Integral Transformations Exercise 7.1
1. if N is the unit outward drawn normally to any closed surface, show that \(\int_V\) div N dV=S [Hint: Apply Gauss’s theorem for the unit vector N]
Solution: By Gauss’s Theorem \(\int_{\mathbf{V}}(\nabla \cdot \mathbf{N}) d \mathbf{V}=\int_{\mathbf{V}} \mathbf{N} \cdot \mathbf{N} d \mathbf{S}=\int_{\mathbf{V}} d \mathbf{S}\)
2. Apply the divergence theorem to evaluate ∫ \(\int_S\) (x+z)dy dz +(y+z)dz dx +(x+y)dx dy where is the surface of the sphere x2+y2+z2=4.
Solution:
Given \(\iint_S(x+z) d y d z+(y+z) d z d x+(z+x) d x d y\)
Here \(\mathbf{F}_1 \stackrel{s}{=} x+z, \mathbf{F}_2=y+z, \mathbf{F}_3=x+y\)
∴ \(\frac{\partial \mathbf{F}_1}{\partial x}=1, \frac{\partial \mathbf{F}_2}{\partial y}=1, \frac{\partial \mathbf{F}_3}{\partial z}=0\)
∴ \(\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}=1+1+0=2\)
By Gauss’s theorem \(\iint_S \mathbf{F}_1 d y d z+\mathbf{F}_2 d z d x+\mathbf{F}_3 d x d y\)
= \(\iiint_V\left(\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}\right) d x d y d z=\iiint 2 d x d y d z=2 \int_V d \mathbf{V}=2 \mathbf{V}\)
= \(2\left[\frac{4}{3} \pi(2)^3\right]=\frac{64 \pi}{3}\) [because For the sphere radius =2]
3. Evalute ∫ \(\int_S\)x dydz + y dzdx + z dxdy where S- is the surface x²+y²+z²=1.
Solution:
Here \(\mathbf{F}_1=x, \mathbf{F}_2=y, \mathbf{F}_3=z\)
∴ \(\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}=1+1+1=3\)
⇒ \(\mathrm{div} \mathbf{F}=3\)
∴ \(\iint_S x d y d z+y d z d x+z d x d y=\int_V \mathrm{div} \mathbf{F} d \mathbf{V}\)
⇒ \(\int_V 3 d \mathbf{V}=3 \mathbf{V}=3 \frac{4}{3} \pi=4 \pi\)……..[radius =1]
4. Find \(\int_S\) F.N dS where F= 2x2-y2j+4xzk and S is the region in the first octant bounded by y2+z2=9 and x=0, x=2
Solution:
⇒ \(\int_S \mathbf{F} . \mathbf{N} d \mathbf{S}=\int_V \text{div} \mathbf{F} d \mathbf{V}\)
= \(\int_V\left[\frac{\partial}{\partial x}\left(2 x^2\right)+\frac{\partial}{\partial y}\left(-y^2\right)+\frac{\partial}{\partial z}(4 x z)\right] d x d y d z\)
= \(\int_{x=0}^2 \int_{y=0}^3 \int_{z=0}^{\sqrt{\left(9-y^2\right)}}(8 x-2 y) d x d y d z=\int_{x=0}^2 \int_{y=0}^3(8 x-2 y) \sqrt{\left(9-y^2\right)} d x d y\)
= \(\int_0^2 8 x\left[\frac{1}{2} y \sqrt{\left(9-y^2\right)}+\frac{1}{2} \cdot 9 \text{Sin}^{-1} \frac{y}{3}\right]_0^3+\left[\frac{2}{3}\left(9-y^2\right)^{3 / 2}\right]_0^3 d x\)
= \(\int_0^2 8 x\left(\frac{9}{2} \cdot \frac{\pi}{2}\right)+\left(0-\frac{2}{3} \cdot 27\right) d x=\left[18 \pi \cdot \frac{x^2}{2}-18 x\right]_0^2=36 \pi-36\)
5. Evalute \(\int_S\) F.N dS where F= 2x2yi-y2j+4xzk taken over the region in the first octant bounded by y2+z2=9 and x=2.
Solution:
Given \(\mathbf{F}=2 x^2 y \mathbf{i}-y^2 \mathbf{j}+4 x z^2 \mathbf{k}\) . ∴ \(\text{div} \mathbf{F}=4 x y-2 y+8 x z\)
∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\int_V \text{div} \mathbf{F} d \mathbf{V}=\int_{x=0}^2 \int_{y=0}^3 \cdot \int_{z=0}^{\sqrt{9-y^2}}(4 x y-2 y+8 x z) d x d y d z\)
= \(\int_{x=0}^2 \int_{y=0}^3[4 x y z-2 y z+4 x z^2 \int_0^{\sqrt{9-y^2}} d x^{\prime} d y\)
= \(\int_{x=0}^2 \int_{y=0}^3\left[4 x y \sqrt{9-y^2}-2 y \sqrt{9-y^2}+4 x\left(9-y^2\right) d x\right] d y\)
= \(\int_{y=0}^3\left[2 x^2 y \sqrt{9-y^2}-2 y \sqrt{9-y^2}+2 x^2\left(9-y^2\right)\right]_{x=0}^2 d y\)
= \(\int_0^3\left[8 y \sqrt{9-y^2}-4 y \sqrt{9-y^2}+8\left(9-y^2\right)\right] d y\)
= \(\int_0^3\left[4 y \sqrt{9-y^2}+8\left(9-y^2\right)\right] d y\)
= \(\left[-\frac{4}{3}\left(9-y^2\right)^{3 / 2}+8\left(9 y-\frac{y^3}{3}\right)\right]^3=36+144=180\)
6. Find \(\int_S\)(4xi-2y2j+z2k .N.dS where S is the region bounded by x2+y=4, z=0 and z=3.
Solution:
Given \(\mathbf{F}=4 x \mathbf{i}-2 y^2 \mathbf{j}+z^2 \mathbf{k}\)
∴ \(\text{div} \mathbf{F}=(4-4 y+2 z)\)
⇒ \(\int_S \mathrm{~F} \cdot \mathrm{N} d \mathrm{~S}=\int_V \text{div} \mathrm{F} d \mathrm{~V}=\int_{x=-2}^2 \int_{y=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{z=0}^3(4-4 y+2 z) d x d y d z\)
= \(\int_{x=-2}^2 \int_{y=-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\left[4 z-4 y z+z^2\right]_{z=0}^3 d x d y\)
= \(\int_{x=-2}^2 \int_{y=-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(21-12 y) d x d y=\int_{x=-2}^2 42 \sqrt{4-x^2} d x=84 \pi\)
7. Using the divergence theorem, show that the volume V of the region bounded by surface S is V = ∫\(\int_S\)x dx dy = ∫\(\int_S\)y dx dz = ∫\(\int_S\)z dx dy =\(\frac{1}{3}\) ∫\(\int_S\)x dx dy + y dx dz + z dx dy
Solution:
By Gauss’s theorem \(\iint_S x d y d z=\int_V \frac{\partial}{\partial x}(x) d \mathbf{V}=\int_V d \boldsymbol{V}=\mathbf{V}\)
Similarly \(\iint_S y d z d x=\int_S^S z d x d y=\mathbf{V}\)
∴ Adding \(3 \mathbf{V}=\iint_S(x d y d z+y d z d x+z d x d y)\)
⇒ \(\mathbf{V}=\frac{1}{3} \iint_S(x d y d z+y d z d x+z d x d y)\)
8. If F= 2xyi-yzj+x2k find \(\int_S\)F.NdS where S is the entire surface of the cube bounded by the coordinate planes and the planes x=a, y=a, z=a by the application of Gauss’s theorem.
Solution:
= \(\int_0^a \int_0^a \int_0^a(2 y-z) d x d y d z=\int_0^a \int_0^a(2 y-z)[x]_0^a d y . d z\)
= \(a \int_0^a\left[y^2-y z\right]_{y=0}^a d z=a \int_0^a\left(a^2-a z\right) d z=a\left[a^2 z-a \frac{z^2}{2}\right]_0^a=\frac{1}{2} a^4\)
9. If F= 4xz\(\overline{\mathrm{i}}\)-y\(\overline{\mathrm{j}}\) =yz\(\overline{\mathrm{k}}\) find \(\int_S\)F.Nds by divergence theorem where S is the surface of the cube bounded by x=0, x=1, y=0,y=1,z=0, z=1.
Solution:
⇒ \(\int_V \text{div} \mathbf{F} d \mathbf{V}=\int_0^1 \int_0^1 \int_0^1(4 z-y) d x d y d z=\int_0^1 \int_0^1(4 z-y) d y d z\)
= \(\int_0^1\left[4 z y-\frac{y^2}{2}\right]_{\nu=0}^1 d z=\int_0^1\left(4 z-\frac{1}{2}\right) d z=\frac{3}{2}\)