91Ó°ÊÓ

Write the given vector function as,

$v=r^2\cos \mathrm{θ}\text{ }\stackrel{\mathrm{Âì}}{r}+r^2\cos \mathrm{Ï•}\text{ }\stackrel{\mathrm{Âì}}{\mathrm{θ}}−r^2\cos \mathrm{θ}\sin \mathrm{Ï•}\text{ }\stackrel{\mathrm{Âì}}{\mathrm{Ï•}}$

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

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

$∇F\left(r,\text{ }\mathrm{θ},\text{ }\mathrm{ϕ}\right)=\frac{1}{r^2}\frac{∂\left(r^2{F}_1\right)}{∂r}+\frac{1}{r\sin \mathrm{θ}}\frac{∂\left(\sin \mathrm{θ}{F}_2\right)}{∂\mathrm{θ}}+\frac{1}{r\sin \mathrm{θ}}\frac{∂F_3}{∂\mathrm{ϕ}}$

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

The given function is$v=r^2\cos \mathrm{θ}\text{ }\stackrel{\mathrm{Âì}}{r}+r^2\cos \mathrm{Ï•}\text{ }\stackrel{\mathrm{Âì}}{\mathrm{θ}}−r^2\cos \mathrm{θ}\sin \mathrm{Ï•}\text{ }\stackrel{\mathrm{Âì}}{\mathrm{Ï•}}$ . The divergence of vector v is computed as follows:

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

Thus, the divergence of the function is $∇â‹\dots v=4r\cos \mathrm{θ}$.

The volume of integration is octant of radius R, where $\mathrm{θ}$ and $\mathrm{ϕ}$ ranges from 0 to $\frac{\mathrm{π}}{2}$.

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

$\begin{array}{c}\underset{}{\overset{}{∫}}\left(∇v\right)â‹\dots \text{ }d\mathrm{Ï„}=\underset{0}{\overset{R}{∫}}\underset{0}{\overset{\mathrm{Ï€}/2}{∫}}\underset{0}{\overset{\mathrm{Ï€}/2}{∫}}4r\cos \mathrm{θ}\text{ }{r}^{2}dr\text{ }\sin \mathrm{θ}\text{ }d\mathrm{θ}\text{ }d\mathrm{Ï•}\\ =4\underset{0}{\overset{2}{∫}}{r}^{3}dr\underset{0}{\overset{\mathrm{Ï€}/2}{∫}}\cos \mathrm{θ}\sin \mathrm{θ}d\mathrm{θ}\underset{0}{\overset{\mathrm{Ï€}/2}{∫}}d\mathrm{Ï•}\\ =4\left(\frac{R^4}{4}\right){\left(\frac{{\sin }^2\mathrm{θ}}{2}\right)}_0^{\mathrm{Ï€}/2}\left(\frac{\mathrm{Ï€}}{2}\right)\\ =\frac{\mathrm{Ï€}{R}^4}{4}\end{array}$ ……. (1)

The right side of the gauss divergence theorem is $âˆ\circledR vâ‹\dots da$.the surface area has the top and bottom area. Thus the right side can be written as:

$âˆ\circledR vâ‹\dots da=\underset{\left(i\right)}{âˆ\circledR }vâ‹\dots da+\underset{\left(ii\right)}{âˆ\circledR }vâ‹\dots da++\underset{\left(iii\right)}{âˆ\circledR }vâ‹\dots da$. For surface (i), $\underset{\left(i\right)}{âˆ\circledR }vâ‹\dots da=âˆ\circledR v_râ‹\dots da$, $d{a}_1=R^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\text{ }\stackrel{\mathrm{Âì}}{r}$.

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

$\begin{array}{c}\underset{\left(\text{i}\right)}{âˆ\circledR }vâ‹\dots da=\underset{0}{\overset{\frac{\mathrm{Ï€}}{2}}{∫}}\underset{0}{\overset{\frac{\mathrm{Ï€}}{2}}{∫}}\left(R^2\cos \mathrm{θ}\right)R^2\sin \mathrm{θ}d\mathrm{Ï•}\\ =R^4\underset{z=0}{\overset{\frac{\mathrm{Ï€}}{2}}{∫}}\cos \mathrm{θ}\text{ }\sin \mathrm{θ}\underset{0}{\overset{\frac{\mathrm{Ï€}}{2}}{∫}}d\mathrm{Ï•}\\ =R^4{\left(\frac{{\sin }^2\mathrm{θ}}{2}\right)}_0^{\frac{\mathrm{Ï€}}{2}}\left(\frac{\mathrm{Ï€}}{2}\right)\\ =\frac{\mathrm{Ï€}{R}^4}{4}\end{array}$

For left surface, $\underset{\left(ii\right)}{âˆ\circledR }vâ‹\dots da=âˆ\circledR v_{\mathrm{Ï•}}â‹\dots da$, $d{a}_2=r\text{ }dr\text{ }d\mathrm{θ}\text{ }\stackrel{\mathrm{Âì}}{\mathrm{θ}}$, where $\mathrm{Ï•}=0$.

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

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

For back surface, $\underset{\left(iii\right)}{âˆ\circledR }vâ‹\dots da=âˆ\circledR v_{\mathrm{Ï•}}â‹\dots da$,$d{a}_2=r\text{ }dr\text{ }d\mathrm{θ}\text{ }\stackrel{\mathrm{Âì}}{\mathrm{Ï•}}$ , where $\mathrm{Ï•}=\frac{\mathrm{Ï€}}{2}$.

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

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

Solve further as,

$\begin{array}{c}\underset{\left(\text{ii}\right)}{âˆ\circledR }vâ‹\dots da=−\left(\frac{R^4}{4}\right){\left(\sin \mathrm{θ}\right)}_0^{\frac{\mathrm{Ï€}}{2}}\\ =−\frac{1}{4}{R}^4\end{array}$

For bottom surface, $\underset{\left(iv\right)}{âˆ\circledR }vâ‹\dots da=âˆ\circledR v_{\mathrm{θ}}â‹\dots da$, $d{a}_2=r\text{ }dr\text{ }d\mathrm{Ï•}\text{ }\stackrel{\mathrm{Âì}}{\mathrm{θ}}$, where $\mathrm{θ}=\frac{\mathrm{Ï€}}{2}$.

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

$\begin{array}{c}\underset{\left(\text{iv}\right)}{âˆ\circledR }vâ‹\dots da=\underset{0}{\overset{R}{∫}}\underset{0}{\overset{\frac{\mathrm{Ï€}}{2}}{∫}}\left(r^2\cos \mathrm{Ï•}\right)r\text{ }dr\text{ }d\mathrm{Ï•}\text{ }\stackrel{\mathrm{Âì}}{\mathrm{θ}}\\ =\underset{0}{\overset{R}{∫}}{r}^3\text{ }dr\underset{0}{\overset{\frac{\mathrm{Ï€}}{2}}{∫}}\cos \mathrm{Ï•}\text{ }d\mathrm{Ï•}\text{ }\stackrel{\mathrm{Âì}}{\mathrm{Ï•}}\\ =−\frac{1}{4}{R}^4\end{array}$

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

$\begin{array}{c}\underset{}{âˆ\circledR }vâ‹\dots da=\underset{\left(i\right)}{∫}vâ‹\dots da+\underset{\left(ii\right)}{∫}vâ‹\dots da+\underset{\left(iii\right)}{∫}vâ‹\dots da+\underset{\left(iv\right)}{∫}vâ‹\dots da\\ =\frac{\mathrm{Ï€}{R}^4}{4}−0−\frac{1}{4}{R}^4+\frac{1}{4}{R}^4\\ =\frac{\mathrm{Ï€}{R}^4}{4}\end{array}$ …… (2)

From equations (1) and (2), the left and right side of gauss divergence theorem is satisfied and is equal to$\frac{\mathrm{Ï€}{R}^4}{4}$ .