# VC5.13

By Stokes'theorm,

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

Since the integration is performed in the xy-plane,we have $r=xi+yj\,$ so that $dr=dxi+dyj\,$. The path of integration C consists of lines say DA,AB,BE,ED,where A,B,E,D are (a,0),(a,b),(-a,-b),(-a,0) respectively.

Therefore,$\oint _{C}F\cdot dr\,$

=$\oint _{C}[(x^{2}+y^{2})i-2xyj]\cdot (dxi+dyj)=\oint _{C}[(x^{2}+y^{2})dx-2xydy]\,$ --(2)

On DA,y=0,dy=0,x varies from -a to a, On AB,x=a,dx=0,y varies from a to -a, On BE,y=b,dy=0 and x varies from a to -a. and on ED,x=-a,dx=0 and y varies from b to 0..

Using (2),LHS of (1)

=$\int _{{DA}}[(x^{2}+y^{2})dx-2xydy]+\int _{{AB}}[(x^{2}+y^{2})dx-2xydy]+\int _{{BE}}[(x^{2}+y^{2})dx-2xydy]+\int _{{ED}}[(x^{2}+y^{2})dx-2xydy]\,$

=$\int _{{-a}}^{{a}}x^{2}dx+\int _{0}^{b}(-2ay)dy+\int _{{a}}^{{-a}}(x^{2}+b^{2})dx+\int _{b}^{0}(2ay)dy\,$

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

=${\frac {2a^{3}}{3}}-ab^{2}-{\frac {2a^{3}}{3}}-2ab^{2}-ab^{2}=-4ab^{2}\,$

Now,${\mathrm {curl}}F={\begin{vmatrix}i&j&k\\{\frac {\partial }{\partial x}}&{\frac {\partial }{\partial y}}&{\frac {\partial }{\partial z}}\\x^{2}+y^{2}&-2xy&0\end{vmatrix}}=-4yk\,$ on simplification.

Hence $n=k\,$

RHS=$\oint _{R}(-4yk)\cdot kdxdy=-4\int _{{x=-a}}^{{a}}\int _{{y=0}}^{{b}}ydxdy\,$

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

Hence the theorm is verified.