Q58P - Introduction to Electrodynamics

Question

Check Stokes' theorem for the function $v=yi$, using the triangular surface shown in Fig. 1.51. [Answer:  $a^2$],

Step-by-Step Solution

Answer

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

Step 1: Describe the given information

The given function is $v=yi$ . Stokes theorem has to be verified for the function$v=yi$ , over the given triangular path as shown below:

Step 2: Definestokes theorem

The integral of curlof a function $f\left(x,y,z\right)$  over an open surface area is equal to the line integral of the function $\int \left(\nabla \times v\right)\cdot ds=\underset{l}{\oint }v\cdot dl$ .

Step 3: Compute the left side of strokes theorem

Compute the curl of vector  v as follows:

 $\begin{array}{c}\nabla \times \mathbf{v}=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& 0& y\end{array}\right|\\ =\left(1\right)i+0+0\\ =i\end{array}$

 

The area vector is, using the area of triangle, isobtained as

$\begin{array}{c}\overrightarrow{da}=\left(\frac{1}{2}a\right)\left(2a\right)i\\ =a^{2}i\end{array}$ 

 

The left side of stokes theorem is computed as follows:

 $\begin{array}{c}\underset{S}{\overset{}{\int }}\left(\nabla \times \mathbf{v}\right)\cdot \overrightarrow{da}=i\cdot \left(a^{2}i\right)\\ =a^2\end{array}$                              …….. (1)

Step 4: Compute the right side of strokes theorem

The differential length vector is given by $\overrightarrow{dl}=dxi+dyj+dzk$. The right part of the strokes theorem is calculated as:

$\begin{array}{c}\underset{}{\overset{}{\int }}v\cdot dl=\underset{}{\overset{}{\int }}\left(yk\right)\left(dxi+dyj+dzk\right)\\ =\underset{}{\overset{}{\int }}ydz\end{array}$ 

 

Along the path (i), in x-z plane, $z=a-x$ , thus $dx=-dz$ and$y=0$ . Hence the above integral becomes,

$\begin{array}{c}\int v\cdot dl=\int \left(0\right)dz\\ =0\end{array}$ .

 

Along the path (ii), in x-y plane, $dz=0$ hence the line integral becomes,

 $\begin{array}{c}\int v\cdot dl=\int ydz\\ =0\end{array}$

 

Along the path (iii), in y-z plane, $z=a-\frac{y}{2}$ , thus $dz=-\frac{1}{2}dy$  Hence the line integral becomes, 

 $\begin{array}{c}\int v\cdot dl=\underset{2a}{\overset{0}{\int }}ydz\\ =\underset{2a}{\overset{0}{\int }}y\left(-\frac{1}{2}dy\right)\\ =-\frac{1}{2}\underset{2a}{\overset{0}{\int }}ydy\end{array}$

 

Solve further as

 $\begin{array}{c}\int v\cdot dl=-\frac{1}{2}{\left(\frac{y^2}{2}\right)}_{2a}^0\\ =-\frac{1}{4}\left(0-4a^2\right)\\ =a^2\end{array}$

 

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

$\begin{array}{c}\int v\cdot dl=0+a^2+0\\ =a^2\end{array}$                                …….. (2)

 

From equation (1) and (2), the left and right side gives same result. Hence strokes theorem is verified.