Question

(a) Check the divergence theorem for the function ${\mathit{v}}_{\mathbf{1}}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathbf{}\hat{\mathbf{r}}$, using as your volume the sphere of radius R, centred at the origin.

(b) Do the same for ${\mathit{v}}_{\mathbf{2}}\mathbf{=}\mathbf{\left(}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{\right)}\hat{\mathbf{r}}$. (If the answer surprises you, look back at Prob. 1.16)

Short answer

(a) The left and right side of the gauss divergence theorem is equal. Thus, the gauss divergence theorem is proved for $v_1=r^2\hat{\mathrm{r}}$.

(b) The left and right side of the gauss divergence theorem is not equal. Thus, the gauss divergence theorem is not proved for $v_2=\left(\frac{1}{{\mathrm{r}}^2}\right)\hat{\mathrm{r}}$.

Step-by-step solution

Verify the divergence theorem for function v1 (r)  in part (a)

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

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

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

$$\begin{gathered} \int \left(\nabla .v_1\right)dv=\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{R}{\int }}\left(4r\right)r^2\sin \theta drd\theta d\varphi \\ =4\left(\frac{R^2}{4}\right)\underset{0}{\overset{\pi }{\int }}\sin \theta d\theta \underset{0}{\overset{2\pi }{\int }}d\varphi \\ =4\pi R^2 \end{gathered}$$

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, $da=r^2\sin \theta d\theta d\varphi$.

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

$\begin{array}{rcl}\int v_1.da& =& \int (r^2\sin \theta drd\theta d\varphi )r^2\\ & =& \underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\pi }{\int }}{r}^4\sin \theta d\theta d\varphi \\ & =& 4\pi R^2\end{array}$

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

Verify the divergence theorem for function v2=(1r2)r^ in part (b)

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:

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

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

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 $da=r^2\sin \theta d\theta d\varphi$.

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

$\begin{array}{rcl}\int v_1.da& =& \int (r^2\sin \theta drd\theta d\varphi )\left(\frac{1}{r^2}\right)\\ & =& \underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\pi }{\int }}\sin \theta d\theta d\varphi \\ & =& 4\pi \end{array}$

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).