Chapter 1 Vector Analysis: Problem 1.55

Problem 1.55: Check Stokes’ theorem using the function $v=ay\overrightarrow{x}+bx\overrightarrow{y}$ ($a$ and $b$ are constants) and the circular path of radius $R$, centered at the origin in the $xy$ plane.

 Solution:

Stokes Theorem States That for a vector $\overrightarrow{v}$,

$${\int }_S(\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{v})\cdot d\overrightarrow{a}={\oint }_P\overrightarrow{v}\cdot d\overrightarrow{l}$$ 

Given function is $$\overrightarrow{v}=ay\hat{x}+bx\hat{y}$$ where $a$ and $b$ are constants. 

The path given is a circular path of radius $R$ in the $xy$ plane.

 

 

 

 

 

 

 

 In Cartesian coordinate System $$\overrightarrow{da}=dxdy\hat{z}$$

 and 

$$\overrightarrow{dl}=dx\hat{x}+dy\hat{y}$$

$$(\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{v})=\left|\begin{array}{ccc}\hat{x}& \hat{y}& \hat{z}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ ay& bx& 0\end{array}\right|$$ $$=\hat{x}(\frac{\partial }{\partial y}(0)-\frac{\partial }{\partial z}(x))-\hat{y}(\frac{\partial }{\partial x}(0)-\frac{\partial }{\partial z}(ay))+\hat{z}(\frac{\partial }{\partial x}(bx)-\frac{\partial }{\partial y}(ay))$$

$$=\hat{z}(b-a)$$ 

And hence 

$${\int }_S(\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{v})\cdot \overrightarrow{da}={\int }_S(b-a)\hat{z}\cdot dxdy\hat{z}$$

Since $(b-a)$ is constant 

$${\int }_S(\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{v})\cdot \overrightarrow{da}=(b-a){\int }_{S}dxdy$$ 

For the given surface 

$${\int }_{S}dxdy=\pi R^2$$ 

Hence, 

$${\int }_S(\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{v})\cdot \overrightarrow{da}=\pi R^2...(i)$$

$${\oint }_C\overrightarrow{v}.\overrightarrow{dl}={\oint }_C(ay\hat{x}+bx\hat{y})\cdot (dx\hat{x}+dy\hat{y})$$ $${\oint }_C\overrightarrow{v}.\overrightarrow{dl}={\oint }_C(aydx+bxdy)$$ 

For computing the line integral, Let's divide the circle in upper half $C_1$ and lower half $C_2$. For upper half, $$y=\sqrt{R^2-x^2}$$ and for lower half $$y=-\sqrt{R^2-x^2}$$ 

Let's compute the line integral for upper half first, 

$$dy=\frac{1}{2\sqrt{R^2-x^2}}(-2x)dx=\frac{-x}{\sqrt{R^2-x^2}}dx$$ So, $${\int }_{C_1}\overrightarrow{v}.\overrightarrow{dl}={\int }_{C_1}(a\sqrt{R^2-x^2}dx+bx(\frac{-x}{\sqrt{R^2-x^2}}))dx$$ $${\int }_{C_1}\overrightarrow{v}.\overrightarrow{dl}={\int }_{C_1}\frac{(a(R^2-x^2)-b{x}^2)}{\sqrt{R^2-x^2}}dx$$ $${\int }_{C_1}\overrightarrow{v}.\overrightarrow{dl}={\int }_{C_1}\frac{a{R}^2-(a+b)x^2}{\sqrt{R^2-x^2}}dx$$ Since the Counterclockwise direction is considered to be positive, 

$\therefore x$ varies from $R$ to $-R$. $${\int }_{C_1}\overrightarrow{v}.\overrightarrow{dl}={\int }_R^{-R}(\frac{a{R}^2}{\sqrt{R^2-x^2}}-\frac{(a+b)x^2}{\sqrt{R^2-x^2}})dx$$ $${\int }_{C_1}\overrightarrow{v}.\overrightarrow{dl}=\{a{R}^2.si{n}^{-1}(\frac{x}{R})-(a+b)[\frac{-x}{2}\sqrt{R^2-x^2}+\frac{R^2}{2}si{n}^{-1}(\frac{x}{R})]{\}}_R^{-R}$$ $${\int }_{C_1}\overrightarrow{v}.\overrightarrow{dl}=\{\frac{1}{2}(a-b)R^{2}si{n}^{-1}(\frac{x}{R}){\}}_R^{-R}\because {\int }_R^{-R}\frac{-x}{2}\sqrt{R^2-x^2}=0$$ $${\int }_{C_1}\overrightarrow{v}.\overrightarrow{dl}=\frac{1}{2}{R}^2(a-b)si{n}^{-1}(\frac{-R}{R})-\frac{1}{2}{R}^2(a-b)si{n}^{-1}(\frac{R}{R})$$ $${\int }_{C_1}\overrightarrow{v}.\overrightarrow{dl}=\frac{1}{2}{R}^2(a-b)(\frac{-\pi }{2}-\frac{\pi }{2})$$ $${\int }_{C_1}\overrightarrow{v}.\overrightarrow{dl}=\frac{1}{2}{R}^2(b-a)\pi$$ For the second integral, $$dy=-\frac{1}{2\sqrt{R^2-x^2}}(-2x)dx=\frac{x}{\sqrt{R^2-x^2}}dx$$ $${\int }_{C_2}\overrightarrow{v}.\overrightarrow{dl}={\int }_{C_2}(a(-\sqrt{R^2-x^2})dx+bx(\frac{x}{\sqrt{R^2-x^2}})dx)$$ $${\int }_{C_2}\overrightarrow{v}.\overrightarrow{dl}={\int }_{C_2}\frac{-a{R}^2+(a+b)x^2}{\sqrt{R^2-x^2}}dx$$ Here $x$ varies from $-R$ to $R$ and hence $${\int }_{C_2}\overrightarrow{v}.\overrightarrow{dl}={\int }_{-R}^R\frac{-a{R}^2+(a+b)x^2}{\sqrt{R^2-x^2}}dx$$ Reversing the limits we get, $${\int }_{C_2}\overrightarrow{v}.\overrightarrow{dl}={\int }_R^{-R}\frac{a{R}^2-(a+b)x^2}{\sqrt{R^2-x^2}}dx={\int }_{C_1}\overrightarrow{v}.\overrightarrow{dl}$$ and hence $${\oint }_C\overrightarrow{v}\cdot \overrightarrow{dl}=\frac{1}{2}{R}^2(b-a)\pi +\frac{1}{2}{R}^2(b-a)\pi$$ $${\oint }_C\overrightarrow{v}\cdot \overrightarrow{dl}=\pi R^2(b-a)$$ Hence stokes theorem is proved. 

 

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