Stoke's Theorem Engineering Maths, Btech first year

Stoke's Theorem

This theorem gives relationship between Line Integral and Surface Integral.
If S be an open surface bounded by closed curve C and $\overrightarrow{F}$ is a continuous dufferentiable vector point function then: $${\oint }_C\overrightarrow{F}.d\overrightarrow{R}={\iint }_S(curl(\overrightarrow{F})).\hat{N}ds$$

Sample Problem 1

Verify Stoke's Theorem for $\overrightarrow{F}=(x^2+y^2)\hat{i}-2xy\hat{j}$ taken around the rectangle bounded by the lines $$x=\pm a,y=0,y=b$$

Solution

Tracing the curves by using given equations: $$x=\pm a,y=0,y=b$$

By Stoke's Theorem: $${\oint }_{ABCD}\overrightarrow{F}.d\overrightarrow{R}={\iint }_{ABCD}(curl(\overrightarrow{F})).\hat{N}ds$$ L.H.S. : $${\oint }_{ABCD}\overrightarrow{F}.d\overrightarrow{R}={\int }_{AB}\overrightarrow{F}.d\overrightarrow{R}+{\int }_{BC}\overrightarrow{F}.d\overrightarrow{R}+{\int }_{CD}\overrightarrow{F}.d\overrightarrow{R}+{\int }_{DA}\overrightarrow{F}.d\overrightarrow{R}$$ For curve AB :
The equation is: $$x=a\Rightarrow dx=0$$ Points are A(a,0) to B(a,b) $${\int }_{AB}\overrightarrow{F}.d\overrightarrow{R}={\int }_{AB}[(x^2+y^2)\hat{i}-2xy\hat{j}].(dx\hat{i}+dy\hat{j}+dz\hat{k})$$ $$={\int }_{AB}(x^2+y^2)dx-2xydy$$ $$={\int }_0^b(a^2+y^2)0-2aydy$$ $$={\int }_0^b-2aydy$$ $$=\frac{-2a{b}^2}{2}=\boxed{{\displaystyle -a{b}^2}}$$ Along BC :
$$y=b\Rightarrow dy=0$$ Points : B(a,b) to C(-a,b) $${\int }_{BC}\overrightarrow{F}.d\overrightarrow{R}={\int }_{BC}(x^2+y^2)dx-2xydy$$ $$={\int }_a^{-a}(x^2+b^2)dx$$ $$={[\frac{x^3}{3}+b^{2}x]}_a^{-a}$$ $$=[\frac{-a^3}{3}-b^{2}a]-[\frac{a^3}{3}+b^{2}a]$$ $$=\boxed{{\displaystyle \frac{-2a^3}{3}-2a{b}^2}}$$ Along CD :
$$x=-a\Rightarrow dx=0$$ Points : C(-a,b) to D(-a,0) $${\int }_{CD}\overrightarrow{F}.d\overrightarrow{R}={\int }_{CD}(x^2+y^2)dx-2xydy$$ $$={\int }_b^0(a^2+y^2)0+2aydy$$ $$={\int }_b^{0}2aydy$$ $$=\boxed{{\displaystyle -a{b}^2}}$$ Along DA :
$$y=0\Rightarrow dy=0$$ Points: D(-a,0) to A(a,0) $${\int }_{DA}\overrightarrow{F}.d\overrightarrow{R}={\int }_{DA}(x^2+y^2)dx-2xydy$$ $$={\int }_{-a}^a{x}^{2}dx$$ $$={\left[\frac{x^3}{3}\right]}_{-a}^a$$ $$=\frac{a^3}{3}-\frac{-a^3}{3}$$ $$=\boxed{{\displaystyle \frac{2a^3}{3}}}$$ $${\oint }_{ABCD}\overrightarrow{F}.d\overrightarrow{R}=-a{b}^2-\frac{2a^3}{3}-2a{b}^2-a{b}^2+\frac{2a^3}{3}$$ $$=\boxed{{\displaystyle -4a{b}^2}}$$ R.H.S. : $$curl(\overrightarrow{F})=\mathrm{\nabla }\times \overrightarrow{F}$$ $$=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})\times [(x^2+y^2)\hat{i}-2xy\hat{j}]$$ $$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \\ x^2+y^2& -2xy& 0\end{array}\right|$$ $$=\hat{i}(0+0)-\widehat{0-0}+\hat{k}(-2y-2y)$$ $$\Rightarrow curl\overrightarrow{F}=-4y\hat{k}$$ Since surface ABCD lying in X-Y plane so it must be projected on X-Y plane $$\Rightarrow \hat{N}=\hat{k}$$ $$\Rightarrow ds=\frac{dxdy}{|\hat{N}\hat{k}|}=dxdy$$ $${\iint }_{ABCD}(curl(\overrightarrow{F})).\hat{N}ds$$ $$=\iint (-4y\hat{k}).\hat{N}dxdy$$ $$=\iint -4ydxdy$$

$$\Rightarrow {\int }_0^b{\int }_{-a}^a-4ydxdy$$ $$={\int }_0^b{\left[4yx\right]}_{-a}^{a}dy$$ $$={\int }_0^{b}8yady={\left[4y^{2}a\right]}_0^b$$ $$=\boxed{{\displaystyle -4b^{2}a}}$$ = R.H.S., Hence Proved :)