Chapter 1 - More Problems on This Chapter - Problem - Page 55: 55

$(b-a)\pi R^2$

Given function is $\vec{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 Stokes' Theorem states that $\int_S (\vec{\nabla} \times \vec{v})\cdot \vec{da}= \oint_C \vec{v}\cdot \vec{dl} $ In Cartesian coordinate System $\vec{da} = dxdy\hat{z}$ and $\vec{dl} = dx\hat{x} + dy\hat{y}$ $ (\vec{\nabla} \times \vec{v})= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ ay & bx & 0 \end{vmatrix}$ $\qquad \quad \space = \hat{x}(\frac{\partial}{\partial{y}} (0)- \frac{\partial}{\partial{z}} (x)) - \hat{y}(\frac{\partial}{\partial{x}} (0) - \frac{\partial}{\partial{z}} (ay))+\hat{z}(\frac{\partial}{\partial{x}} (bx) - \frac{\partial}{\partial{y}} (ay))$ $\qquad \quad \space = \hat{z}(b-a)$ And hence $\int_S (\vec{\nabla} \times \vec{v})\cdot \vec{da}=\int_S (b-a)\hat{z}\cdot dxdy\hat{z}$ Since $(b-a)$ is constant $\int_S (\vec{\nabla} \times \vec{v})\cdot \vec{da}=(b-a)\int_S dxdy$ For the given surface $\int_S dxdy = \pi R^2$ Hence, $\int_S (\vec{\nabla} \times \vec{v})\cdot \vec{da}=\pi R^2 \qquad ...(i)$ $\oint_C \vec{v}.\vec{dl} = \oint_C (ay\hat{x} + bx\hat{y})\cdot (dx\hat{x}+dy\hat{y})$ $\oint_C \vec{v}.\vec{dl} = \oint_C (aydx + bxdy)$ 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, $\int_{C_1}\vec{v}.\vec{dl}=\int_{C_1} (a\sqrt{R^2-x^2}dx+bx(\frac{-x}{\sqrt{R^2-x^2}})dx)$ $\int_{C_1}\vec{v}.\vec{dl}=\int_{C_1} \frac{(a(R^2-x^2)-{bx^2})}{\sqrt{R^2-x^2}}dx$ $\int_{C_1}\vec{v}.\vec{dl}=\int_{C_1} \frac{aR^2-(a+b)x^2}{\sqrt{R^2-x^2}}dx$ Since the Counterclockwise direction is considered to be positive, $\therefore x $ varies from $R$ to $-R$. $\int_{C_1}\vec{v}.\vec{dl}=\int_{R}^{-R} (\frac{aR^2}{\sqrt{R^2-x^2}} - \frac{(a+b)x^2}{\sqrt{R^2-x^2}} )dx$ $\int_{C_1}\vec{v}.\vec{dl}=\big\{ {aR^2}.sin^{-1}({x\over R}) - (a+b)\big [{-x\over 2} \sqrt{R^2-x^2}+{R^2\over 2} sin^{-1}({x\over R})\big ]\big \}_R^{-R}$ $\int_{C_1}\vec{v}.\vec{dl}=\big\{ \frac{1}{2}(a-b)R^2sin^{-1}({x\over R})\big \}_R^{-R} \qquad \because \int_R^{-R} \frac{-x}{2}\sqrt{R^2-x^2}=0$ $\int_{C_1}\vec{v}.\vec{dl}=\frac{1}{2}R^2(a-b)sin^{-1}({-R\over R}) - \frac{1}{2}R^2(a-b)sin^{-1}({R\over R})$ $\int_{C_1}\vec{v}.\vec{dl}=\frac{1}{2}R^2(a-b)({-\pi \over 2}-{\pi \over 2})$ $\int_{C_1}\vec{v}.\vec{dl}=\frac{1}{2}R^2(b-a)\pi$ For the second integral, $dy = -\frac{1}{2\sqrt{R^2-x^2}}(-2x)dx = \frac{x}{\sqrt{R^2-x^2}}dx$ $\int_{C_2}\vec{v}.\vec{dl}=\int_{C_2} (a(-\sqrt{R^2-x^2})dx+bx(\frac{x}{\sqrt{R^2-x^2}})dx)$ $\int_{C_2}\vec{v}.\vec{dl}=\int_{C_2} \frac{-aR^2+(a+b)x^2}{\sqrt{R^2-x^2}}dx$ Here $x$ varies from $-R$ to $R$ and hence $\int_{C_2}\vec{v}.\vec{dl}=\int_{-R}^{R} \frac{-aR^2+(a+b)x^2}{\sqrt{R^2-x^2}}dx$ Reversing the limits we get, $\int_{C_2}\vec{v}.\vec{dl}=\int_{R}^{-R} \frac{aR^2-(a+b)x^2}{\sqrt{R^2-x^2}}dx = \int_{C_1}\vec{v}.\vec{dl}$ and hence $\oint_C \vec{v}\cdot \vec{dl} =\frac{1}{2}R^2(b-a)\pi + \frac{1}{2}R^2(b-a)\pi $ $\oint_C \vec{v}\cdot \vec{dl} =\pi R^2(b-a)$ Hence stokes theorem is proved.