91Ó°ÊÓ

The given function is$v=y\text{ }i$ . Stokes theorem has to be verified for the function$v=y\text{ }i$ , over the given triangular path as shown below:

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$∫\left(∇×v\right)â‹\dots \text{ }ds=\underset{l}{âˆ\circledR }vâ‹\dots \text{ }dl$.

Compute the curl of vector v as follows:

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

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

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

The left side of stokes theorem is computed as follows:

$\begin{array}{c}\underset{S}{\overset{}{∫}}\left(∇×\mathbf{v}\right)â‹\dots \stackrel{→}{da}=iâ‹\dots \left(a^{2}i\right)\\ =a^2\end{array}$ …….. (1)

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

$\begin{array}{c}\underset{}{\overset{}{∫}}vâ‹\dots \text{ }dl=\underset{}{\overset{}{∫}}\left(y\text{ }k\right)\left(dx\text{ }i+dy\text{ }j+dz\text{ }k\right)\\ =\underset{}{\overset{}{∫}}y\text{ }dz\end{array}$

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

$\begin{array}{c}∫\text{ }vâ‹\dots dl=∫\left(0\right)\text{ }dz\text{ }\\ =0\end{array}$.

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

$\begin{array}{c}∫vâ‹\dots \text{ }dl=∫y\text{ }dz\\ =0\end{array}$

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

$\begin{array}{c}∫\text{ }vâ‹\dots dl=\underset{2a}{\overset{0}{∫}}y\text{ }dz\text{ }\\ =\underset{2a}{\overset{0}{∫}}y\left(−\frac{1}{2}dy\right)\\ =−\frac{1}{2}\underset{2a}{\overset{0}{∫}}ydy\end{array}$

Solve further as

$\begin{array}{c}∫\text{ }vâ‹\dots dl=−\frac{1}{2}{\left(\frac{y^2}{2}\right)}_{2a}^0\\ =−\frac{1}{4}\left(0−4a^2\right)\\ =a^2\end{array}$

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

$\begin{array}{c}∫vâ‹\dots \text{ }dl=0+a^2+0\\ =a^2\end{array}$ …….. (2)

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