Q54P - Introduction to Electrodynamics

Question

Check the divergence theorem for the function

 $\mathit{v}\mathbf{=}\left(r^{2}cos\theta \right)\stackrel{\mathbf{\wedge }}{\mathbf{r}}\mathbf{+}\left(r^{2}cos\varphi \right)\stackrel{\mathbf{\wedge }}{\mathbf{\theta }}\mathbf{+}{\mathit{r}}^{\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathit{s}\mathit{i}\mathit{n}\mathit{\varphi }\stackrel{\mathbf{\wedge }}{\mathbf{\varphi }}$

using as your volume one octant of the sphere of radius R (Fig. 1.48). Make sure you include the entire surface. [Answer: $\mathit{\pi }{\mathit{R}}^{\mathbf{4}}\mathbf{/}\mathbf{4}$ ]

Step-by-Step Solution

Answer

The left and right side of gauss divergence theorem is satisfied and is equal to $\frac{\pi R^4}{4}$  .

Step 1: Define the given information

Write the given vector function as,

 $v=r^2\cos \theta \stackrel{\Lambda }{r}+r^2\cos \varphi \stackrel{\Lambda }{\theta }-r^2\cos \theta \sin \varphi \stackrel{\Lambda }{\varphi }$

Step 2: Define the divergence in spherical coordinates

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,$\int \left(\nabla \cdot v\right)\cdot d\tau =\underset{s}{\oint }v\cdot da$ .

The divergence of vector function $F\left(r,\theta ,\varphi \right)$  in spherical coordinates is

 $\nabla F\left(r,\theta ,\varphi \right)=\frac{1}{r^2}\frac{\partial \left(r^2{F}_1\right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(\sin \theta F_2\right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial F_3}{\partial \varphi }$

 

Here, $r,\theta ,\varphi$ are the spherical coordinates.

Step 3: Compute divergence of the function F

The given function is $v=r^2\cos \theta \stackrel{\Lambda }{r}+r^2\cos \varphi \stackrel{\Lambda }{\theta }-r^2\cos \theta \sin \varphi \stackrel{\Lambda }{\varphi }$  . The divergence of vector v is computed as follows:

 $\begin{array}{c}\nabla \cdot v=\frac{1}{r^2}\frac{\partial \left(r^2{v}_r\right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(\sin \theta v_{\theta }\right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial \left(v_{\varphi }\right)}{\partial \varphi }\\ =\frac{1}{r^2}\frac{\partial \left(r^2{r}^2\cos \theta \right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(\sin \theta r^2\cos \varphi \right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial \left(-r^2\cos \theta \sin \varphi \right)}{\partial \varphi }\\ =4r\cos \theta +r\cos \varphi \cot \theta -r\cot \theta \cos \varphi \\ =4r\cos \theta \end{array}$

 

 

Thus, the divergence of the function is $\nabla \cdot v=4r\cos \theta$ .

Step 4: Compute the left side of gauss divergence theorem

The volume of integration is octant of radius R, where  $\theta$ and  $\varphi$ ranges from 0 to $\frac{\pi }{2}$ .

 

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

 $\begin{array}{c}\underset{}{\overset{}{\int }}\left(\nabla v\right)\cdot d\tau =\underset{0}{\overset{R}{\int }}\underset{0}{\overset{\pi /2}{\int }}\underset{0}{\overset{\pi /2}{\int }}4r\cos \theta r^{2}dr\sin \theta d\theta d\varphi \\ =4\underset{0}{\overset{2}{\int }}{r}^{3}dr\underset{0}{\overset{\pi /2}{\int }}\cos \theta \sin \theta d\theta \underset{0}{\overset{\pi /2}{\int }}d\varphi \\ =4\left(\frac{R^4}{4}\right){\left(\frac{{\sin }^2\theta }{2}\right)}_0^{\pi /2}\left(\frac{\pi }{2}\right)\\ =\frac{\pi R^4}{4}\end{array}$          ……. (1)

Step 5: Compute the right side of gauss divergence theorem

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

 

$\oint v\cdot da=\underset{\left(i\right)}{\oint }v\cdot da+\underset{\left(ii\right)}{\oint }v\cdot da++\underset{\left(iii\right)}{\oint }v\cdot da$ . For surface (i), $\underset{\left(i\right)}{\oint }v\cdot da=\oint v_r\cdot da$ , $d{a}_1=R^2\sin \theta d\theta d\varphi \stackrel{\Lambda }{r}$ .

 

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

 $\begin{array}{c}\underset{\left(\text{i}\right)}{\oint }v\cdot da=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(R^2\cos \theta \right)R^2\sin \theta d\varphi \\ =R^4\underset{z=0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \sin \theta \underset{0}{\overset{\frac{\pi }{2}}{\int }}d\varphi \\ =R^4{\left(\frac{{\sin }^2\theta }{2}\right)}_0^{\frac{\pi }{2}}\left(\frac{\pi }{2}\right)\\ =\frac{\pi R^4}{4}\end{array}$

 

For left surface, $\underset{\left(ii\right)}{\oint }v\cdot da=\oint v_{\varphi }\cdot da$ , $d{a}_2=rdrd\theta \stackrel{\Lambda }{\theta }$ , where $\varphi =0$ .

 

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

 $\begin{array}{c}\underset{\left(\text{ii}\right)}{\oint }v\cdot da=\underset{0}{\overset{R}{\int }}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(r^2\cos \theta \sin \varphi \right)rdrd\theta \stackrel{\Lambda }{\varphi }\\ =\underset{0}{\overset{R}{\int }}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(r^2\cos \theta \sin \left(0\right)\right)rdrd\theta \stackrel{\Lambda }{\varphi }\\ =0\end{array}$

 

For back surface, $\underset{\left(iii\right)}{\oint }v\cdot da=\oint v_{\varphi }\cdot da$ ,$d{a}_2=rdrd\theta \stackrel{\Lambda }{\varphi }$  , where $\varphi =\frac{\pi }{2}$ .

 

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

 $\begin{array}{c}\underset{\left(\text{ii}\right)}{\oint }v\cdot da=\underset{0}{\overset{R}{\int }}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(-r^2\cos \theta \sin \varphi \right)rdrd\theta \stackrel{\Lambda }{\varphi }\\ =-\underset{0}{\overset{R}{\int }}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(r^2\cos \theta \sin \left(\frac{\pi }{2}\right)\right)rdrd\theta \stackrel{\Lambda }{\varphi }\\ =-\underset{0}{\overset{R}{\int }}\underset{0}{\overset{\frac{\pi }{2}}{\int }}{r}^3\cos \theta drd\theta \stackrel{\Lambda }{\varphi }\\ =-\underset{0}{\overset{R}{\int }}{r}^{3}dr\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta d\theta \end{array}$

 

Solve further as,

 $\begin{array}{c}\underset{\left(\text{ii}\right)}{\oint }v\cdot da=-\left(\frac{R^4}{4}\right){\left(\sin \theta \right)}_0^{\frac{\pi }{2}}\\ =-\frac{1}{4}{R}^4\end{array}$

 

For bottom surface, $\underset{\left(iv\right)}{\oint }v\cdot da=\oint v_{\theta }\cdot da$ , $d{a}_2=rdrd\varphi \stackrel{\Lambda }{\theta }$ , where $\theta =\frac{\pi }{2}$ .

 

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

 $\begin{array}{c}\underset{\left(\text{iv}\right)}{\oint }v\cdot da=\underset{0}{\overset{R}{\int }}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(r^2\cos \varphi \right)rdrd\varphi \stackrel{\Lambda }{\theta }\\ =\underset{0}{\overset{R}{\int }}{r}^{3}dr\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \varphi d\varphi \stackrel{\Lambda }{\varphi }\\ =-\frac{1}{4}{R}^4\end{array}$

 

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

 $\begin{array}{c}\underset{}{\oint }v\cdot da=\underset{\left(i\right)}{\int }v\cdot da+\underset{\left(ii\right)}{\int }v\cdot da+\underset{\left(iii\right)}{\int }v\cdot da+\underset{\left(iv\right)}{\int }v\cdot da\\ =\frac{\pi R^4}{4}-0-\frac{1}{4}{R}^4+\frac{1}{4}{R}^4\\ =\frac{\pi R^4}{4}\end{array}$                          …… (2)

 

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