91影视

The divergence of function ${\mathit{v}}_{\mathbf{1}\mathbf{}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$ is defined as$\mathbf{鈭�}\mathbf{.}{\mathit{v}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{r}}\mathbf{\left(}{\mathit{r}}^{\mathbf{2}}{\mathit{v}}_{\mathbf{r}}\mathbf{\right)}$.The function $v_1=r^2\hat{r}$. Now, find the divergence of $v_1=r^2\hat{r}$as:

$$\begin{gathered} 鈭�.v_1=\frac{1}{{\mathrm{r}}^2}\frac{鈭�}{鈭�\mathrm{r}}(r^2\hat{r}) \\ =\frac{1}{r^2}4r^3 \\ =4r \end{gathered}$$

The volume integral is taken over the surface of radius R. Now, compute the left part of gauss divergence theorem as:

$$\begin{gathered} 鈭�\left(鈭�.v_1\right)dv=\underset{0}{\overset{2蟺}{鈭�}}\underset{0}{\overset{蟺}{鈭�}}\underset{0}{\overset{R}{鈭�}}\left(4r\right)r^2\sin 胃drd胃d蠒 \\ =4\left(\frac{R^2}{4}\right)\underset{0}{\overset{蟺}{鈭�}}\sin 胃d胃\underset{0}{\overset{2蟺}{鈭�}}d蠒 \\ =4蟺R^2 \end{gathered}$$

Taking the area integration over the surface area of sphere of radius R, to calculate the right side of Gauss divergence theorem, the differential area becomes, $da=r^2\sin 胃d胃d蠒$.

The right side of Gauss divergence theorem is computed as follows:

$\begin{array}{rcl}鈭�v_1.da& =& 鈭�(r^2\sin 胃drd胃d蠒)r^2\\ & =& \underset{0}{\overset{2蟺}{鈭�}}\underset{0}{\overset{蟺}{鈭�}}{r}^4\sin 胃d胃d蠒\\ & =& 4蟺R^2\end{array}$

Thus, the left and right side of gauss divergence theorem are equal. Hence, the theorem is verified for the function in part (a).

The function is given by $v_2=\left(\frac{1}{{\mathrm{r}}^2}\right)\hat{\mathrm{r}}$. Now find the divergence of $v_2=\left(\frac{1}{{\mathrm{r}}^2}\right)\hat{\mathrm{r}}$as:

$$\begin{gathered} 鈭�.v_1=\frac{1}{{\mathrm{r}}^2}\frac{鈭�}{鈭�\mathrm{r}}{r}^2\left(\frac{1}{r^2}\hat{r}\right) \\ =\frac{1}{r^2}\frac{鈭�}{鈭�r}\left(1\right) \\ =0 \end{gathered}$$

The volume integral is taken over the surface of radius R. As$鈭�.v_2=0$, then the left part of gauss divergence theorem is computed as:$鈭�(鈭�-v_1)dv=0$

Taking the area integration over the surface area of sphere of radius R, to calculate the right side of Gauss divergence theorem, then differential area becomes $da=r^2\sin 胃d胃d蠒$.

The right side of Gauss divergence theorem is computed as follows:

$\begin{array}{rcl}鈭�v_1.da& =& 鈭�(r^2\sin 胃drd胃d蠒)\left(\frac{1}{r^2}\right)\\ & =& \underset{0}{\overset{2蟺}{鈭�}}\underset{0}{\overset{蟺}{鈭�}}\sin 胃d胃d蠒\\ & =& 4蟺\end{array}$

Thus, the left and right side of gauss divergence theorem are not obtained to be equal. Hence, the theorem is not verified for the function in part (b).