91Ó°ÊÓ

The given path is circular of radius R is shown as follows:

The vector v is given as $v=ayi +bx j$.

The integral of curl of a functionf (x, y, z) over an open surface area is equal to the line integral of the function $\underset{\mathbf{s}}{\mathbf{∫}}\left(∇×v\right)\mathbf{·}\mathbf{ }\mathit{d}\mathit{s}\mathbf{=}\underset{\mathbf{l}}{\mathbf{âˆ\circledR }}\mathit{v}\mathbf{·}\mathbf{ }\mathit{d}\mathit{l}$ .The right side of the gauss divergence theorem is the line integral , that is, $\underset{l}{âˆ\circledR }v·dl$

The diagram of the open surface area possessed by a circle of radius of R units is shown below:

Let the vector v be defined as $v=ayi +bx j$and the $∇$operator is defined as

$∇=\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k$

The divergence of vector v is computed as follows:

$\begin{array}{rcl}∇×v& =& \left|\begin{array}{ccc}i& j& k\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ ay& bx& 0\end{array}\right|\\ & =& \left[\frac{∂}{∂y}\left(0\right)-\frac{∂}{∂z}\left(bx\right)\right]i-\left[\frac{∂}{∂x}\left(0\right)-\frac{∂}{∂z}\left(ay\right)\right]j+\left[\frac{∂}{∂x}\left(bx\right)-\frac{∂}{∂y}\left(ay\right)\right]k\\ & =& \left(b-a\right)k\end{array}$

For the circular path of radius R, the area vector is $\stackrel{→}{da}= \mathrm{π}{R}^2 k$. The left part of the strokes theorem is calculated as:

$\begin{array}{rcl}\underset{S}{\overset{}{∫}}\left(∇×\mathbf{v}\right)·\stackrel{→}{da}& =& \underset{S}{∫}\left(b-a\right)k·\left( \mathrm{π}{R}^2 k\right)\\ & =& \underset{S}{∫} \mathrm{π}{R}^2 \left(b-a\right)\\ & =&  \mathrm{π}{R}^2 \left(b-a\right)\\ & & \end{array}$

The differential length vector is given by $\stackrel{→}{dl}= dx i+dy j$. Here,

$$\begin{gathered} x=R\cos \mathrm{θ} \\ y=R\sin \mathrm{θ} \end{gathered}$$

Differentiation above equations with respect to $\mathrm{θ}$.

$$\begin{gathered} dx=-R\sin \mathrm{θ}d\mathrm{θ} \\ dy=R\cos \mathrm{θ} d\mathrm{θ} \end{gathered}$$

Thus the displacement vector becomes

$\stackrel{→}{dl}= -R\sin \mathrm{θ}d\mathrm{θ} i+R\cos \mathrm{θ} d\mathrm{θ} j$

Hence the right side line integral in the stokes theorem becomes,

$\begin{array}{rcl}\underset{}{\overset{}{∫}}vdl& =& \underset{0}{\overset{2\mathrm{π}}{∫}}\left(ay i+bx j\right)\left( -R\sin \mathrm{θ}d\mathrm{θ} i+R\cos \mathrm{θ} d\mathrm{θ} j\right)\\ & =& -\underset{0}{\overset{2\mathrm{π}}{∫}}a{R}^2{\sin }^2\mathrm{θ} d\mathrm{θ}+b\underset{0}{\overset{2\mathrm{π}}{∫}}{R}^2{\cos }^2\mathrm{θ} d\mathrm{θ}\\ & =& -\frac{a{R}^2}{2}\underset{0}{\overset{2\mathrm{π}}{∫}}\left(1-\cos 2\mathrm{θ}\right)d\mathrm{θ}+\frac{b{R}^2}{2}\underset{0}{\overset{2\mathrm{π}}{∫}}\left(1+\cos 2\mathrm{θ}\right)d\mathrm{θ}\\ & =& -\frac{a{R}^2}{2}\left(2\mathrm{π}\right)+\frac{b{R}^2}{2}\left(2\mathrm{π}\right)\\ & & \end{array}$

Solve further as,

$\underset{}{\overset{}{∫}}vdl=\mathrm{π}{R}^2\left(b-a\right)$

Thus the left and right sides give the same result. Hence strokes theorem is verified.