Q40P - Introduction to Electrodynamics

Question

Compute the divergence of the function $\mathit{v}\mathbf{=}\mathbf{(}\mathbf{r}\mathbf{ }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{ }\mathbf{\theta }\mathbf{)}\hat{\mathbf{r}}\mathbf{+}\mathbf{(}\mathit{r}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\mathit{\theta }\mathbf{)}\hat{\mathbf{\theta }}\mathbf{+}\mathit{r}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathit{\varphi }\hat{\mathbf{\varphi }}$

Check the divergence theorem for this function, using as your volume the inverted hemispherical bowl of radius R, resting on the xy plane and centered at the origin (Fig. 1.40).

      

                                   

Step-by-Step Solution

Answer

The divergence value for the given function is obtained as $\frac{5}{3}\pi R^3$  . The right and left side of the gauss divergence theorem are computed and obtained to be equal to $\frac{5}{3}\pi R^3$.

Step 1: Define the divergence in spherical coordinates

Consider the vector point function.

$F (x,y,z)=F_1(x,y,z)+F_2(x,y,z)+F_3(x,y,z)$,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 $F(r,\theta ,\varphi )$  in spherical coordinates is

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

 

Here,   are the spherical coordinates.

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

Step: 2 Compute divergence of the function F

The integral of derivative of a function $\mathit{f}\mathbf{ }\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{)}$  over an open surface area is equal to the volume integral of the function.

The given function is  $v=(\mathrm{r} \cos \mathrm{\theta })\hat{\mathrm{r}}+(r sin \theta )\hat{\mathrm{\theta }}+r sin\theta cos\varphi \hat{\mathrm{\varphi }}$ and the del operatoe 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:

  $$\begin{gathered} \nabla .v=F(r,\theta ,\varphi )=\frac{1}{r^2}\frac{\partial \left(r^{2}r\cos \theta \right)}{\partial r}+\frac{1}{r \sin \theta }\frac{\partial (\sin \theta r \sin \theta )}{\partial \theta }+\frac{1}{r \sin \theta }\frac{\partial (r\sin \theta r\cos \varphi }{\partial \varphi } \\ =\frac{1}{r^2}(3r^2 \cos \theta )+\frac{1}{r \sin \theta }(2r \sin \theta \cos \theta )+\frac{1}{r \sin \theta }\left(r \sin \theta \left)\right(-\sin \varphi \right) \\ =3\cos \theta +2\cos \theta -\sin \varphi \\ =5\cos \theta -\sin \varphi \end{gathered}$$

Step 3: Compute the left side of gauss divergence theorem

The surface integral of vector $\stackrel{\rightharpoonup }{v}$, along the path (i) is computed as:

Solve further as, 

Step 4: Compute the right side of gauss divergence theorem

The right side of the gauss divergence theorem is $\oint v.da$  .the surface area has the top and bottom area. Thus the right side can be written as: $\oint v.da={\oint }_{top}v.da+{\oint }_{bottom}v.da$ . For top surface $\oint { }_{top}v.da=\oint v_r.da$  , which is equal to $\pi R^3$ .

 

For bottom surface, ${\oint }_{bottom}v.da$ , can be computed as.

$$\begin{gathered} {\oint }_{bottom}v.da=\oint v_{\theta }.da \\ =\oint r sin\frac{\theta }{2}\left(rdr\right)d\varphi \\ =\frac{2\pi }{3}{R}^3 \end{gathered}$$

The value of${\oint }_{bottom}v.da$ can be computed as follows:

$$\begin{gathered} \oint v.da=\pi R^3+\frac{2}{3}\pi R^3 \\ =\frac{5}{3}\pi R^3 \end{gathered}$$

The right and left side of the gauss divergence theorem are obtained to be equal to $\frac{5}{3}\pi R^3$.

Thus the divergence value is $\frac{5}{3}\pi R^3$  .