# VC5.16

Here the path of integration C consists of the straight lines say OA,AB,BC and CD where the coordinates of A,B,C are (a,0),(a,b) and (0,b) respectively.Let S be the plane region bounded by closed curve C,then by Stokes' theorm,we must have

$\oint _{C}F\cdot dr=\iint _{S}{\mathrm {curl}}F\cdot ndS\,$ --(1)

Since the integration is performed in the xy-plane,we have $r=xi+yj\,$ so that $dr=dxi+dyj\,$

LHS of (1)=$\oint _{C}(x^{2}i+2xyj)\cdot (dxi+dyj)\,$

=$\oint _{C}(x^{2}dx+xydy)\,$

=$\int _{{OA}}(x^{2}dx+xydy)+\int _{{AB}}(x^{2}dx+xydy)+\int _{{BC}}(x^{2}dx+xydy)+\int _{{CO}}(x^{2}dx+xydy)\,$ --(2)

Along OA,y=0,dy=0 and x varies from 0 to a. Along AB,x=a,dx=0 and y varies from 0 to b. Along BC,y=b,dy=0 and x varies from a to 0. Along CO,x=0,dx=0 and y varies from b to 0.

Therefore,from (2),LHS=$\int _{0}^{a}x^{2}dx+\int _{0}^{b}aydy+\int _{a}^{0}x^{2}dx+\int _{b}^{0}(0)dy\,$

=$[{\frac {x^{3}}{3}}]_{0}^{a}+a[{\frac {y^{2}}{2}}]_{0}^{b}+[{\frac {x^{3}}{3}}]_{a}^{0}+0\,$

=${\frac {a^{3}}{3}}+{\frac {ab^{2}}{2}}-{\frac {a^{3}}{3}}={\frac {ab^{2}}{2}}\,$ --(3)

Now,${\mathrm {curl}}F={\begin{vmatrix}i&j&k\\{\frac {\partial }{\partial x}}&{\frac {\partial }{\partial y}}&{\frac {\partial }{\partial z}}\\x^{2}&xy&0\end{vmatrix}}\,$

=$0i-0j+[{\frac {\partial }{\partial x}}(xy)-{\frac {\partial }{\partial y}}(x^{2})]k\,$

Also,n=unit vector normal to surface S=k.

Therefore ${\mathrm {curl}}F\cdot n=yk\cdot k=y\,$

Now,RHS=$\iint _{S}ydxdy=\int _{{x=0}}^{{a}}\int _{{y=0}}^{{b}}ydxdy\,$

=$\int _{0}^{a}[{\frac {y^{2}}{2}}]_{0}^{b}dx={\frac {b^{2}}{2}}\int _{0}^{a}dx={\frac {b^{2}}{2}}[x]_{0}^{a}={\frac {ab^{2}}{2}}\,$ --(4)

From (3) and (4),Stokes'theorm is verified.