x^2-y^2}\)

y varies from 0 to\(\) and x varies from 0 to a.

 

Hence the required integral is \(\int_0^a \int_0^{\sqrt{a^3-x^2}} \cdot \int_0^{\sqrt{a^2-x^2-y^2}} x y z\)  dz dy dx

= \(\int_0^a \int_0^{\sqrt{a^2-x^2}}\left[\frac{x y z^2}{2}\right]_0^{\sqrt{a^2-x^2-y^2}} d y d x\)

= \(\int_0^a \int_0^{\sqrt{a^2-x^2}} \frac{x y\left(a^2-x^2-y^2\right)}{2} d y d x\)

= \(\frac{1}{2} \int_0^a \int_0^{\sqrt{a^2-x^2}}\left(x y a^2-x^3 y-x y^3\right) d y d x\)

= \(\frac{1}{2} \int_0^a\left[\frac{x y^2 a^2}{2}-\frac{x^3 y^2}{2}-\frac{x y^4}{4}\right]_0^{\sqrt{a^2-x^2}} d x\)

= \(\frac{1}{2} \int_0^a\left\{\frac{x\left(a^2-x^2\right) a^2-x^3\left(a^2-x^2\right)}{2}-\frac{x\left(a^2-x^2\right)^2}{4}\right\} d x\)

= \(\frac{a^6}{48}\)

= \(\int_0^a \int_0^{\sqrt{a^2-x^2}}\left[\frac{x y z^2}{2}\right]_0^{\sqrt{a^2-x^2-y^2}} d y d x\)

= \(\int_0^a \int_0^{\sqrt{a^2-x^2}} \frac{x y\left(a^2-x^2-y^2\right)}{2} d y d x\)

= \(\frac{1}{2} \int_0^a \int_0^{\sqrt{a^2-x^2}}\left(x y a^2-x^3 y-x y^3\right) d y d x\)

= \(\frac{1}{2} \int_0^a\left[\frac{x y^2 a^2}{2}-\frac{x^3 y^2}{2}-\frac{x y^4}{4}\right]_0^{\sqrt{a^2-x^2}} d x\)

= \(\frac{1}{2} \int_0^a\left\{\frac{x\left(a^2-x^2\right) a^2-x^3\left(a^2-x^2\right)}{2}-\frac{x\left(a^2-x^2\right)^2}{4}\right\} d x\)=\(\frac{a^6}{48}\)

12. Evaluate \(\iiint_V\left(x^2+y^2+z^2\right)\) dx dy dz where V is the volume of the cube bounded by the coordinate planes and the planes x=y = z = a.

Solution: Hence a column parallel to the z-axis is bounded by the planes z=0 and z=a.

Here the region S above which the volume V stands is the region in the xy-plane bounded by the lines x=0,x=a,y=0,y=a.

Hence the given integral

= \(\int_0^a \int_0^a \int_0^a\left(x^2+y^2+z^2\right) d x d y d z\)

= \(\int_0^a \int_0^a\left[x^2 z+y^2 z+\frac{z^3}{3}\right] d x d y\)

= \(\int_0^a \int_0^a\left(x^2 a+y^2 a+\frac{1}{3} a^3\right) d x d y=\int_0^a\left[x^2 a y+\frac{1}{3} y^3 a+\frac{1}{3} a^3 y\right] d x\)

= \(\int_0^a\left(x^2 a^2+\frac{1}{3} a^4+\frac{1}{3} a^4\right) d x=\left[\frac{1}{3} x^3 a^2+\frac{1}{3} a^4 x+\frac{1}{3} a^4 x\right]=a^5\)

 

13. Evaluate \(\iiint_V(2 x+y)\)dxdy dz, where V is the closed region bounded by the cylinder z = 4- x² and the planes  x= 0, y = 0, y = 2 and z = 0.

Solution: Here a column parallel to the z-axis is bounded by the plane z=0 and the surface z=4-x of the cylinder.

This cylinder z=4-x meets the z-axis,x=0,y=0, at(0,0,4) and the x-axis , y=0, z=0 at (2,0,0) in the given region.

 

 

Therefore, it is evident that the limits of integration for z are from 0 to 4-x, for from 0 to 2 and for x from 0 to 2.

Hence the given integral

= \(\int_{x=0}^2 \int_{y=0}^2 \int_{z=0}^{4-x^2}(2 x+y) d x d y d z=\int_{x=0}^2 \int_{y=0}^2(2 x+y)[z]_0^{4-x^2} d x d y\)

= \(\int_{x=0}^2 \int_{y=0}^2(2 x+y)\left(4-x^2\right) d x d y=\int_{x=0}^2 \int_{y=0}^2\left[8 x-2 x^3+\left(4-x^2\right) y\right] d x d y\)

14. Use the substitution x+y + z = u,y + z = uv, z = uvw to evaluate the integral \(\iiint[x y z(1-x-y-z)]^{1 / 2}\)dx dy dz taken over the tetrahedral volume enclosed by the planes x=0, y = 0, z = 0 and x+y + z= 1.

Solution:

Here x = \(u(1-v), y=u v(1-u v), z=u v w\).

∴ \(\frac{\partial(x, y, z)}{\partial(u, v, w)}\)

= \(\left|\begin{array}{ccc}
1-v & -u & 0 \\
v(1-w) & u(1-w) & -u v \\
v w & u w & u v
\end{array}\right|=u^2 v\)

The tetrahedral volume is covered by taking the limits for u from 0 to 1, v from 0 to and w from 0 to 1.

Required integral = \(\int_0^1 \int_0^1 \int_0^1\left[u^3 v^{2 w}(1-u)(1-v)(1-w)\right]^{1 / 2} u^2 v d u d v d w\)

= \(\int_0^1 u^{7 / 2}(1-u)^{1 / 2} d u \int_0^1 v^2(1-v)^{1 / 2} d v \int_0^1 w^{1 / 2}(1-w)^{1 / 2} d w \text {. }\)

Putting u = \(\sin ^2 \theta, \nu=\sin ^2 \phi, w=\sin ^2 t\) the integral becomes

∴ \(\int_0^{\pi / 2} 2 \sin ^2 \theta \cos ^2 \theta d \theta \int_0^{\pi / 2} 2 \sin ^5 \phi \cos \phi d \phi \int_0^{\pi / 2} 2 \sin ^2 t \cos ^2 t d t=\frac{\pi^2}{1920}\)

15. Evaluate ∫∫∫ xyz dx dy dz over the positive octant of the sphere x² + y² + z² = a² by transforming it into spherical coordinates.

Solution:  x= r sinθ cos Φ, y= r sin θ sin Φ, z= r cosθ.

dx dy dz \(=\frac{\partial(x, y, z)}{\partial(r, \theta, \phi)}\) dr dθ dΦ=-r² sin dr dθ dΦ

To cover the positive octant of the sphere r varies from 0 to 0 varies from 0 to π/2 and θ varies from π /2  to 0.

Required integral = \(\int_{r=0}^a \int_{\phi=0}^{\pi / 2} \int_{\theta=\pi / 2}^0 r^3 \sin ^2 \theta \cos \theta \sin \phi \cos \phi\left(-r^2 \sin ^2 \theta\right) d r d \theta d \phi\)

= \(\int_{r=0}^a \int_{\theta=0 \phi=0}^{\pi / 2} \int^{\pi / 2} r^5 \sin \phi \cos \phi \sin ^3 \theta \cos \theta d r d \theta d \phi\)

= \(\int_0^a r^5 d r \int_0^{\pi / 2} \sin \phi \cos \phi d \phi \int_0^{\pi / 2} \sin ^2 \theta \cos \theta d \theta=\frac{a^6}{48}\)

16. Evaluate ∫∫∫ (x²+y² + z² )dx dy dz taken over the volume enclosed by the sphere x²+y² + z²= 1.

Solution: \(\iint_R\)∫(x²+y² + z² )dx dy dz

R is the region bounded by the sphere x²+y² + z² +  = 1

Change to spherical polar co-ordinates, we have

x = r sin θ  cos Φ , y=sin θ  sin Φ, z = r cos θ  , dx dy dz = r² sinθ  dr dθ  dΦ

I = \(\iiint_{R^{\prime}} r^2 \cdot r^2 \sin \theta d r d \theta d \phi\)

Over the region \(R^1: r\) varies from 0 to 1, θ varies from 0 to \(\pi, \phi\) varies from 0 to \(2 \pi\)

I = \(\int_{\phi=0}^{2 \pi} \int_{\theta=0}^\pi \int_0^1 r^4 \sin \theta d r d \theta d \phi\)

= \(\int_{\phi=0}^{2 \pi} \int_{\theta=0}^\pi \frac{r^5}{5} \sin \theta d \theta d \phi\)

= \(\frac{1}{5} \int_{\phi=0}^{2 \pi} \int_{\theta=0}^\pi \sin \theta d \theta d \phi\)

= \(\frac{1}{5} \int_{\phi=0}^{2 \pi}(-\cos \theta)_0^\pi d \phi=\int_{\phi=0}^{2 \pi} \frac{2}{5} d \phi\)

= \(\left(\frac{2}{5} \phi\right)_0^{2 \pi}=\frac{4 \pi}{5}\)

17. Evaluate \(\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} \frac{d z d y d x}{\sqrt{1-x^2-y^2-z^2}}\) dz dy dx coordinates. 0 by changing to a spherical polar

Solution:
Here the regeion of integration is bounded by z=0, z=\(\sqrt{1-x^2-y^2}\): y=0, y=\(\sqrt{1-x^2}\); x=0, x=1 which is the sphere x+y+z=1 is the positive octant.

To change spherical polar coordinates,

Put x=r sin θ cos Φ, y= r sin θ sin Φ, z= r cos θ  so that x²+y²+z²=r²

∴ f(x,y,z) \(=\frac{1}{\sqrt{1-x^2-y^2-z^2}}\)=\(\frac{1}{\sqrt{1-r^2}}\)

Note that r varies from 0 to 1,θ varies from 0 t0 π/2 and Φ varies from 0 to π/2.

Hence the given triple integral is equivalent to \(\int_0^{\pi / 2} \int_0^{\pi / 2} \int_0^1 r^2 \sin \theta \cdot \frac{1}{\sqrt{1-r^2}} d r d \theta d \phi\)

= \(\int_0^{\pi / 2} \int_0^{\pi / 2} \int_0^1(\frac{1}{\sqrt{1-r^2}}-\sqrt{1-r^2})) \sin \theta d r d \theta d \phi\)

= \(\int_0^{\pi / 2} \int_0^{\pi / 2} \sin \theta\left[\text{Sin}^{-1} r-\left(\frac{r \sqrt{1-r^2}}{2}+\frac{1}{2} \text{Sin}^{-1} r\right)\right]_0^1 d \theta d \phi\)

= \(\int_0^{\pi / 2} \int_0^{\pi / 2} \sin \theta \cdot \frac{\pi}{4} d \theta d \phi=\frac{\pi}{4} \int_0^{\pi / 2}[-\cos \theta]_0^{\pi / 2} d \phi\)

= \(\frac{\pi}{4} \int_0^{\pi / 2} d \phi=\frac{\pi}{4} \cdot \frac{\pi}{2}=\frac{\pi^2}{8}\)

18. Evaluate ∫∫∫ z² dx dy dz taken over the volume bounded by the surfaces x² +y² = a², x² +y² = z and z = 0.

Solution: The limits of z=0 are z= x+y   the limits of y and the limits of x are x=-a to x=a.

∴ I = \(\iiint z^2 d x d y d z=\int_{x=-a}^{x=a}\left[\int_{y=-\sqrt{a^2-x^2}}^{y=\sqrt{a^2-x^2}}\left\{\int_{z=0}^{z=x^2+y^2} z^2 d z\right\} d y\right] d x\)

= \(\int_{x=-a}^{x=a}\left\{\int_{y=-\sqrt{a^2-x^2}}^{y=\sqrt{a^2-x^2}}\left[\frac{z^3}{3}\right]_{z=0}^{z=x^2+y^2} d y\right\} d x\)

= \(\int_{x=-a}^{x=a}\left[\int_{y=-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \frac{\left(x^2+y^2\right)^3}{3}\right] d x\)

= \(\int_{x=-a}^{x=a} \iint_{y=-\sqrt{a^2-x^2}}^{y=x^2} \frac{1}{3}\left(x^6+3 x^4 y^2+3 x^2 y^4+y^6\right) d y d x\)

= \(\frac{2}{3} \int_{x=-a}^{x=a}\left[x^6 y+3 x^4 \frac{y^3}{3}+3 x^2 \frac{y^5}{5}+\frac{y^7}{7}\right] d x\)

= \(\frac{4}{3} \int_0^a\left[x^6+3 x^4 \frac{\left(a^2-x^2\right)}{3}+\frac{3 x^2\left(a^2-x^2\right)}{5}+\frac{\left(a^2-x^2\right)^3}{7}\right] \sqrt{a^2-x^2} d x\)

Put \(x=a \sin \theta\).

Then \(d x=a \cos \theta d \theta. x=0, a \Rightarrow \theta=0, \pi / 2\).

∴ I = \(\frac{4}{3} \int_0^{\pi / 2}\left[a^6 \sin ^6 \theta+a^6 \sin ^4 \theta \cos ^2 \theta+\frac{3}{5} a^6 \sin ^2 \theta \cos ^4 \theta+\frac{a^6}{7} \cos ^6 \theta\right] a \cos \theta a \cos \theta d \theta\)

= \(\frac{4 a^8}{3} \int_{\theta=0}^{\theta=\pi / 2}\left(\sin ^6 \theta \cos ^2 \theta+\sin ^4 \theta \cos ^4 \theta+\frac{3}{5} \sin ^2 \theta \cos ^6 \theta+\frac{1}{7} \cos ^8 \theta\right) d \theta\)

= \(\frac{4 a^8}{3}[\frac{1}{8} \times \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}+\frac{3}{8} \times \frac{1}{6} \times \frac{3}{4} \times \frac{1}{2}\)

x \(\frac{\pi}{2}+\frac{3}{5} \times \frac{5}{8} \times \frac{3}{6} \times \frac{1}{4} \times \frac{1}{2} \times \frac{\pi}{2}+\frac{1}{7} \times \frac{7}{8}\)

x \(\frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}]\)

= \(\frac{4 a^8}{3}\left[\frac{15 \pi+9 \pi+9 \pi+15 \pi}{8 \times 6 \times 4 \times 2 \times 2}\right]=\frac{48 \pi a^8}{3(8 \times 6 \times 2 \times 2)}=\frac{\pi a^8}{12}\)

19. Using double integral find the area enclosed by the curves y = 2x² and y² = 4x.

Solution:
The region of integration is the region bounded by y² = 4x and y = 2x²

To find A, solve the equations y² = 4x and y = 2x²

y = 2x² ⇒ y² = 4x⁴⇒4x⁴ = 4 ⇒ x(x³-l) = 0

⇒x = 0, 1.

∴A is (1,1)

Required area =\(\iint_R d x d y\)

Take a strip PQ parallel toy axis with P lies only = 2x², Q lies on y² = 4x ⇒ y = 2\(\sqrt{x}\)

The limits of are y = 2x² to y = 2\(\sqrt{x}\) and the limits ofx are x = 0 to x = 1.

Required Area = \(\int_0^1 \int_{2 x^2}^{2 \sqrt{x}} d x d y=\int_0^1 [y x_{2 x^2}^{2 \sqrt{x}} d x=\int_0^1\left[2 \sqrt{x}-2 x^2\right] d x\)

= \(\left[\frac{2 x^{3 / 2}}{3 / 2}-2 \frac{x^3}{3}\right]_0^1=\frac{4}{3}-\frac{2}{3}=\frac{2}{3}.\)

20. Find the smaller of the areas bounded by y = 2- x and x² +y² = 4 using double integral.

Solution:
Region R is the upper part of the.

Required area A = \(\iint_R d x d y\)

To find limits for y, take a strip PQ parallel to the they-axis with P lies on y = 2 -x and Q lies on the circle x²+y² = 4.

y limits are \(y=2-x\) to \(y=\sqrt{4-x^2}\) and x limits are x=0 to x=2.

∴ A = \(\int_0^2 \int_{2-x}^{\sqrt{4-x^2}} d x d y=\int_0^2\left[\int_{2-x}^{\sqrt{4-x^2}} d y\right] d x=\int_0^2[y] d x\)

= \(\int_0^{\sqrt{4-x^2}}\left[\sqrt{4-x^2}-(2-x)\right] d x\)

= \(\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \text{Sin}^{-1} \frac{x}{2}-2 x+\frac{x^2}{2}\right]\)

= \(0+2\left(\text{Sin}^{-1} 1-\text{Sin}^{-1} 0\right)-2 \cdot 2+\frac{4}{2}\)

=2.π/2-4+2=π-2.

21. Find the area bounded by the parabola y² = 4-x and y² = 4 -4x as a double integral and evaluate it.

Solution: Given y² = 4− x =− (x- 4) is a parabola with vertex (4, 0) and towards the negative x-axis, axis of symmetry the x-axis. y² = 4 − 4x = − 4(x − 1) is a parabola with vertex (1,0) and towards the negative x-axis, axis of symmetry the x-axis.

To find the points of intersection, solve y² = 4- x and y² = 4- 4x.

∴ 4 −x = 4- 4x ⇒ 3x = 0 ⇒ x = 0 and y² = 4- x ⇒  y² = 4  ⇒ y = ±4 and the points of intersection are (0, 2), (0,- 2).

 

 

 

 

 

 

 

 

The region is the shaded region in the figure.

Both curves are symmetric about x-axis.

Required area A=2 (Area above the x-axis)=2∫\(\int_R\)dx dy

It is convenient to take strip PQ  parallel to the x-axis with P lies on y² = 4− 4x and Q lies on y² = 4−x.

Now y² = 4− 4x ⇒ x = 1 −y² /4 and y² = 4− x ⇒ x = 4 −y² and the limits of y are y = 0,y = 2.

Required area, A = \(2 \int_0^2\left[\int_{1-y^2 / 4}^{4-y^2} d x\right] d y=2 \int_0^2[x] y_{1-y^2 / 4}^{4-y^2} d y=2 \int_0^2\left[4-y^2-\left(1-\frac{y^2}{4}\right)\right] d y \)

= \(2 \int_0^2\left(3-\frac{3}{4} y^2\right) d y=2\left[3 y-\frac{3}{4} \frac{y^3}{3}\right]=2\left[3 \times 2-\frac{8}{4}\right]=2[6-2]=8\)

22. Find the area bounded by x² = 4y and x− 2y + 4 = 0 using double integral.

Solution: Solving the given curves x² = 4y, x −2y + 4 = 0;

We get x² =4 \(\left(\frac{x+4}{2}\right)\) ⇒. x² −  2x- 8 = 0

⇒ (x- 4)(x + 2) = 0 ⇒ x = 4 or −2.

Ifx = 4 theny = 4; Ifx =− 2 then y = 1.

The points of intersection are (−2, 1) and (4, 4).

Take strip PQ parallel to y-axis with P lies on x2 = 4y and Q lies on x − 2y + 4 = 0.

∴ The limits of y are \(y=\frac{x^2}{4}\) to \(y=\frac{x+4}{2}\) and the limits of x are x=-2 to x=4.

Required area = \(\int_{-2}^4 \int_{x^2 / 4}^{(x+4) / 2} d x d y=\int_{x=-2}^{x=4}\left[\int_{y=x^2 / 4}^{y=(x+4) / 2} d y\right] d x=\int_{-2}^4[y] \int_{y=x / 4}^{y=(x+4) / 2} d x\)

= \(\int_{-2}^4\left(\frac{x+4}{2}-\frac{x^2}{4}\right) d x=\left[\frac{x^2}{4}+2 x-\frac{x^3}{12}\right]_{-2}^4=\left(4+8-\frac{16}{3}\right)-\left(1-4+\frac{2}{3}\right)=15-\frac{18}{3}=9\) .

23. Find the smaller area bounded \(\frac{x^2}{9}+\frac{y^2}{4}\) = 1 and \(\frac{x}{3}+\frac{y}{2}\) = 1 using double integral.

Solution:

The curves \(\frac{x^2}{9}+\frac{y^2}{4}\)=1, \(\frac{x}{3}+\frac{y}{2}\) =1 intersect at A(3,0), B(0,2).

Take strip PQ parallel to y-axis with P lies on \(\frac{x^2}{9}+\frac{y^2}{4}\) =1 and Q lies on \(\frac{x}{3}+\frac{y}{2}\) =1.

The limits of integration are y-varies from

y = \(2(1-x / 3)\) to \(y=2 \sqrt{1-x^2 / 9}\) and x varies from x=0 to x=3.

= \(\int_0^3\left[2 \sqrt{1-\frac{x^2}{9}}-2\left(1-\frac{x}{3}\right)\right] d x\)

= \(2\left[\frac{x}{6} \sqrt{1-\frac{x^2}{9}}+\frac{3}{2} \text{Sin}^{-1} \frac{x}{3}-x+\frac{x^2}{6}\right]\)

= \(\frac{3 \pi}{2}-6+3=\frac{3 \pi}{2}-3=\frac{3(\pi-2)}{2}\)

24. Find the smaller area bounded by y² = 4x, x +y = 3 and x-axis using double integral

Solution: Solving y= 4x², x+y=3, we get y²= 4(3-y) = y²+4y-12=0

= (y+6)(y-2) =0 = y=-6 or 2

Take strip parallel to x-axis with P lies on y²=4x and Q lies on x+y=3.

To find the smaller area bounded by y=4x²,x+y=3 and x-axis, the limits of integration are x varies from x=y²/4 x=3-y and y varies from y=0 to y=2.

The required area = \(\int_0^2 \int_{y^2 / 4}^{3-3} d y d x=\int_0^2[\int_{y^2 / 4}^{3-y} d x d y\)

= \(\int_0^2[x] \cdot d y=\int_0^{3-y}\left[3-y-\frac{y^2}{4}\right] d y=\left[3 y-\frac{y^2}{2}-\frac{y^3}{12}\right]\)

= \(6-2-\frac{2}{3}=\frac{10}{3}\)

25. Find the area of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\)= 1 by double integration.

Solution: The area of the ellipse = 4 (Area in the first quadrant)

Take strip PQ parallel toy-axis with P on x-axis and Q on the ellipse.

The limits of integration are y varies from y = 0 to y = \(b \sqrt{1-x^2 / a^2}\) and x varies from x = 0 to x = a.

∴ Area of the ellipse = \(4 \int_0^a \int_0^{b \sqrt{1-x^2 / a^2}} d x d y\)

= \(4 \int_0^a[y]_0^{b \sqrt{1-x^2 / a^2}} d x=4 \int_0^a b \sqrt{1-\frac{x^2}{a^2}} d x\)

= \(\frac{4 b}{d} \int_0^a \sqrt{a^2-x^2} d x=\frac{4 b}{a}\left[\frac{x}{a} \sqrt{a^2-x^2}+\frac{a^2}{2} \text{Sin}^{-1} \frac{x}{a}\right]\)

= \(\frac{4 b}{a}\left(\frac{a^2}{2} \frac{\pi}{2}\right)=\pi a b\) .

26. Using double integration find the area of the parallelogram whose vertices are A (1,0), B (3, 1), C (2, 2), and D (0, 1).

Solution: The given points A (1, 0), B (3, 1), C (2, 2) and D (0, 1) are the vertices of a parallelogram ABCD.

The required area is the area of the parallelogram ABCD. Area of the parallelogram ABCD

We shall find the equations of AB and AD.

 

Equation of \(\stackrel{\leftrightarrow}{A B}\)is \(\frac{y-0}{0-1}\)=\(\frac{x-1}{1-3}\)  ⇒ y=1/2(x-1)……. (1)

Equation of \(\stackrel{\leftrightarrow}{A D}\)  is \(\frac{y-0}{0-1}\)= \(\frac{x-1}{1-0}\)……. (2)

Area of \(\triangle A B D=\iint_{A B D} d x d y\).

Take a strip PQ parallel to the x-axis with P on (2) and Q is on (1).

∴ x=-y+1 and x=2 y+1 and y varies from 0 to 1.

= \(\int_0^1 3 y d y=3\left[\frac{y^2}{2}\right]=\frac{3}{2}\) .

∴ Area of the parallelogram A B C D is = \(2 \times \frac{3}{2}=3\)

27. Find the area bounded between ∫r = 2 cos θ and r = 4 cos θ .

Solution: Area A = \(\iint_R r d r d \theta\)where the region R is the region between the circles r = 2 cos θ and r = 4 cos θ

The area is the region outside the circle r = 2 cos θ and inside, the circle r = 4 cos θ.

We first integrate w.r.to r and so, we take the radius vector OPQ.

 

When PQ is moved to cover the area A, r varies from r = 2 cosθ to r = 4 cos θ, and 0 varies from 0 =- π/2 to θ = π/2.

∴ Area  A = \(\int_{-\pi / 2}^{\pi / 2} \int_{2 \cos \theta}^{4 \cos \theta} r d r d \theta\)

= \(\int_{-\pi / 2}^{\pi / 2}\left[\frac{r^2}{2}\right]_{2 \cos \theta}^{4 \cos \theta} d \theta\)

= \(\frac{1}{2} \int_{-\pi / 2}^{\pi / 2}\left(4^2 \cos ^2 \theta-2^2 \cos ^2 \theta\right) d \theta\)

= \(6 \int_{-\pi / 2}^{\pi / 2} \cos ^2 \theta d \theta=6 \times 2 \int_0^{\pi / 2} \cos ^2 \theta d \theta=12 \frac{1}{2} \frac{\pi}{2}=3 \pi \text {. }\)

28. Find the area of one loop of the lemniscate r² = a² cos 2θ.

Solution:

Given r² = a² cos 2 θ

Area of the loop = \(\iint_R r d r d \theta\), where R is the region as in the figure.

Since the loop is symmetric about the initial line, the required area is twice the area above the initial line. First, we integrate w.r.to r.

In the region, take a radical strip OP, its ends are r=0 and r= a\(\sqrt{\cos 2 \theta}\)

When the strip is moved to cover the region R, θ varies from 0 to π/4

Required area A = \(2 \int_0^{\pi / 4} \int_0^{a \sqrt{\cos 2 \theta}} r d r d \theta=2 \int_0^{\pi / 4}\left[\left(\frac{r^2}{2}\right)\right]_0^{a \sqrt{\cos 2 \theta}} d \theta\)

= \(\int_0^{\pi / 4} a^2 \cos 2 \theta d \theta\)

= \(a^2 \int_0^{\pi / 4} \cos 2 \theta d \theta=a^2\left[\frac{\sin 2 \theta}{2}\right]\)

= \(\frac{a^2}{2}\left(\sin \frac{\pi}{2}-\sin 0\right)=\frac{a^2}{2}\)

29. Find the area of a the loop of the curve r = a sin 3θ.

Solution:

Given r = a sin 3θ.

The area of the \(\iint_R r d r d \theta\)

But the loop is formed by two consecutive values of θ  when r = 0.

When r = 0, a sin 3θ  = 0⇒ 3θ = 0 or π⇒0 = 0 or π /3 and r varies from r = 0 to r = a sin 3θ

Area of the loop = \(\int_0^{\pi / 3} \int_0^{a \sin 3 \theta} r d r d \theta\)

= \(\int_0^{\pi / 3}\left[\frac{r^2}{2}\right]_0^{a \sin 3 \theta} d \theta=\frac{1}{2} \int_0^{\pi / 3} a^2 \sin ^2 3 \theta d \theta\)

= \(\frac{a^2}{2} \int_0^{\pi / 3} \frac{1-\cos 6 \theta}{2} d \theta=\frac{a^2}{4}\left[\theta-\frac{\sin 6 \theta}{6}\right]_0^{\pi / 3}\)

= \(\frac{a^2}{4}\left[\frac{\pi}{3}-\frac{\sin 2 \pi-\sin 0}{6}\right]=\frac{\pi a^2}{12}\)

30. Find the area of the cardioid r = a(1+ cos θ).

Solution: Given r = a (1 + cos θ)

Area = ∫\(\int_R r d r d \theta\)

 

Now r varies from 0 to a (1 + cosθ ) and θ varies from −π to π

Required area = \(\int_{\theta=-\pi}^{\theta=\pi}\left[\int_{r=0}^{r=a(1+\cos \theta)} r d r\right] d \theta\)

= \(\int_{\theta=-\pi}^{\theta=\pi}\left[\frac{r^2}{2}\right]_{r=0}^{r=a(1+\cos \theta)} d \theta\)

= \(\int_{-\pi}^\pi \frac{a^2}{2}(1+\cos \theta)^2 d \theta=a^2 \int_0^\pi\left(1+2 \cos \theta+\cos ^2 \theta\right) d \theta\)

= \(a^2 \int_0^\pi\left[1+2 \cos \theta+\frac{1+\cos \theta}{2}\right] d \theta\)

= \(a^2 \int_0^\pi\left[\frac{3}{2}+2 \cos \theta+\frac{1}{2} \cos 2 \theta\right] d \theta\)

= \(a^2\left[\frac{3 \theta}{2}+2 \sin \theta+\frac{1}{4} \sin 2 \theta\right]=\frac{3 \pi a^2}{2} \)

31. Find the area which is inside the circle r = 3a cos θ and outside the cardioid r =a(1+cos θ).

Solution:

Given r = 3a cos θ  → (1) and r = a (1 + cos θ)  → (2)

Required area A = ∫∫r dr dθ

Eliminating r from (1) and (2), we get 3a cos θ = a (1 + cos θ) ⇒ 2 cos θ = 1

⇒  cos θ = 1/2 ⇒  θ=−π/3 or π/3.

 

The required area is the region shown in the figure. Since both curves are symmetrical about the initial line, the required area is twice the area above the initial line.

In this region take a radial strip OPP where P lies on (2) and P’ lies on (1).

When it moves, it will cover the required area.

∴  r varies from a (1 + cos θ) to 3a cos θ and 0 varies from 0 to π/3

Required area = \(2 \int_0^{\pi / 3} \int_{r=a(1+\cos \theta)}^{r=3 a \cos \theta} r d r d \theta\)

= \(2 \int_0^{\pi / 3}\left[\frac{r^2}{2}\right]_{a(1+\cos \theta)}^{3 a \cos \theta} d \theta\)

= \(\int_0^{\pi / 3}\left[9 a^2 \cos ^2 \theta-a^2(1+\cos \theta)^2\right] d \theta\)

= \(a^2 \int_0^{\pi / 3}\left[9 \cos ^2 \theta-\left(1+2 \cos \theta+\cos ^2 \theta\right)\right] d \theta\)

= \(a^2 \int_0^{\pi / 3}\left[8 \cos ^2 \theta-1-2 \cos \theta\right] d \theta=a^2 \int_0^{\pi / 3}\left[8\left\{\frac{1+\cos 2 \theta}{2}\right\}-1-2 \cos \theta\right] d \theta\)

= \(a^2\left[4\left(\theta+\frac{\sin 2 \theta}{2}\right)-\theta-2 \sin \theta\right]_0^{\pi / 3}\)

= \(a^2\left[4\left(\frac{\pi}{3}+\frac{\sin \frac{2 \pi}{3}}{2}\right)-\frac{\pi}{3}-2 \sin \frac{\pi}{3}-0\right]\)

= \(a^2\left[\frac{4 \pi}{3}+2 \frac{\sqrt{3}}{2}-\frac{\pi}{3}-2 \frac{\sqrt{3}}{2}\right]=a^2\left[\frac{4 \pi}{3}-\frac{\pi}{3}\right]=\pi a^2\)

32. Find the area common to r = \(a \sqrt{2}\) and r- 2a cosθ .

Solution: Given r = a\(\sqrt{2}\) → (1) and r = 2a cosθ  → (2)

(1) is a circle with centre (0, 0) and radius a\(\sqrt{2}\)

(2) is a circle with centre (a, 0) and radius a.

Solve (1) and (2) to find the point of intersection.

∴ a\(\sqrt{2}\) 2a cosθ ⇒ cosθ \(\frac{1}{\sqrt{2}}\) ⇒ θ=π/4.

Since the circles are symmetrical about the initial line OX, the required area = 2 [area OABC]- 2[area OAB+ area OBC]

In OAB, take a strip OP. When OP moves it covers the area OAR Ends of OP we, r = 0 and r = a\(\sqrt{2}\).

∴ r varies from 0 to a\(\sqrt{2}\) and 0 varies from 0 toπ/4 . In the area, OBC, take a strip OQ.

When OQ moves it covers the area OBC. Ends of OQ are, r = 0 and r = 2a cos θ

Required area = \(2\left[\int_{0^{-}}^{\pi / 4} \int_0^{\sqrt{2}} r d r d \theta+\int_{\pi / 4}^{\pi / 2} \int_0^{2 a \cos \theta} r d r d \theta\right]\)

= \(2 \int_0^{\pi / 4}\left[\frac{r^2}{2}\right]_0^{a \sqrt{2}} d \theta+2 \int_{\pi / 4}^{\pi / 2}\left[\frac{r^2}{2}\right]_0^{2 a \cos \theta} d \theta\)

= \(\int_0^{\pi / 4} 2 a^2 d \theta+\int_{\pi / 4}^{\pi / 2} 4 a^2 \cos ^2 \theta d \theta\)

= \(2 a^2[\theta]+4 a^2 \int_{\pi / 4}^{\pi / 4}\left(\frac{1+\cos 2 \theta}{2}\right) d \theta\)

= \(2 a^2 \frac{\pi}{4}+2 a^2\left[\theta+\frac{\sin 2 \theta}{2}\right]_{\pi / 4}^{\pi / 2}\)

= \(\frac{\pi a^2}{2}+2 a^2\left[\frac{\pi}{2}-\frac{\pi}{4}+\frac{1}{2}\left(\sin \pi-\sin \frac{\pi}{2}\right)\right]\)

= \(\frac{\pi a^2}{2}+2 a^2\left[\frac{\pi}{4}-\frac{1}{2}\right]=\frac{\pi a^2}{2}+\frac{\pi a^2}{2}-a^2=a^2(\pi-1)\)

33. Find the area inside the circle r = a sin θ but lying outside the cardioid r = a (1 – cos θ).

Solution: Given r= a sin θ →(1) and r= a(1- cos θ) →(2)

Area=∫∫ r dr dθ

Eliminating r from (1) and (2), we get a sin o a=(1- cos θ)= sin θ+ cos θ=1

sin² θ+ cos² θ + 2 sin θ cos θ =1⇒ 1+2 sin 2θ=1= sin 2θ=0

2θ =0, π⇒ θ=0 or π/2

 

Area = \(\int_0^{\pi / 2} \int_{a(1-\cos \theta)}^{a \sin \theta} r d \theta=\int_0^{\pi / 2}\left[\frac{r^2}{2}\right]_{a(1-\cos \theta)}^{a \sin \theta} d \theta\)

= \(\frac{1}{2} \int_0^{\pi / 2}\left[a^2 \sin ^2 \theta-a^2(1-\cos \theta)^2\right] d \theta\)

= \(\frac{a^{2^{\pi / 2}}}{2} \int_0^2\left[\sin ^2 \theta-\left(1-2 \cos \theta+\cos ^2 \theta\right)\right] d \theta\)

= \(\frac{a^2}{2} \int_0^{2 / 2}\left\{-1+2 \cos \theta-\left(\cos ^2 \theta-\sin ^2 \theta\right)\right\} d \theta\)

= \(\frac{a^2}{2}\left[\int_0^{\pi / 2}[-1+2 \cos \theta\} d \theta-\int_0^{\pi / 2}\left\{\cos ^2 \theta-\sin ^2 \theta\right] d \theta\right]\)

= \(\frac{a^2}{2} \int_0^{\pi / 2}(-1+2 \cos \theta) d \theta\)

= \(\frac{a^2}{2}[-\theta+2 \sin \theta]=\frac{\pi / 2}{0}=\frac{a^2}{2}\left[-\frac{\pi}{2}+2 \sin \frac{\pi}{2}\right]\)

= \(\frac{a^2}{2}\left[-\frac{\pi}{2}+2\right]=\frac{a^2}{4}[4-\pi] \text {. }\)

 

34. Find the area of the surface of the sphere of radius r.

Solution:
Taking the origin as the centre and radius r, the equation of the sphere is x² +y² + z²= r². Let us consider the surface of the sphere in the first octant. It will be 1/8 of the surface of the sphere. The orthogonal projection of this surface area on the XOY plane is the quadrant of the circle x² +y² − r² in that plane.

Hence this surface area = \(\iint\left\{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1\right\}^{1 / 2} d y d x\) taken over the area of the quarter of the circle \(x^2+y^2=r^2\) on the positivequadrant.

Now \(\frac{\partial z}{\partial x}=-\frac{x}{z}, \frac{\partial y}{\partial z}=-\frac{y}{z}\).

Surface area of the sphere = \(8 \iint\left(\frac{x^2}{z^2}+\frac{y^2}{z^2}+1\right)^{1 / 2} d x d y\)

= \(8 \iint \frac{\left(x^2+y^2+z^2\right)^{1 / 2}}{z} d x d y=8 \iint \frac{r}{z} d x d y=8 \int_0^r \int_0^{\sqrt{r^2-x^2}} \frac{r d x d y}{\sqrt{r^2-x^2-y^2}}\)

= \(8 \times \frac{\pi r^2}{2}=4 \pi r^2\)

35. Find the area of the surface of the sphere x² +y² + Z² = 9a² cut off by the cylinder x² +y² =3ax.

Solution: The Projection of the required area S on the xy-plane is the circle x²+y²=3ax.

 

On the sphere z = \(\sqrt{9 a^2-x^2-y^2}\),

⇒ \(\frac{\partial z}{\partial x}=-\frac{x}{\sqrt{9 a^2-x^2-y^2}}, \frac{\partial z}{\partial y}=-\frac{y}{\sqrt{9 a^2-x^2-y^2}}\)

S = \(\iint_R \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} d x d y\)

where R is the region enclosed by the circle \(x^2+y^2=3 a x\).

= \(\iint_R \sqrt{1+\frac{x^2}{9 a^2-x^2-y^2}+\frac{y^2}{9 a^2-x^2-y^2}} d x d y=\iint_R \frac{3 a}{\sqrt{9 a^2-x^2-y^2}} d x d y\)

= \(3 a \iint_R \frac{r d r d \theta}{\sqrt{9 a^2-r^2}}\) changing to polars.

The polar equation of the circle is r= 3a cos θ.

To cover the area of this circle t varies from 0 to 3a cos θ  and θ from −π/2 to π/2.

∴ S = \(3 a \int_{-\pi / 2}^{\pi / 2} \int_0^{3 a \cos \theta} \frac{r d r d \theta}{\sqrt{9 a^2-r^2}}\)

= \(3 a \int_{-\pi / 2}^{\pi / 2}\left[-\sqrt{9 a^2-r^2}\right]{ }_0^{3 a \cos \theta}=9 a^2 \int_{-\pi / 2}^{\pi / 2}\left[1-\sqrt{1-\cos ^2 \theta}\right] d \theta\)

= \(18 a^2 \int_0^{\pi / 2}\left[1-\sqrt{1-\cos ^2 \theta}\right] d \theta=18 a^2 \int_0^{\pi / 2}[1-\sin \theta] d \theta=18 a^2\left(\frac{\pi}{2}-1\right)=9 a^2(\pi-2)\)

36. Find the surface area of the cylinder x² +y² =ax cut off by the sphere x² +y² + Z² = a² 

Solution: The equation of the sphere is x² + y² + z²= a² → (1)

The equation of the cylinder is x² +y²= ax  →(2)

The surface area of the cylinder cut off by the sphere is required.

Projecting the surface on the xz-plane, we get the required surface area of S.

∴ S= 2 \(\iint_{D_1} \sqrt{\left(\frac{\partial y}{\partial x}\right)^2+\left(\frac{\partial y}{\partial z}\right)^2}\) +1dx dz

Where D₁ is the region obtained by eliminating y² from (1) and (2).

∴ z²+ax=a²→(3)

The surface is x²+y²=ax

Differentiating partially w. r. to x and z, treating y as function of x and z, we get

2x+2y \(\frac{\partial y}{\partial x}=a \Rightarrow \frac{\partial y}{\partial x}=\frac{a-2 x}{2 y}\) and

2y \(\frac{\partial y}{\partial z}=0 \Rightarrow \frac{\partial y}{\partial z}=0\)

∴ \(\left(\frac{\partial y}{\partial x}\right)^2+\left(\frac{\partial y}{\partial z}\right)^2+1\)

= \(\frac{(a-2 x)^2}{4 y^2}+1=\frac{(a-2 x)^2+4 y^2}{4 y^2}\)

= \(\frac{a^2-4 a x+4 x^2+4\left(a x-x^2\right)}{4 y^2}=\frac{a^2}{4 y^2}\)

∴ \( \sqrt{\left(\frac{\partial y}{\partial x}\right)^2+\left(\frac{\partial y}{\partial z}\right)^2+1}=\frac{a}{2 y}=\frac{a}{2 \sqrt{a x-x^2}}\)

We have \(z^2+a x=a^2 \Rightarrow z^2=a^2-a x \Rightarrow z= \pm \sqrt{a^2-a x}\)

∴ S = \(2 \iint_{D_1}^{2 \sqrt{a x-x^2}} d x d z=a \int_0^a \int_{-\sqrt{a^2-a x}}^{\sqrt{a^2-a x}} \frac{1}{\sqrt{a x-x^2}} d x d z\)

= \(a \int_0^a\left[\int_{-\sqrt{a^2-a x}}^{\sqrt{a^2-a x}} \frac{1}{\sqrt{a x-x^2}} d z\right] d x\)

= \(a \int_0^a\left[\frac{1}{\sqrt{a x-x^2}}[z]-\sqrt{a^2-a x}\right] d x\)

= \(a \int_0^{\sqrt{a^2-a x}}\left\{\frac{1}{\sqrt{a x-x^2}}\left[\sqrt{a^2-a x}+\sqrt{a^2-a x}\right]\right\} d x\)

= \(2 a \int_0^a \frac{\sqrt{a^2-a x}}{\sqrt{a x-x^2}} d x=2 a \int_0^a \sqrt{\frac{a(a-x)}{x(a-x)}} d x\)

= \(2 a \int_0^a \sqrt{\frac{a}{x}} d x=2 a \sqrt{a} \int_0^a x^{-1 / 2} d x\)

= \(2 a \sqrt{a}\left[\frac{x^{1 / 2}}{1 / 2}\right]=4 a \sqrt{a}\left(a^{1 / 2}-0\right)=4 a^2\)

37. Find the portion of the cone x² +y² = 4z² lying above the xy-plane and inside the cylinder x² +y² = 3y.

Solution:
The projection of the required area on the x,y plane is the circle x² +y² =3y.

Given cone is x²+y²= 4z² ⇒  z = 1/2\(\sqrt{x^2+y^2}\).

 

⇒ \(\frac{\partial z}{\partial x}=\frac{1}{2} \frac{x}{\sqrt{x^2+y^2}}, \frac{\partial z}{\partial y}=\frac{1}{2} \frac{y}{\sqrt{x^2+y^2}}\)

∴ S = \(\iint_k \sqrt{1+\frac{1}{4} \frac{x^2}{x^2+y^2}+\frac{1}{4} \frac{y^2}{x^2+y^2}} d x d y\)

= \(\frac{\sqrt{5}}{2} \iint_R d x d y\), where R is the circle \(x^2+y^2=3 y\)

= \(\frac{\sqrt{5}}{2}\)(area of the circle \(x^2+y^2=3 y\))=\(\frac{\sqrt{5}}{2} \pi\left(\frac{3 a}{2}\right)^2=\frac{9 \sqrt{5}}{8} \pi a^2\)

38. The centre of a sphere of radius r is on the surface of a right cylinder, the radius of whose base is r/2. Find the area of the surface of the cylinder intercepted by the sphere.

Solution: The equation of the sphere is x² +y² +z² =r² and the equation of the cylinder is

⇒ \(\left(x-\frac{r}{2}\right)^2+y^2=\left(\frac{r}{2}\right)^2 \Rightarrow x^2+y^2=r x \text {. }\)

Projecting on the zx-plane, we have

S = \(\iint_R \sqrt{1+\left(\frac{\partial y}{\partial x}\right)^2+\left(\frac{\partial y}{\partial z}\right)^2} d x d z \text {. }\)

On the cylinder \(y^2=r x-x^2\).

∴ \(\frac{\partial y}{\partial x}=\frac{r-2 x}{2 y}, \frac{\partial y}{\partial z}=0 \text {. }\)

The projection of the area on the zx-plane is the curve by eliminating y from x² +y² =rx and x² +y² +z² =r², i.e, z² +rx=r².

Hence the required area = \(2 \iint_R \sqrt{1+\frac{(r-2 x)^2}{4 y^2}} d x d z\)

= \(2 \iint \frac{\sqrt{4 y^2+4 x^2-4 r x+r^2}}{2 y} d x d z=2 r \iint \frac{d x d z}{2 y}=r \iint \frac{d x d z}{\sqrt{r x-x^2}}\)

= \(r \int_0^r \int_{-\sqrt{r^2-r x}}^{\sqrt{r^2-r x}} \frac{d x d z}{\sqrt{r x-x^2}}=2 r \int_0^r\left[\int_0^{\sqrt{r^2-r x}} \frac{d z}{\sqrt{r x-x^2}}\right] d x\)

= \(2 r \int_0^r \frac{1}{\sqrt{r x-x^2}}[z] \int_0^{\sqrt{r^2-r x}} d x\)

= \(2 r \int_0^r \frac{\sqrt{r^2-r x}}{\sqrt{r x-x^2}} d x=2 r \int_0^r \frac{\sqrt{r}}{\sqrt{x}} d x=4 r^2\)

39. Find the volume of the ellipsoid \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\) =1 by using a double integral.

Solution: Note that the” ellipsoid is symmetrical about axes and hence the required volume is 8 times the volume of the ellipsoid in the positive octant.

The region above which the volume lies is bounded by x = 0, x = a, y = 0 andy=b \(\sqrt{1-\frac{x^2}{a^2}}\).

Hence the required volume of the ellipsoid

= \(8 \int_0^a \int_0^{b \sqrt{1-x^2 / a^2}} z d y d x=8 \int_0^a \int_0^b c \sqrt{1-x^2 / a^2} \frac{x^2}{a^2}-\frac{y^2}{b^2} d y d x\)

= \(8 \int_0^t \int_0^t c \sqrt{\left.\frac{t^2}{b^2}-\frac{y^2}{b^2}\right]} d y d x \text {, where } b^2\left(1-\frac{x^2}{a^2}\right)=t^2\)

= \(8 \int_0^a \int \frac{c}{b} \sqrt{t^2-y^2} d y d x=8 \int_0^a \frac{c}{b}\left[y \frac{\sqrt{t^2-y^2}}{2}+\frac{t^2}{2} \text{Sin}^{-1} \frac{y}{t}\right]_0^t d x\)

40. Find the volume bounded by the cylinder X² +y² = 4 and the planes y + z = 3 and z = 0 by using double integral.

Solution: From the it is clear that to get the required volume z=3-y is to be integrated over the circle x² +y² =4 in the XOY plane.

Note that x varies from –\(\sqrt{4-y^2}\) to \(\sqrt{4-y^2}\) and y varies from-2 to +2.

 

∴ Required volume

= \(\int_{-2}^2 \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} z d x d y=\int_{-2}^2 \int_{-\sqrt{4-y^2}:}^{\sqrt{4-y^2}}(3-y) d x d y\)

= \(\int_{-2}^2(3-y)[x]_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} d y=2 \int_{-2}^2(3-y) \sqrt{4-y^2} d y\)

= \(6 \int_{-2}^2 \sqrt{4-y^2} d y-2 \int_{-2}^2 y \sqrt{4-y^2} d y=12 \int_0^2 \sqrt{4-y^2} d y-0\)

= \(12 \int_0^{\pi / 2} 4 \cos ^2 \theta d \theta=48 \cdot \frac{1}{2} \cdot \frac{\pi}{2}=12 \pi \text {. }\)

41. Find the volume of the tetrahedron bounded by coordinate planes and the plane \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\)=1

Solution: The region of integration is the region bounded by \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\)=1, x=0,y=0,z=0.

Its projection in the xy-plane is the ΔOAB bounded by x=0, y=0 and \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\)=1.

Volume V = \(\iiint_D \int d x d y d z\)

= \(\int_0^a \int_0^{b\left(1-\frac{x}{a}\right) c\left(1-\frac{x}{a}-\frac{y}{b}\right)} \int_0^a d z d y d x\)

= \(\int_0^b \int_0^{b\left(1-\frac{x}{a}\right)}[z]_0^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)} d y d x\)

= \(\int_0^{a^b\left(1-\frac{x}{a}\right)} \int_0^c c\left(1-\frac{x}{a}-\frac{y}{b}\right) d y d x\)

= \(c \int_0^a\left[\left(1-\frac{x}{a}\right) y-\frac{y^2}{2 b}\right]_0^{b\left(1-\frac{x}{a}\right)} d x\)

= \(c \int_0^a\left[\left(1-\frac{x}{a}\right) b\left(1-\frac{x}{a}\right)-\frac{1}{2 b} b^2\left(1-\frac{x}{a}\right)^2\right] d x\)

= \(\frac{b c}{2} \int_0^a\left(1-\frac{x}{a}\right)^2 d x=\frac{b c}{2}\left[\frac{\left(1-\frac{x}{a}\right)^3}{-\frac{1}{a} 3}\right]^a\)

= \(\frac{-a b c}{6}[0-1]=\frac{a b c}{6}\)

42. Find the volume of the sphere x² + y² + Z² = a².

Solution: The sphere x² +y² +z² =a² is symmetric about the coordinate planes

The volume of the sphere=8 (Volume of the sphere in the first octant).
x² +y² +z² =a² ⇒ z² =a² -x² -y² ⇒ z=± \(\sqrt{a^2-x^2-y^2}\)

In the first octant z varies from z=0 to z \(\sqrt{a^2-x^2-y^2}\)

The section of the sphere by the xy-plane z=0 is the circle x²+y²=a²

⇒y²=a²-x^2⇒ y=±\(\sqrt{a^2-x^2}\).

∴ y varies from 0 to \(\sqrt{a^2-x^2}\) and x varies from 0 to a.

∴ Volume of the sphere = \(8 \int_0^a \int_0^{\sqrt{a^2-x^2} \sqrt{a^2-x^2-y^2}} \int_0^a d x d y d z\)

= \(8 \int_0^{\sqrt{a^2-x^2}}\left\{\int_0^{\sqrt{a^2-x^2}}\left[\int_0^d d z\right] d y\right\} d x\)

= \(8 \int_0^a\left\{\int_0^{\sqrt{a^2-x^2}}[z]_{z=0}^{z=\sqrt{a^2-x^2-y^2}} d y\right\} d x=8 \int_0^a\left[\int_0^{\sqrt{a^2-x^2}} \sqrt{a^2-x^2-y^2} d y\right] d x\)

= \(8 \int_0^a\left[\frac{y}{2} \sqrt{a^2-x^2-y^2}+\frac{a^2-x^2}{2} \text{Sin}^{-1} \frac{y}{\sqrt{a^2-x^2}}\right] d x\)

= \(8 \int_0^{y=\sqrt{a^2-x^2}} \frac{a^2-x^2}{2} \frac{\pi}{2} d x\)

= \(2 \pi \int_0^a\left(a^2-x^2\right) d x=2 \pi\left[a^2 x-\frac{x^3}{3}\right]=2 \pi\left(a^3-\frac{a^3}{3}\right)=\frac{4 \pi a^3}{3}\)

43. Find the volume bounded by the cylinder x² +y²  = 4 and the planes y + z-4 and z = 0 by using double integral.

Solution: The required volume of the cylinder x²+y²=4, cut off between the planes z=0 and y+z =4 is V ∫\(\int_D\)∫ dx dy dz.

∴  z varies from z=0 to z=4-y

∴ The projection of the region in the plane is x²+y²=4 =y=± \(\sqrt{4-x^2}\)

∴ y varies from \(-\sqrt{4-x^2}\) to \(+\sqrt{4-x^2}\) and x-varies from -2 to 2

Volume V = \(\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}[z]_0^{4-y} d y d x=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(4-y) d y d x\)

= \(\int_{-2}^2\left[4 y-\frac{y^2}{2}\right]_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} d x\)

= \(\int_{-2}^2\left\{4\left[\sqrt{4-x^2}-\left(-\sqrt{4-x^2}\right)-\frac{1}{2}\left[4-x^2-\left(4-x^2\right)\right]\right\} d x=8 \int_{-2}^2 \sqrt{4-x^2} d x\right.\)

= \(8 \cdot 2 \int_0^2 \sqrt{4-x^2} d x=16\left[\frac{x \sqrt{4-x^2}}{2}+\frac{4}{2} \text{Sin}^{-1} \frac{x}{2}\right]=16\left[0+2 \text{Sin}^{-1} 1-(0+0)\right]\)

= \(16 \cdot 2 \frac{\pi}{2}=16 \pi\)

44. Find the volume common to the cylinders x² + y²= a² and x²+ z² = a².

Solution:  Given cylinder x²+y²=a² →(1)

⇒ y²=a²-x² or y=±  \(\sqrt{a^2-x^2}\) and

x²+z² =a² →(2)= z²=a²-x²  0r  z= ±\(\sqrt{a^2-x^2}\)

The required volume can be covered as follows:

z: from-\(\sqrt{a^2-x^2}\) to \(\sqrt{a^2-x^2}\) : y: from – \(\sqrt{a^2-x^2}\) to \(\sqrt{a^2-x^2}\) : x: from  − a to a.

Thus the volume V enclosed by the cylinders

V = \(\int_{-a-\sqrt{a^2-x^2}}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} d z d x=2 \int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}[z]_0^{\sqrt{a^2-x^2}} d y d x\)

= \(4 \int_{-a}^{a \sqrt{a^2-x^2}} \int_0^{\sqrt{2}} \sqrt{\left(a^2-x^2\right)} d y d x\)

= \(4 \int_{-a}^a \sqrt{a^2-x^2}[y] d x=8 \int_0^{\sqrt{a^2-x^2}}\left(a^2-x^2\right) d x\)

= \(8(a^2 x-\frac{x^3}{3} \int_0^a=8 (a^3-\frac{a^3}{3})\)

= \(8 a^3\left(1-\frac{1}{3}\right)=\frac{16 a^3}{3}\) cubic units

45. Find the volume of the portion of the sphere x² +y² + z² = a² lying inside the cylinder x² + y² = ax.

Solution: We solve the problem by transforming it into cylindrical coordinates.

Putx-p cos Φ , y = ρ sin Φ, z = z.

Required volume = ∫∫∫ρ dz dρ dΦ.

The equation of the sphere is ρ² + z² = a² and the equation of the cylinder is  ρ= a cos Φ.

The volume inside the cylinder bounded by the sphere

= 2 x volume shown shaded in the.

Note that z varies from 0 to\(\sqrt{a^2-\rho^2}\),ρ varies from 0 to a cosΦ and Φ varies from 0 to π.

Required volume

= \(2 \int_0^\pi \int_0^{a \cos \phi} \int_0^{\sqrt{a^2-\rho^2}} \rho d z d \rho d \phi\)

= \(2 \int_0^\pi \int_0^{a \cos \phi} \rho \sqrt{a^2-\rho^2} d \rho d \phi\)

= \(2 \int_0^\pi\left[-\frac{\left(a^2-\rho^2\right)^{3 / 2}}{3}\right]_0^{a \cos \phi} d \phi\)

= \(\frac{2 a^3}{3} \int_0^\pi\left(1-\sin ^3 \phi\right) d \phi\)

= \(\frac{2 a^3}{9}(3 \pi-4) .\)