Q43P - Introduction to Electrodynamics

Question

Question:  (a) Find the divergence of the function  

$\mathit{v}\mathbf{=}\mathit{s}\mathbf{ }\mathbf{(}\mathbf{2}\mathbf{+}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\varphi }\mathbf{)}\hat{\mathbf{s}}\mathbf{+}\mathbf{\left(}\mathit{s}\mathit{s}\mathit{i}\mathit{n}\mathit{\varphi }\mathit{c}\mathit{o}\mathit{s}\mathit{\varphi }\mathbf{\right)}\hat{\mathbf{\varphi }}\mathbf{+}\mathbf{3}\mathit{z}\hat{\mathbf{z}}$

(b) Test the divergence theorem for this function, using the quarter-cylinder (radius 2, height 5) shown in Fig. 1.43.

(c) Find the curl of v.

Step-by-Step Solution

Answer

 

(a) The divergence of the function is $\nabla ·v=8$ .

 

(b)The gauss divergence theorem, using the quarter-cylinder, is verified.

 

(c)The curl of v is $\nabla ·v=0$ .

Step 1: Define the divergence in spherical coordinates

The integral of the derivative of a function over an open surface area is equal to the volume integral of the function.

 $\mathbf{\oint }\mathbf{(}\mathbf{\nabla }\mathbf{·}\mathit{v}\mathbf{)}\mathbf{·}\mathit{d}\mathit{\tau }\mathbf{=}\mathbf{\oint }\mathit{v}\mathbf{ }\mathit{d}\mathit{a}\mathbf{.}$

Consider the vector point function $F(x,y,z)=F_1(x,y,z)\mathrm{i}+F_2(x,y,z)\mathrm{j}+F_3(x,y,z)\mathrm{k} ,$ where $F_1,F_2,F_3$ are components of $F(x,y,z)$
 .

The divergence of function $F(x,y,z)$ is computed as follows:

$\nabla F(x,y,z)=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}$

 

Here, $\frac{\partial F_1}{\partial x},\frac{\partial F_2}{\partial y},\frac{\partial F_3}{\partial z}$ are the partial derivatives of function $F(x,y,z)$ with respect to $x,y,z$ .

 

The divergence of vector function in spherical coordinates is

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

 

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

Step: 2 Compute divergence of the function F

(a)

The given function is $v=s(2+{\sin }^2\theta )s+\left(s\sin \theta \cos \theta \right)\varphi +3z\hat{z}$ , and the del [S1] operator is defined as . The divergence of vector v is computed as follows:

 $$\begin{gathered} \nabla ·v=\nabla F(r,\theta ,\varphi )=\frac{1}{s}\frac{\partial \left(s{v}_s\right)}{\partial s}+\frac{1}{s}\frac{\partial \left(v_{\varphi }\right)}{\partial \theta }+\frac{1}{r \sin \theta }\frac{\partial \left(v_z\right)}{\partial z} \\ =\frac{1}{s}\frac{\partial \left(s\right(s\left(2\right)+{\sin }^2\theta \left)\right))}{\partial s}+\frac{1}{s}\frac{\partial \left(s\sin \varphi \cos \varphi \right)}{\partial \varphi }+\frac{\partial \left(3z\right)}{\partial z} \\ =\frac{1}{s}\frac{\partial \left(s^2\right(2)+{\sin }^2\theta ))}{\partial s}+\frac{1}{s}\frac{\partial \left(s\sin \varphi \cos \varphi \right)}{\partial \varphi }+\frac{\partial \left(3z\right)}{\partial z} \\ =4+2{\sin }^2\varphi +{\cos }^2\varphi -{\sin }^2+3 \end{gathered}$$ .

Solving further

$$\begin{gathered} \nabla .v=4+({\sin }^2\varphi +{\cos }^2\varphi )-3 \\ =4+\left(1\right)+3 \\ =8 \end{gathered}$$

 

Thus, the divergence of the function is $\nabla .v=8$

Step 3: Determine the total areal vector

(b)

The cylinder mentioned has a height of 5 units and a radius of 2 units, as shown below:

The surface integral of vector , along the path (i) is computed as follows: 

$$\begin{gathered} \int (\nabla v)·d\tau ={\int }_0^5{\int }_0^{n/2}{\int }_0^{2}8sdsd\varphi dz \\ =8{\int }_0^{2}sds{\int }_0^{n/2}d\varphi {\int }_0^{5}dz \\ =8\left(2\right)\left(\frac{\pi }{5}\right)5 \\ =40\pi \end{gathered}$$ 

                     ……. (1)

 

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

 $$\begin{gathered} \oint v·da={\oint }_{\left(i\right)}v·da+{\oint }_{\left(ii\right)}v·da++{\oint }_{\left(iii\right)}v·\mathrm{da}.\mathrm{for} \mathrm{surface} \left(i\right) {\oint }_{\left(i\right)}v·=\oint v_r·da,a=and \\ d{a}_1=sd\varphi dz\hat{z} \end{gathered}$$

 

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

$$\begin{gathered} {\oint }_{\left(i\right)}v·da={\int }_{z-0}^5{\int }_{\varphi -0}^{\frac{n}{2}}{s}^2(2+{\sin }^2\varphi )d\phi dz \\ ={\int }_{z-0}^5{{\left(2\varphi +\left(\frac{\varphi }{2}-\frac{\sin 2\varphi }{4}\right)\right)}_0}^{\frac{\pi }{2}}dz \\ =4\left(\frac{5\pi }{2}\right){\left(z\right)}_0^5 =25\pi \end{gathered}$$

 

For surface (ii) ${\oint }_{\left(i\right)}v·da=\oint v_z·da$ , and $d{a}_2=sd\varphi dz\hat{z} ,$ where z = 5 .

 

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

$$\begin{gathered} {\oint }_{\left(ii\right)}v·da={\int }_{z-0}^2{\int }_{\varphi -0}^{\frac{n}{2}}\left(3z\right)sd\varphi dz \\ =15\pi \end{gathered}$$

 

For surface (iii) ${\oint }_{\left(iii\right)}v·da=\oint v·da,$ and $d{a}_3=sdsd\varphi \hat{z},$ where z = 0 .

 

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

$$\begin{gathered} {\oint }_{\left(iii\right)}v·da={\int }_{z-0}^2{\int }_{\varphi -0}^{\frac{n}{2}}(-3z)sd\varphi dz \\ =0 \end{gathered}$$

 

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

 $$\begin{gathered} {\oint }_{\left(i\right)}v·da={\oint }_{\left(i\right)}v·da+{\oint }_{\left(ii\right)}v·da+{\oint }_{\left(iii\right)}v·da \\ =25\pi +15\pi +0 \\ =40\pi \end{gathered}$$

                          …… (2)

 

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

Step 6: Compute the curl of vector v

(c)

The curl of vector is calculated as follows:

 $$\begin{gathered} \nabla \times v=\left\{\frac{1}{s}\left(\frac{\partial v_z}{\partial \varphi }\right)-\frac{\partial v_{\varphi }}{\partial \varphi }\right\}\hat{s}\left(\frac{\partial v_z}{\partial z}-\frac{\partial v_z}{\partial s}\right)\hat{\varphi }+\frac{1}{s}\left\{\frac{1}{\partial s}\left(s{v}_{\varphi }\right)-\frac{\partial v_s}{\partial \varphi }\right\}\hat{z} \\ =\left\{\frac{1}{s}\left(\frac{\partial v_z}{\partial \varphi }\right)-\frac{\partial ( s \sin \varphi \cos \varphi )}{\partial \varphi }\right\}\hat{s}+\left(\frac{\partial \left(s\right(2+{\sin }^2\varphi )}{\partial z}-\frac{\partial \left(\right(3z\left)\right)}{\partial s}\right)\hat{\varphi } \\ =+\frac{1}{s}\left\{\frac{1}{\partial s}\left(s\right(s \sin \varphi \cos \varphi \left)\right)-\frac{\partial \left(s\right(2+{\sin }^2\varphi \left)\right)}{\partial \varphi }\right\}\hat{z} \\ =\left\{\frac{1}{s}\times 0-0\right\}\hat{s}+\left\{0+0\right\}\hat{\varphi }\frac{1}{s}\left\{2s \sin \varphi \cos \varphi -2s \sin \varphi \cos \varphi \right\}\hat{z} \\ =0 \end{gathered}$$

 

Therefore, the curl of v is $\nabla \times v=0$