From Example Problems
Jump to: navigation, search

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.

Main Page