Q1.34P - Introduction to Electrodynamics

Question

Test Stokes' theorem for the function $\mathit{v}\mathbf{=}\left(xy\right)\mathit{i}\mathbf{ }\mathbf{+}\left(2yz\right)\mathbf{ }\mathit{j}\mathbf{+}\left(3zx\right)\mathbf{ }\mathit{k}$ , using the triangular shaded area of Fig. 1.34.

Step-by-Step Solution

Answer

The left and right side gives same result. Hence, strokes theorem is verified.

 

Step 1: Define the Gauss divergence theorem

The integral derivativeof a function  $\mathit{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$ over an open surface area is equal to the volume integral of the function, $\mathbf{\int }\mathbf{(}\mathbf{\nabla }\mathbf{·}\mathbf{v}\mathbf{)}\mathbf{·}\mathbf{ }\mathit{d}\mathit{l}\mathbf{=}\underset{\mathbf{s}}{\mathbf{\oint }}\mathit{v}\mathbf{·}\mathbf{ }\mathit{d}\mathit{s}$.The right side of the gauss divergence theorem is the surface integral, that is,  $\underset{s}{\oint }v·da$

 

The diagram of the triangular path is shown below:

 

Step 2: Compute the curl of vector v

Let the vector   be defined as $\mathbf{v}=xy i+2yz j+3zx k$ and the $\nabla$ operator be 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:

 $\begin{array}{rcl}\nabla ·v& =& \left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)·\left(xy i+2yz j+3zx k\right)\\ & =& y+2z+3x\\ & & \end{array}$

 

Now, compute the left part of gauss divergence theorem as:

 $\begin{array}{rcl}\int \left(\nabla ·v\right)dv& =& \underset{0}{\overset{2}{\int }}\underset{0}{\overset{2}{\int }}\underset{0}{\overset{2}{\int }}\left(3x+2z+y\right)dxdydz\\ & =& \underset{0}{\overset{2}{\int }}\underset{0}{\overset{2}{\int }}\left(\frac{3}{2}\times 4+2z\times 2+y\times 2\right)dydz\\ & =& \underset{0}{\overset{2}{\int }}\underset{0}{\overset{2}{\int }}\left(6+4z+2y\right)dydz\\ & =& \underset{0}{\overset{2}{\int }}\left(12+8z+4\right)dz\\ & & \end{array}$ 

Solve further as:

 

 $\begin{array}{rcl}\int \left(\nabla ·v\right)dv& =& {\left(12z+4z^2+4z\right)}_0^2\\ & =& \left(24+4\times 4+4z\right)-\left(0\right)\\ & =& 48\\ & & \end{array}$

Step 2:  Compute the left side of strokes theorem

 

The area vector is given by $\overrightarrow{da}= dydz i$ as the open surface area lies in y-z plane. The left part of the strokes theorem is calculated as:

 $\begin{array}{rcl}\underset{S}{\overset{}{\int }}\left(\nabla \times \mathbf{v}\right)·\overrightarrow{da}& =& \underset{S}{\int }\left(-2yi -3zj -xk\right)·\left(dy dz i\right)\\ & =& \underset{S}{\int }-2y dydz\\ & & \end{array}$

The equation of line (i) is $y+z=2$ . Along this line,z varies 0 to 2 and y varies from 0 to $2-z$ . Thus the integral $\underset{S}{\overset{}{\int }}\left(\nabla \times \mathbf{v}\right)·\overrightarrow{da}=\underset{S}{\int }-2y dydz$  can be written as:

$\begin{array}{rcl}\underset{S}{\overset{}{\int }}\left(\nabla \times \mathbf{v}\right)·\overrightarrow{da}& =& \underset{0}{\overset{2}{\int }}dz\underset{0}{\overset{2-z}{\int }}-2y dy\\ & =& \underset{0}{\overset{2}{\int }}dz{\left(\frac{-2y^2}{2}\right)}_0^{2-z}\\ & =& \underset{0}{\overset{2}{\int }}dz{\left(-y^2\right)}_0^{2-z}\\ & =& \underset{0}{\overset{2}{\int }}dz\left(-{\left(2-z\right)}^2-\left(0\right)\right)\\ & & \end{array}$

Solve further as,

 $\begin{array}{rcl}\underset{S}{\overset{}{\int }}\left(\nabla \times \mathbf{v}\right)·\overrightarrow{da}& =& \underset{0}{\overset{2}{\int }}-\left(4+z^2-4z\right)dz\\ & =& -{\left(4z+\frac{z^3}{3}-2z^2\right)}_0^2\\ & =& -\left(\left(8+\frac{8}{3}-8\right)-\left(0\right)\right)\\ & =& -\frac{8}{3}\\ & & \end{array}$

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

 $\begin{array}{rcl}\underset{}{\overset{}{\int }}v· da& =& \underset{}{\overset{}{\int }}\underset{}{\overset{}{\int }}\left(2xzi+\left(x+2\right)j+y\left(z^3-3\right)k\right)\left(dx dy k\right)\\ & =& \underset{0}{\overset{2}{\int }}\underset{0}{\overset{2}{\int }}y\left(z^2-3\right) dx dy\\ & =& \underset{0}{\overset{2}{\int }}\underset{0}{\overset{2}{\int }}y\left(0-3\right) dx dy\\ & =& \underset{0}{\overset{2}{\int }}\underset{0}{\overset{2}{\int }}-3y dx dy\\ & & \end{array}$

Step 3: Compute the right side of strokes theorem

The differential length vector is given by $\overrightarrow{dl}= dx i+dy j+dz k$ . The right part of the strokes theorem is calculated as:

 $\begin{array}{rcl}\underset{}{\overset{}{\int }}v· dl& =& \underset{}{\overset{}{\int }}\left(2xzi+\left(x+2\right)j+y\left(z^3-3\right)k\right)\left(dx i+dy j+dz k\right)\\ & =& \underset{}{\overset{}{\int }}\left(xy\right) dx+\left(2yz\right) dy +\left(3zx\right) dz\\ & & \end{array}$

Along the path (i),$z=x=0$  , thus  $dx= dz=0$ and $y=0$ . Hence the above integral becomes, $\underset{}{\overset{}{\int }}v· dl=\underset{}{\overset{}{\int }}\left(2yz\right) dy$ .

Along the path (ii), $x=0$ , $z=2-y$  thus  $dx=0$ and $dz=-dy$ . Hence the above integral becomes,

 $\begin{array}{rcl}\underset{}{\overset{}{\int }}v· dl& =& \underset{2}{\overset{0}{\int }}2y\left(2-y\right) dy \\ & =& -\underset{0}{\overset{2}{\int }}4y-2y^2 dy \\ & =& {\left(-2y^2-\frac{2}{3}{y}^3\right)}_0^2\\ & =& -\left(8-\frac{16}{3}\right)\\ & =& -\frac{8}{3}\\ & & \end{array}$

Along the path (ii)i, $x=y=0$ , thus $dx=dy=0$  and z varies deom 2 to 0.. Hence the integral  $\underset{}{\overset{}{\int }}v· dl$ becomes 0 along path (iii).

Step 3: Draw a conclusion

The integral of all the three parts are added to give:

$\begin{array}{rcl}\underset{}{\overset{}{\int }}v· dl& =& 0-\frac{8}{3}+0\\ & =& \frac{8}{3}\\ & & \end{array}$

Thus the left and right side gives same result. Hence strokes theorem is verified.