Marksparks

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

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

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

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

= 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.

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

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

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

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

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

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:

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 \(=\left\{\boldsymbol{F} \times \Sigma \mathbf{i} \frac{\partial}{\partial x}\right\}\)

=∑(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 \((g \nabla f)=g \nabla^2 f+\nabla g \cdot \nabla f\)

∴ \({div}(f \nabla g)-{div}(g \nabla f)=f \nabla^2 g-g \nabla^2 f\)