91影视

Write the given vector function as,

$v=r^2\cos 胃\text{鈥�}\stackrel{蚂}{r}+r^2\cos 蠒\text{鈥�}\stackrel{蚂}{胃}鈭�r^2\cos 胃\sin 蠒\text{鈥�}\stackrel{蚂}{蠒}$

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,$鈭�\left(鈭�鈰�v\right)鈰�\text{鈥�}d蟿=\underset{s}{鈭�}v鈰�\text{鈥�}da$.

The divergence of vector function $F\left(r,\text{鈥�}胃,\text{鈥�}蠒\right)$ in spherical coordinates is

$鈭�F\left(r,\text{鈥�}胃,\text{鈥�}蠒\right)=\frac{1}{r^2}\frac{鈭�\left(r^2{F}_1\right)}{鈭�r}+\frac{1}{r\sin 胃}\frac{鈭�\left(\sin 胃F_2\right)}{鈭�胃}+\frac{1}{r\sin 胃}\frac{鈭�F_3}{鈭�蠒}$

Here, $r,\text{鈥�}胃,\text{鈥�}蠒$are the spherical coordinates.

The given function is$v=r^2\cos 胃\text{鈥�}\stackrel{蚂}{r}+r^2\cos 蠒\text{鈥�}\stackrel{蚂}{胃}鈭�r^2\cos 胃\sin 蠒\text{鈥�}\stackrel{蚂}{蠒}$ . The divergence of vector v is computed as follows:

$\begin{array}{c}鈭�鈰�v=\frac{1}{r^2}\frac{鈭�\left(r^2{v}_r\right)}{鈭�r}+\frac{1}{r\sin 胃}\frac{鈭�\left(\sin 胃\text{鈥�}{v}_{胃}\right)}{鈭�胃}+\frac{1}{r\sin 胃}\frac{鈭�\left(v_{蠒}\right)}{鈭�蠒}\\ =\frac{1}{r^2}\frac{鈭�\left(r^2{r}^2\cos 胃\right)}{鈭�r}+\frac{1}{r\sin 胃}\frac{鈭�\left(\sin 胃\text{鈥�}{r}^2\cos 蠒\right)}{鈭�胃}+\frac{1}{r\sin 胃}\frac{鈭�\left(鈭�r^2\cos 胃\text{鈥�}\sin 蠒\right)}{鈭�蠒}\\ =4r\cos 胃+r\cos 蠒\cot 胃鈭�r\cot 胃\cos 蠒\\ =4r\cos 胃\end{array}$

Thus, the divergence of the function is $鈭�鈰�v=4r\cos 胃$.

The volume of integration is octant of radius R, where $胃$ and $蠒$ ranges from 0 to $\frac{蟺}{2}$.

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

$\begin{array}{c}\underset{}{\overset{}{鈭�}}\left(鈭�v\right)鈰�\text{鈥�}d蟿=\underset{0}{\overset{R}{鈭�}}\underset{0}{\overset{蟺/2}{鈭�}}\underset{0}{\overset{蟺/2}{鈭�}}4r\cos 胃\text{鈥�}{r}^{2}dr\text{鈥�}\sin 胃\text{鈥�}d胃\text{鈥�}d蠒\\ =4\underset{0}{\overset{2}{鈭�}}{r}^{3}dr\underset{0}{\overset{蟺/2}{鈭�}}\cos 胃\sin 胃d胃\underset{0}{\overset{蟺/2}{鈭�}}d蠒\\ =4\left(\frac{R^4}{4}\right){\left(\frac{{\sin }^2胃}{2}\right)}_0^{蟺/2}\left(\frac{蟺}{2}\right)\\ =\frac{蟺R^4}{4}\end{array}$ 鈥︹��. (1)

The right side of the gauss divergence theorem is $鈭�v鈰�da$.the surface area has the top and bottom area. Thus the right side can be written as:

$鈭�v鈰�da=\underset{\left(i\right)}{鈭�}v鈰�da+\underset{\left(ii\right)}{鈭�}v鈰�da++\underset{\left(iii\right)}{鈭�}v鈰�da$. For surface (i), $\underset{\left(i\right)}{鈭�}v鈰�da=鈭�v_r鈰�da$, $d{a}_1=R^2\sin 胃d胃d蠒\text{鈥�}\stackrel{蚂}{r}$.

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

$\begin{array}{c}\underset{\left(\text{i}\right)}{鈭�}v鈰�da=\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\left(R^2\cos 胃\right)R^2\sin 胃d蠒\\ =R^4\underset{z=0}{\overset{\frac{蟺}{2}}{鈭�}}\cos 胃\text{鈥�}\sin 胃\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}d蠒\\ =R^4{\left(\frac{{\sin }^2胃}{2}\right)}_0^{\frac{蟺}{2}}\left(\frac{蟺}{2}\right)\\ =\frac{蟺R^4}{4}\end{array}$

For left surface, $\underset{\left(ii\right)}{鈭�}v鈰�da=鈭�v_{蠒}鈰�da$, $d{a}_2=r\text{鈥�}dr\text{鈥�}d胃\text{鈥�}\stackrel{蚂}{胃}$, where $蠒=0$.

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

$\begin{array}{c}\underset{\left(\text{ii}\right)}{鈭�}v鈰�da=\underset{0}{\overset{R}{鈭�}}\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\left(r^2\cos 胃\sin 蠒\right)r\text{鈥�}dr\text{鈥�}d胃\text{鈥�}\stackrel{蚂}{蠒}\\ =\underset{0}{\overset{R}{鈭�}}\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\left(r^2\cos 胃\sin \left(0\right)\right)r\text{鈥�}dr\text{鈥�}d胃\text{鈥�}\stackrel{蚂}{蠒}\\ =0\end{array}$

For back surface, $\underset{\left(iii\right)}{鈭�}v鈰�da=鈭�v_{蠒}鈰�da$,$d{a}_2=r\text{鈥�}dr\text{鈥�}d胃\text{鈥�}\stackrel{蚂}{蠒}$ , where $蠒=\frac{蟺}{2}$.

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

$\begin{array}{c}\underset{\left(\text{ii}\right)}{鈭�}v鈰�da=\underset{0}{\overset{R}{鈭�}}\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\left(鈭�r^2\cos 胃\sin 蠒\right)r\text{鈥�}dr\text{鈥�}d胃\text{鈥�}\stackrel{蚂}{蠒}\\ =鈭�\underset{0}{\overset{R}{鈭�}}\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\left(r^2\cos 胃\sin \left(\frac{蟺}{2}\right)\right)r\text{鈥�}dr\text{鈥�}d胃\text{鈥�}\stackrel{蚂}{蠒}\\ =鈭�\underset{0}{\overset{R}{鈭�}}\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}{r}^3\cos 胃\text{鈥�}dr\text{鈥�}d胃\text{鈥�}\stackrel{蚂}{蠒}\\ =鈭�\underset{0}{\overset{R}{鈭�}}{r}^{3}dr\text{鈥�}\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\cos 胃\text{鈥夆赌�}d胃\text{鈥�}\end{array}$

Solve further as,

$\begin{array}{c}\underset{\left(\text{ii}\right)}{鈭�}v鈰�da=鈭�\left(\frac{R^4}{4}\right){\left(\sin 胃\right)}_0^{\frac{蟺}{2}}\\ =鈭�\frac{1}{4}{R}^4\end{array}$

For bottom surface, $\underset{\left(iv\right)}{鈭�}v鈰�da=鈭�v_{胃}鈰�da$, $d{a}_2=r\text{鈥�}dr\text{鈥�}d蠒\text{鈥�}\stackrel{蚂}{胃}$, where $胃=\frac{蟺}{2}$.

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

$\begin{array}{c}\underset{\left(\text{iv}\right)}{鈭�}v鈰�da=\underset{0}{\overset{R}{鈭�}}\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\left(r^2\cos 蠒\right)r\text{鈥�}dr\text{鈥�}d蠒\text{鈥�}\stackrel{蚂}{胃}\\ =\underset{0}{\overset{R}{鈭�}}{r}^3\text{鈥�}dr\underset{0}{\overset{\frac{蟺}{2}}{鈭�}}\cos 蠒\text{鈥�}d蠒\text{鈥�}\stackrel{蚂}{蠒}\\ =鈭�\frac{1}{4}{R}^4\end{array}$

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

$\begin{array}{c}\underset{}{鈭�}v鈰�da=\underset{\left(i\right)}{鈭�}v鈰�da+\underset{\left(ii\right)}{鈭�}v鈰�da+\underset{\left(iii\right)}{鈭�}v鈰�da+\underset{\left(iv\right)}{鈭�}v鈰�da\\ =\frac{蟺R^4}{4}鈭�0鈭�\frac{1}{4}{R}^4+\frac{1}{4}{R}^4\\ =\frac{蟺R^4}{4}\end{array}$ 鈥︹�� (2)

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