Check Stokes' theorem using the function $\mathit{v}\mathbf{=}\mathit{a}\mathit{y}\mathit{i}\mathbf{ }\mathbf{+}\mathit{b}\mathit{x}\mathbf{ }\mathit{j}$(aand bare constants) and the circular path of radius R,centered at the origin in the xyplane. [Answer:$\mathit{\pi }{\mathit{R}}^{\mathbf{2}}\left(b-a\right)$ ],

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 $\underset{\mathbf{s}}{\mathbf{\int }}\left(\nabla \times v\right)\mathbf{·}\mathbf{ }\mathit{d}\mathit{s}\mathbf{=}\underset{\mathbf{l}}{\mathbf{\oint }}\mathit{v}\mathbf{·}\mathbf{ }\mathit{d}\mathit{l}$ .The right side of the gauss divergence theorem is the line integral , that is, $\underset{l}{\oint }v·dl$

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 $\nabla$operator 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:

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

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

$\begin{array}{rcl}\underset{S}{\overset{}{\int }}\left(\nabla \times \mathbf{v}\right)·\overrightarrow{da}& =& \underset{S}{\int }\left(b-a\right)k·\left( \pi R^2 k\right)\\ & =& \underset{S}{\int } \pi R^2 \left(b-a\right)\\ & =& \pi R^2 \left(b-a\right)\\ & & \end{array}$

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

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

Differentiation above equations with respect to $\theta$.

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

Thus the displacement vector becomes

$\overrightarrow{dl}= -R\sin \theta d\theta i+R\cos \theta d\theta j$

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

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

Solve further as,

$\underset{}{\overset{}{\int }}vdl=\pi R^2\left(b-a\right)$

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