Integral Transformations Solved Problems
Example.1 Show that \(\int_S\)(axi+byj+czk).N.dS=4\(\frac{\pi}{3}\) (a+b+c) where S is the surface of the sphere x2+y2+z2=1.
Solution:
Here \(\mathbf{F}=a x \mathbf{i}+b y \mathbf{j}+c z \mathbf{k} \quad \text{div} \mathbf{F}=\nabla . \mathbf{F}=\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}=a+b+c\)
By Gauss’s Theorem \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\int_V \nabla \cdot \mathbf{F} d \mathbf{V}=\int_V(a+b+c) d \mathbf{V}=(a+b+c) \mathbf{V}=\frac{4 \pi}{3}(a+b+c)\)
V = \(\frac{4 \pi}{3}\), for the given sphere
Example.2 Verify Gauss’s divergence theorem to evaluate (x³-yz)i-2x²yj+zk).NdS over the surface of a cube bounded by the coordinate planes x=y=z=a.
Solution:
Gauss’s theorem states that \(\iint_{\mathrm{S}} \mathbf{F}_1 d y d z+\mathbf{F}_2 d z d x+\mathbf{F}_3 d x d y=\iiint_V\left(\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}\right) d x d y d z\)
From the problem \(\mathbf{F}_1=x^3-y z, \mathbf{F}_2=-2 x^2 y, \mathbf{F}_3=z\)
⇒ \(\frac{\partial \mathbf{F}_1}{\partial x}=3 x^2, \frac{\partial \mathbf{F}_2}{\partial y}=-2 x^2, \frac{\partial \mathbf{F}_3}{\partial z}=1\)
∴ RHS = \(\iiint_V\left(3 x^2-2 x^2+1\right) d x d y d z\)
= \(\int_0^a \int_0^a \int_0^a\left(x^2+1\right) d x d y d z\)
= \(\int_0^a \int_0^a\left[\frac{x^3}{3}+x\right]_0^a d y d z=\frac{1}{3} a^5+a^3\)
Verification: Let us calculate the value of \(\int \mathbf{F} \cdot \mathbf{N} d \mathbf{S}\) over the six faces of the cube directly
1. For the face PQAR \(\mathbf{N}=\mathbf{i}, d \mathbf{S}=d z d y\) and x=a
∴ \(\int_{S_1} \mathbf{F} . \mathbf{N} d \mathbf{S}=\iint_{S_1}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right] . \mathbf{i} d z d y\)
= \(\int_{z=0}^a \int_{y=0}^a\left(x^3-y z\right) d z d y=\int_{z=0}^a \int_{y=0}^a\left(a^3-y z\right) d z d y=a^3\left[[y]_0^a[z]_0^a\right]-\left[\frac{y^2}{2}\right]_0^a\left[\frac{z^2}{2}\right]_0^a=a^5-\frac{1}{4} a^4\)
2. For the face OBSC \(\mathbf{N}=-\mathbf{i}, d \mathbf{S}=d y d z\) and x=0
∴ \(\int_{S_2} \mathbf{F} . \mathbf{N} d \mathbf{S}=-\int_{y=0}^a \int_{z=0}^a\left(x^3-y z\right) d y d z=\int_0^a \int_0^a y z d y d z=\left[\frac{y^2}{2}\right]_0^a\left[\frac{z^2}{2}\right]_0^a=\frac{1}{4} a^4\)
because x=0 for this face
3. For the face \(\text{PQBS} \mathbf{N}=\mathbf{j}, d \mathbf{S}=d z d x\) and y=a
∴ \(\int_{S_3} \mathbf{F} . \mathbf{N} d \mathbf{S}=-\int_{x=0}^a \int_{z=0}^a\left(2 x^2 y\right) d x d z=-2 a \int_0^a \int_0^a x^2 d x d z\)
because y = a for this face
= \(-2 a\left[\frac{x^3}{3}\right]_0^a[z]_0^a=-\frac{2}{3} a^5\)
4. For the face OARC \(\mathbf{N}=-\mathbf{j}, d \mathbf{S}=d z d x\) and y=0
⇒ \(\int_{S_4} \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_{S_4} 2 x^2 y d x d z=0\) because on this face y=0
5. For the face PRCS \(\mathbf{N}=\mathbf{k}, d \mathbf{S}=d x d y\) and z=a
∴ \(\int_S \mathbf{F} . \mathbf{N} d \mathbf{S}=\int_{x=0}^a \int_{y=0}^a z d x d y=a \int_0^a \int_0^a d x d y\)
because z=a on this face
= \(a[x]_0^a[y]_0^a=a^3\)
6. For the face OBQA \(\mathbf{N}=-\mathbf{k}, d \mathbf{S}=d x d y\) and z=0
∴ \(\int_S \mathbf{F} . \mathbf{N} d \mathbf{S}=-\int_{x=0}^a \int_{y=0}^a z d x d y=0\) because on this face z=0
Hence for the total faces \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=a^5-\frac{1}{4} a^4+\frac{1}{4} a^4-\frac{2}{3} a^5+0+a^3=\frac{1}{3} a^5+a^3\)
Example.3 Apply Gauss’s theorem to prove that \(\int_S\)r.N.dS=3V
Solution: By Gauss’s theorem \(\int_S\)r.N.dS=\(\int_V\) div r dv
div \(\mathbf{r}=\nabla \cdot \mathbf{r}=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right) \cdot(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})=1+1+1=3\)
∴ \(\int_V \text{div} \mathbf{r} d \mathbf{V}=\int_V 3 d \mathbf{V}=3 V\)
Example.4 By transforming into a triple integral, evaluate (x³ dy dz+x²y dz dy+x²z dx dy) where S is the closed surface consisting of the cylinder x²+y²=a² and the circular discs z=0 and z=b.
Solution:
Here \(\mathbf{F}_1=x^3, \mathbf{F}_2=x^2 y, \mathbf{F}_3=x^2 z\)
⇒ \(\frac{\partial \mathbf{F}_1}{\partial x}=3 x^2, \frac{\partial \mathbf{F}_2}{\partial y}=x^2, \frac{\partial \mathbf{F}_3}{\partial z}=x^2\)
∴ \(\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}=3 x^2+x^2+x^2=5 x^2\)
By Gauss’s theorem \(\iint_{\mathrm{S}} \mathbf{F}_1 d y d z+\mathbf{F}_2 d z d x+\mathbf{F}_3 d x d y=\iiint_V\left(\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}\right) d x d y d z\)
∴ \(\iint_{\mathrm{S}} x^3 d y d z+x^2 y d z d x+x^2 z d x d y\)
= \(\iiint_V 5 x^2 d x d y d z=5(4) \int_{x=0}^a \int_{y=0}^{\sqrt{a^2-x^2}} \int_{z=0}^b x^2 d x d y d z\)
= \(20 \int_{x=0}^a \int_{y=0}^{\sqrt{a^2-x^2}} x^2(b) d x d y=20 b \int_0^a x^2 \sqrt{a^2-x^2} \cdot d x\)
Put x = \(a \sin \theta, d x=a \cos \theta d \theta\)
x = \(0 \Rightarrow \theta=0 \quad x=a \Rightarrow \theta=\pi / 2\)
= \(20 b a^4 \int_0^{\pi / 2} \sin ^2 \theta \cdot \cos ^2 \theta d \theta=20 a^4 b \int_0^{\pi / 2}\left(\sin ^2 \theta-\sin ^4 \theta\right) d \theta\)
= \(20 a^4 b\left[\frac{1}{2} \frac{\pi}{2}-\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right]=\frac{5}{4} \pi a^4 b\)
Example.5 Prove that ∫f.curl F dV =∫F×f.dS+∫F.curlf dV
Solution:
Consider the vector function \(\mathbf{F} \times \mathbf{f}\)
By Gauss’s theorem \(\int_S(\mathrm{~F} \times \mathrm{f}) \cdot \overline{\mathrm{N}} d \mathrm{~S}=\int_V \text{div}(\mathrm{F} \times \mathrm{f}) d \mathrm{~V}\)
⇒ \(\int_S(\mathrm{~F} \times \mathbf{f}) \cdot \mathrm{N} d \mathrm{~S}=\int_V \text{div}(\mathrm{F} \times \mathbf{f}) d \mathrm{~V}=\int(\mathbf{F} \times \mathbf{f}) \cdot d \mathbf{S}=\int_V \nabla \cdot(\mathbf{F} \times \mathbf{f}) d \mathbf{V}\)
because \(\mathbf{N} d \mathbf{S}=d \mathbf{S}\)
∴ \(\int(F \times \bar{f}) d \overline{\mathrm{S}}=\int_V(\mathrm{f} . \mathrm{curlF}-\mathrm{F} \text { curlf }) d \mathrm{~V}\)
∴ \(\int \mathbf{f} . \text{curl} \mathbf{F} d \mathbf{V}=\int(\mathbf{F} \times \mathbf{f}) \cdot d \mathbf{S}+\int \mathbf{F} . \text{curl} \mathbf{f} d \mathbf{V}\)
Example.6 Compute \(\int_S\) (ax2+by2+cz2) ds over the sphere x2 +y2+z2=1
Solution:
By divergence theorem \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\int_V \nabla \cdot \mathbf{F} d \mathbf{V}\)
Given \(\mathbf{F}. \mathbf{N}=a x^2+b y^2+c z^2\)
Let \(\phi=x^2+y^2+z^2-1\)
∴ Normal vector N to the surface \(\phi\) is \(\nabla \phi=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right)\left(x^2+y^2+z^2-1\right)=2(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})\)
∴ Unit normal vector \(\mathbf{N}=\frac{2(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})}{2 \sqrt{\left(x^2+y^2+z^2\right)}}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)
∴ \(\mathbf{F} . \mathbf{N}=\mathbf{F} .(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})=a x^2+b y^2+c z^2\)
i.e. \(\mathbf{F}=a x \mathbf{i}+b y \mathbf{j}+c z \mathbf{k}\)
∴ \(\nabla . \mathbf{F}=a+b+c\)
Hence by Gauss theorem \(\oint_S\left(a x^2+b y^2+c z^2\right) d \mathbf{S}=\int_V(a+b+c) d \mathbf{V}\)
=(a+b+c) \(\mathbf{V}=\frac{4 \pi}{3}(a+b+c)\) as \(\mathbf{V}=\frac{4}{3} \pi\) being the volume of the sphere of unit radius.
Example.7 By converting the surface integral to volume integral show that ∫\(\int_S\) x3dy dz + y3dz dx+ z3dx dy= \(\frac{12 \pi}{5}\)where S is the surface of the sphere x2+y2+z2=1
Solution:
Consider the sphere \(x^2+y^2+z^2=a^2\) and
⇒ \(\bar{F}=x^3 \bar{i}+y^3 \bar{j}+z^3 \bar{k}=F_1 \bar{i}+F_2 \bar{j}+F_3 \bar{k}\)
div \(\bar{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=3 x^2+3 y^2+3 z^2=3\left(x^2+y^2+z^2\right)=3 r^2\)
where r is the radius of the sphere.
We have by divergence theorem \(\iint_S \bar{F} \cdot \bar{n} d s=\iiint_{d i v} \bar{F} d \nu\)
i.e. \(\iint_{\text {S }} F_1 d y d z+F_2 d z d x+F_3 d x d y=\iiint_V d i v F d v=3 \iiint_S r^2 d v\)
Changing into a spherical coordinator
x = \(r \sin \theta \cos \phi \quad y=r \sin \theta \sin \phi \quad z=r \cos \theta\)
then \(d V=r^2 \sin \theta d r d \theta d \phi\)
L.H.S. = \(3 \int_{\phi=0}^{2 \pi} \int_{\theta=0}^\pi \int_{r=0}^a r^4 \sin \theta d r d \theta d \phi=3 \int_0^{2 \pi} d \phi \int_0^\pi \sin \theta d \theta \int_0^a r^4 d r\)
= \(3[\phi]_0^{2 \pi}[-\cos \theta]_0^\pi\left[\frac{r^5}{5}\right]_0^a=\frac{12}{5} \pi a^5\)
Taking a=1 we get the value as \(\frac{12}{5} \pi\)
Example.8 Find \(\iint_S \overline{\mathbf{F}} \cdot \overline{\mathbf{N}} d \mathrm{~s}\) where \(\overline{\mathrm{F}}\) =2xy\(\overline{\mathrm{i}}\) +yz2\(\overline{\mathrm{j}}\) +xz\(\overline{\mathrm{k}}\) and S is the surface of the parallelopiped formed by x=0,y=0,z=0,x=0,x=2,y=1,z=3.
Solution:
By Gauss’s theorem \(\iint_S \overline{\mathbf{F}} \cdot \overline{\mathbf{N}} d \mathbf{s}=\iiint_V \nabla \cdot \overline{\mathbf{F}} d \mathbf{v}\)
⇒ \(\nabla. \overline{\mathbf{F}}=\frac{\partial \mathbf{f}_1}{\partial x}+\frac{\partial \mathbf{f}_2}{\partial y}+\frac{\partial \mathbf{f}_3}{\partial z}=2 y+z^2+x\)
⇒ \(\iiint_V(\nabla . \overline{\mathbf{F}}) d \mathbf{v}=\int_{x=0}^2 \int_{y=0}^1 \int_{z=0}^3\left(x+2 y+z^2\right) d x d y d z=\int_{x=0}^2 \int_{y=0}^1\left[x z+2 y z+\frac{z^3}{3}\right]_0^3 d x d y\)
= \(\int_{x=0}^2 \int_{y=0}^1(3 x+6 y+9) d x d y=\int_{x=0}^2\left[3 x y+3 y^2+9 y\right]_{y=0}^1 d x\)
= \(\int_{x=0}^2(3 x+12) d x=\left[\frac{3 x^2}{2}+12 x\right]_0^2=6+24=30\)
Example.9 Valuate by Gauss divergence theorem∫ \(\int_S\)4xz dy dz-y2 dz dx+yz x 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:
By divergence theorem \(\int_S \overline{\mathbf{F}} \cdot \mathbf{N} d \mathbf{S}=\int_V d i v \overline{\mathbf{F}} d \mathbf{V}\)
⇒ \(\iint_S \mathbf{f}_1 d y d z+\mathbf{f}_2 d z d x+\mathbf{f}_3 d x d y=\iiint_V\left(\frac{\partial \mathbf{f}_1}{\partial x}+\frac{\partial \mathbf{f}_2}{\partial y}+\frac{\partial \mathbf{f}_3}{\partial z}\right) d x d y d z\)
From the given problem \(\mathbf{f}_1=4 x z, \mathbf{f}_2=-y^2\) and \(\mathbf{f}_3=y z\)
We get \(\iiint_V\left[\frac{\partial}{\partial x}(4 x z)+\frac{\partial}{\partial y}\left(-y^2\right)+\frac{\partial}{\partial z}(y z)\right] d x d y d z\)
= \(\iiint_{x=0}^1(4 z-2 y+y) d x d y d z=\int_{x=0}^1 \int_{y=0}^1 \int_{z=0}^1(4 z-y) d z d y d x\)
= \(\int_{x=0}^1 \int_{x=0}^1\left[2 z^2-y z z_{z=0} d y d x=\int_{x=0}^1 \int_{y=0}^1(2-y) d y d x\)
= \(\int_{x=0}^1\left[2 y-\frac{y^2}{2}\right]_0^1 d x=\int_{x=0}^1\left(2-\frac{1}{2}\right) d x\right.\)
= \(\frac{3}{2} \int_0^1 d x=\frac{3}{2}[x]_0^1=\frac{3}{2}\)
Example.10 If\(\overline{\mathrm{F}}\) =(2x2-3z)\(\overline{\mathrm{i}}\)-2xy\(\overline{\mathrm{j}}\)-4x\(\overline{\mathrm{k}}\) then evaluate \(\int_S\)div\(\overline{\mathrm{F}}\) dv where V is the closed region bounded by the planes x=0, y=0,z=0 and 2x+2y+z=4.
Solution:
div \(\overline{\mathbf{F}}=\nabla \cdot \overline{\mathbf{F}}=\frac{\partial \mathbf{f}_1}{\partial x}+\frac{\partial \mathbf{f}_2}{\partial y}+\frac{\partial \mathbf{f}_3}{\partial z}=4 x-2 x=2 x\)
Limits of z are 0 to 4-(2 x+2 y)
Limits of y are 0 to 2-x
Limits of x are 0 to 2
∴ \(\int_V \nabla \cdot \overline{\mathbf{F}} d \mathbf{V}=\int_{x=0}^2 \int_{y=0}^{2-x} \int_{z=0}^{4-2 x-2 y} 2 x d x d y d z=\int_{x=0}^2 \int_{y=0}^{2-x}[2 x z]_{z=0}^{4-2 x-2 y} d y d x\)
= \(\int_{x=0}^2 \int_{y=0}^{2-x} 2 x(4-2 x-2 y) d y d x=\int_{x=0}^2 \int_{y=0}^{2-x}\left(8 x-4 x^2-4 x y\right) d y d x\)
= \(\int_{x=0}^2\left[8 x y-4 x^2 y-2 x y^2\right]_{y=0}^{2-x} d x=\int_0^2\left[8 x(2-x)-4 x^2(2-x)-2 x(2-x)^2\right] d x\)
= \(\int_0^2\left(2 x^3-8 x^2+8 x\right) d x=\left[\frac{2 x^4}{4}-\frac{8 x^3}{3}+\frac{8 x^2}{2}\right]_0^2=\frac{1}{2} \cdot 2^4-\frac{8}{3} \cdot 2^3+4 \cdot 2^2=\frac{8}{3}\)