Q59P - Introduction to Electrodynamics

Question

Check the divergence theorem for the function

$\mathit{v}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{\varphi }\mathbf{ }\widehat{\mathbf{r}\mathbf{ }}\mathbf{+}\mathbf{4}{\mathit{r}}^{\mathbf{2}}\mathbf{ }\mathit{c}\mathit{o}\mathit{s}\mathbf{ }\mathit{\theta }\hat{\mathbf{\theta }}\mathbf{+}{\mathit{r}}^{\mathbf{2}}\mathbf{ }\mathit{t}\mathit{a}\mathit{n}\mathbf{ }\mathit{\theta }\hat{\mathbf{\varphi }}$

using the volume of the "ice-cream cone" shown in Fig. 1.52 (the top surface is spherical, with radius R and centered at the origin). [Answer: $\mathit{\pi }{\mathit{R}}^{\mathbf{4}}\mathbf{/}\mathbf{12}\mathbf{ }\mathbf{(}\mathbf{2}\mathit{\pi }\mathbf{+}\mathbf{3}\sqrt{\mathbf{3}}\mathbf{)}$]

Step-by-Step Solution

Answer

From equations (1) and (2), the left and right side of gauss divergence theorem is satisfied.

Step 1: Describe the given information.

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

$\nabla v\left(r, \theta , \varphi \right)=\frac{1}{r^2}\frac{\partial \left(r^2{v}_r\right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(\sin \theta v_{\theta }\right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial v_{\varphi }}{\partial \varphi }$

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

Step 2: Define the divergence in spherical coordinates.

The integral of derivative of a function f (x, y, z) over an open surface area is equal to the volume integral of the function $\mathbf{\int }\left(\nabla ·v\right)\mathbf{·}\mathbf{ }\mathit{d}\mathit{\tau }\mathbf{=}\underset{\mathbf{s}}{\mathbf{\oint }}\mathit{v}\mathbf{·}\mathbf{ }\mathit{d}\mathit{a}$.

 

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

$\nabla v\left(r, \theta , \varphi \right)=\frac{1}{r^2}\frac{\partial \left(r^2{v}_r\right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(\sin \theta v_{\theta }\right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial v_{\varphi }}{\partial \varphi }$

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

Step: 3 Compute divergence of the function F.

The given function is $\text{v}=\left(r^2\sin \theta \right)\stackrel{\wedge }{r}+\left(4r^2\cos \theta \right)\stackrel{\wedge }{\theta }+r^2\tan \theta \stackrel{\wedge }{\varphi }$and the del operator is defined as $\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k$. The divergence of vector v is computed as follows:

$\begin{array}{c}\nabla \cdot v=\frac{1}{r^2}\frac{\partial \left(r^2\left(r^2\sin \theta \right)\right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(\sin \theta \left(4r^2\cos \theta \right)\right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial \left(r^2\tan \theta \right)}{\partial \varphi }\\ =\frac{1}{r^2}\left(4r^3\sin \theta \right)+\frac{1}{r\sin \theta }\left(4r^2\cos 2\theta \right)+\frac{1}{r\sin \theta }\frac{\partial \left(r^2\tan \theta \right)}{\partial \varphi }\\ =\frac{4r}{\sin \theta }\left({\sin }^2\theta +{\cos }^2\theta -{\sin }^2\theta \right)\\ =\frac{{\cos }^2\theta }{\sin \theta }\end{array}$

 

Thus, the divergence of the function is $\nabla \cdot v=\frac{{\cos }^2\theta }{\sin \theta }$.

Step 4: Compute the left side of gauss divergence theorem

The volume mentioned has the radius R units, $\theta$varies from 0 to $\frac{\pi }{6}$,$\varphi$ varies from 0 to $2\mathrm{\pi }$.

 

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

$\begin{array}{c}\underset{}{\overset{}{\int }}\left(\nabla v\right)\cdot d\tau =\underset{0}{\overset{R}{\int }}\underset{0}{\overset{\pi /6}{\int }}\underset{0}{\overset{2\pi }{\int }}\left(4r\frac{{\cos }^2\theta }{\sin \theta }\right)r^2\sin \theta drd\theta d\varphi \\ =\underset{0}{\overset{R}{\int }}4r^{3}dr\underset{0}{\overset{\pi /6}{\int }}{\cos }^2\theta d\theta \underset{0}{\overset{2\pi }{\int }}d\varphi \\ =2\pi R^4{\left(\frac{\theta }{2}+\frac{\sin 2\theta }{4}\right)}_0^{\pi /6}\\ =2\pi R^4\left(\frac{\pi }{12}+\frac{\sin 60}{4}\right)\end{array}$

 

Solve further as,

 $\begin{array}{c}\underset{}{\overset{}{\int }}\left(\nabla v\right)\cdot d\tau =2\pi R^4\left(\frac{\pi +3\sqrt{3/2}}{12}\right)\\ =\frac{\pi R^4}{6}\left(\pi +\frac{3\sqrt{3}}{2}\right)\end{array}$……. (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 of surface of cone has radius R units, $\theta$varies from 0 to $\frac{\pi }{6}$, $\varphi$varies from 0 to $2\mathrm{\pi }$.Thus the right side can be written as:

 

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

 

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

$\begin{array}{c}\underset{}{\oint }v\cdot da=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\underset{0}{\overset{2\pi }{\int }}{R}^4{\sin }^2\theta d\varphi d\theta \\ =2\pi R^4\underset{0}{\overset{\frac{\pi }{6}}{\int }}{\sin }^2\theta d\theta \\ =2\pi R^4{\left[\frac{\theta }{2}-\frac{1}{4}\sin 2\theta \right]}_0^{\frac{\pi }{6}}\\ =2\pi R^4\left[\frac{\pi }{12}-\frac{\sin 60}{4}\right]\end{array}$

 

Solve further as,

$\begin{array}{c}\underset{}{\oint }v\cdot da=2\pi R^4\left[\frac{\pi -3\sqrt{\frac{3}{2}}}{12}\right]\\ =\frac{\pi R^4}{6}\left[\pi -\frac{3\sqrt{3}}{2}\right]\end{array}$

 

The surface area of curved surface of cone has radius varying from 0 to R units,$\theta$is equal to$\frac{\pi }{6}$, $\varphi$varies from 0 to $2\mathrm{\pi }$.

 

For curve surface of cone (ii), $\underset{\left(ii\right)}{\oint }v\cdot da=\oint v_{\theta }\cdot da$,$da=r\sin \theta d\varphi dr\stackrel{\Lambda }{\theta }$ .Thus, the areal vector for surface (ii) is computed as,

$\begin{array}{c}\underset{\left(\text{ii}\right)}{\oint }v\cdot da=\underset{0}{\overset{R}{\int }}\underset{0}{\overset{2\pi }{\int }}4r^3\cos \theta \sin \theta d\varphi dr\\ =\underset{0}{\overset{R}{\int }}{r}^{3}dr\underset{0}{\overset{2\pi }{\int }}\cos \theta \sin \theta d\varphi \\ =\underset{0}{\overset{R}{\int }}{r}^{3}dr\left(\sqrt{3}2\pi \right)\\ =\frac{\sqrt{3}\pi R^4}{2}\end{array}$

 

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

  $\begin{array}{c}\underset{}{\oint }v\cdot da=\underset{\left(i\right)}{\int }v\cdot da+\underset{\left(ii\right)}{\int }v\cdot da\\ =\frac{\pi R^4}{6}\left(\pi -\frac{3\sqrt{3}}{2}\right)+\frac{\sqrt{3}\pi R^4}{2}\\ =\frac{\pi R^4}{6}\left(\pi -\frac{3\sqrt{3}}{2}+3\sqrt{3}\right)\\ =\frac{\pi R^4}{6}\left(\pi +\frac{3\sqrt{3}}{2}\right)\end{array}$                   

                                                                                                 …… (2)

 

From equations (1) and (2), the left and right side of gauss divergence theorem is satisfied.