91Ó°ÊÓ

The divergence of vector function $v\left(r, \mathrm{θ}, \mathrm{ϕ}\right)$in spherical coordinates is

$∇v\left(r, \mathrm{θ}, \mathrm{ϕ}\right)=\frac{1}{r^2}\frac{∂\left(r^2{v}_r\right)}{∂r}+\frac{1}{r\sin \mathrm{θ}}\frac{∂\left(\sin \mathrm{θ}{v}_{\mathrm{θ}}\right)}{∂\mathrm{θ}}+\frac{1}{r\sin \mathrm{θ}}\frac{∂v_{\mathrm{ϕ}}}{∂\mathrm{ϕ}}$

Here, $r, \mathrm{θ}, \mathrm{ϕ}$are the spherical coordinates.

The integral of derivative of a function f (x, y, z) over an open surface area is equal to the volume integral of the function$\mathbf{∫}\left(∇·v\right)\mathbf{·}\mathbf{ }\mathit{d}\mathit{Ï„}\mathbf{=}\underset{\mathbf{s}}{\mathbf{âˆ\circledR }}\mathit{v}\mathbf{·}\mathbf{ }\mathit{d}\mathit{a}$.

The divergence of vector function $v\left(r, \mathrm{θ}, \mathrm{ϕ}\right)$in spherical coordinates is

$∇v\left(r, \mathrm{θ}, \mathrm{ϕ}\right)=\frac{1}{r^2}\frac{∂\left(r^2{v}_r\right)}{∂r}+\frac{1}{r\sin \mathrm{θ}}\frac{∂\left(\sin \mathrm{θ}{v}_{\mathrm{θ}}\right)}{∂\mathrm{θ}}+\frac{1}{r\sin \mathrm{θ}}\frac{∂v_{\mathrm{ϕ}}}{∂\mathrm{ϕ}}$

Here $r, \mathrm{θ}, \mathrm{ϕ}$are the spherical coordinates.

The given function is $\text{v}=\left(r^2\sin \mathrm{θ}\right)\text{ }\stackrel{âˆ\S }{r}+\left(4r^2\cos \mathrm{θ}\right)\text{ }\stackrel{âˆ\S }{\mathrm{θ}}+r^2\tan \mathrm{θ}\text{ }\stackrel{âˆ\S }{\mathrm{Ï•}}$and the del operator is defined as $∇=\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k$. The divergence of vector v is computed as follows:

$\begin{array}{c}∇â‹\dots v=\frac{1}{r^2}\frac{∂\left(r^2\left(r^2\sin \mathrm{θ}\right)\right)}{∂r}+\frac{1}{r\sin \mathrm{θ}}\frac{∂\left(\sin \mathrm{θ}\left(4r^2\cos \mathrm{θ}\right)\right)}{∂\mathrm{θ}}+\frac{1}{r\sin \mathrm{θ}}\frac{∂\left(r^2\tan \mathrm{θ}\right)}{∂\mathrm{Ï•}}\\ =\frac{1}{r^2}\left(4r^3\sin \mathrm{θ}\right)+\frac{1}{r\sin \mathrm{θ}}\left(4r^2\cos 2\mathrm{θ}\right)+\frac{1}{r\sin \mathrm{θ}}\frac{∂\left(r^2\tan \mathrm{θ}\right)}{∂\mathrm{Ï•}}\\ =\frac{4r}{\sin \mathrm{θ}}\left({\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}−{\sin }^2\mathrm{θ}\right)\\ =\frac{{\cos }^2\mathrm{θ}}{\sin \mathrm{θ}}\end{array}$

Thus, the divergence of the function is $∇â‹\dots v=\frac{{\cos }^2\mathrm{θ}}{\sin \mathrm{θ}}$.

The volume mentioned has the radius R units, $\mathrm{θ}$varies from 0 to $\frac{\mathrm{π}}{6}$,$\mathrm{ϕ}$varies from 0 to $2\mathrm{π}$.

The surface integral of vector $\stackrel{→}{v}$, over the volume is computed as:

$\begin{array}{c}\underset{}{\overset{}{∫}}\left(∇v\right)â‹\dots \text{ }d\mathrm{Ï„}=\underset{0}{\overset{R}{∫}}\underset{0}{\overset{\mathrm{Ï€}/6}{∫}}\underset{0}{\overset{2\mathrm{Ï€}}{∫}}\left(4r\frac{{\cos }^2\mathrm{θ}}{\sin \mathrm{θ}}\right)r^2\sin \mathrm{θ}\text{ }drd\mathrm{θ}\text{ }d\mathrm{Ï•}\text{ }\\ =\underset{0}{\overset{R}{∫}}4r^{3}dr\underset{0}{\overset{\mathrm{Ï€}/6}{∫}}{\cos }^2\mathrm{θ}\text{ }d\mathrm{θ}\underset{0}{\overset{2\mathrm{Ï€}}{∫}}d\mathrm{Ï•}\text{ }\\ =2\mathrm{Ï€}{R}^4{\left(\frac{\mathrm{θ}}{2}+\frac{\sin 2\mathrm{θ}}{4}\right)}_0^{\mathrm{Ï€}/6}\\ =2\mathrm{Ï€}{R}^4\left(\frac{\mathrm{Ï€}}{12}+\frac{\sin 60}{4}\right)\end{array}$

Solve further as,

$\begin{array}{c}\underset{}{\overset{}{∫}}\left(∇v\right)â‹\dots \text{ }d\mathrm{Ï„}=2\mathrm{Ï€}{R}^4\left(\frac{\mathrm{Ï€}+3\sqrt{3/2}}{12}\right)\\ =\frac{\mathrm{Ï€}{R}^4}{6}\left(\mathrm{Ï€}+\frac{3\sqrt{3}}{2}\right)\end{array}$……. (1)

The right side of the gauss divergence theorem is $âˆ\circledR vâ‹\dots da$.The surface area of surface of cone has radius R units, $\mathrm{θ}$varies from 0 to $\frac{\mathrm{Ï€}}{6}$, $\mathrm{Ï•}$varies from 0 to $2\mathrm{Ï€}$.Thus the right side can be written as:

For surface (i), $\underset{}{âˆ\circledR }vâ‹\dots da=âˆ\circledR v_râ‹\dots da$, $da=R^2\sin \mathrm{θ}d\mathrm{Ï•}d\mathrm{θ}\stackrel{\mathrm{Âì}}{r}$.

Thus, the areal vector for surface (i) is computed as,

$\begin{array}{c}\underset{}{âˆ\circledR }vâ‹\dots da=\underset{0}{\overset{\frac{\mathrm{Ï€}}{6}}{∫}}\underset{0}{\overset{2\mathrm{Ï€}}{∫}}{R}^4{\sin }^2\mathrm{θ}d\mathrm{Ï•}d\mathrm{θ}\\ =2\mathrm{Ï€}{R}^4\underset{0}{\overset{\frac{\mathrm{Ï€}}{6}}{∫}}{\sin }^2\mathrm{θ}d\mathrm{θ}\\ =2\mathrm{Ï€}{R}^4{\left[\frac{\mathrm{θ}}{2}−\frac{1}{4}\sin 2\mathrm{θ}\right]}_0^{\frac{\mathrm{Ï€}}{6}}\\ =2\mathrm{Ï€}{R}^4\left[\frac{\mathrm{Ï€}}{12}−\frac{\sin 60}{4}\right]\end{array}$

Solve further as,

$\begin{array}{c}\underset{}{âˆ\circledR }vâ‹\dots da=2\mathrm{Ï€}{R}^4\left[\frac{\mathrm{Ï€}−3\sqrt{\frac{3}{2}}}{12}\right]\\ =\frac{\mathrm{Ï€}{R}^4}{6}\left[\mathrm{Ï€}−\frac{3\sqrt{3}}{2}\right]\end{array}$

The surface area of curved surface of cone has radius varying from 0 to R units,$\mathrm{θ}$is equal to$\frac{\mathrm{π}}{6}$, $\mathrm{ϕ}$varies from 0 to $2\mathrm{π}$.

For curve surface of cone (ii), $\underset{\left(ii\right)}{âˆ\circledR }vâ‹\dots da=âˆ\circledR v_{\mathrm{θ}}â‹\dots da$,$da=r\sin \mathrm{θ}d\mathrm{Ï•}dr\text{ }\stackrel{\mathrm{Âì}}{\mathrm{θ}}$.Thus, the areal vector for surface (ii) is computed as,

$\begin{array}{c}\underset{\left(\text{ii}\right)}{âˆ\circledR }vâ‹\dots da=\underset{0}{\overset{R}{∫}}\underset{0}{\overset{2\mathrm{Ï€}}{∫}}4r^3\cos \mathrm{θ}\sin \mathrm{θ}d\mathrm{Ï•}dr\\ =\underset{0}{\overset{R}{∫}}{r}^{3}dr\underset{0}{\overset{2\mathrm{Ï€}}{∫}}\cos \mathrm{θ}\sin \mathrm{θ}d\mathrm{Ï•}\\ =\underset{0}{\overset{R}{∫}}{r}^{3}dr\left(\sqrt{3}2\mathrm{Ï€}\right)\\ =\frac{\sqrt{3}\mathrm{Ï€}{R}^4}{2}\end{array}$

Thus, the total areal vector for the total surface area is computed as,

$\begin{array}{c}\underset{}{âˆ\circledR }vâ‹\dots da=\underset{\left(i\right)}{∫}vâ‹\dots da+\underset{\left(ii\right)}{∫}vâ‹\dots da\\ =\frac{\mathrm{Ï€}{R}^4}{6}\left(\mathrm{Ï€}−\frac{3\sqrt{3}}{2}\right)+\frac{\sqrt{3}\mathrm{Ï€}{R}^4}{2}\\ =\frac{\mathrm{Ï€}{R}^4}{6}\left(\mathrm{Ï€}−\frac{3\sqrt{3}}{2}+3\sqrt{3}\right)\\ =\frac{\mathrm{Ï€}{R}^4}{6}\left(\mathrm{Ï€}+\frac{3\sqrt{3}}{2}\right)\end{array}$

…… (2)

From equations (1) and (2), the left and right side of gauss divergence theorem is satisfied.