From Example Problems
Jump to: navigation, search

Here S is the surface of the sphere x^{2}+y^{2}+z^{2}=1\, lying above the xy-plane.Let C be the boundary PQK of the surface S.Then the curve C in the xy-plane and equations of C are x^{2}+y^{2}=1,z=0\,

The parametric equations of C are x=\cos \theta ,y=\sin \theta ,z=0,0\leq \theta \leq 2\pi \, --(1)

We need to verify Stokes'theorm that is \oint _{C}F\cdot dr=\iint _{S}{\mathrm  {curl}}F\cdot ndS\, --(2)

LHS of (2)=\oint _{C}(yi+zj+xk)\cdot (dxi+dyj+dzk)=\int _{C}(ydx+zdy+xdz)\,

=\oint _{C}ydx\, as on C,z=0 and dz=0.

=\int _{{\theta =0}}^{{2\pi }}y{\frac  {dx}{d\theta }}d\theta \,

=\int _{{0}}^{{2\pi }}(\sin \theta )(-\sin \theta )d\theta \,

=-\int _{{0}}^{{2\pi }}{\frac  {1-\cos \theta )}{2}}d\theta \,

={\frac  {1}{2}}[\theta -{\frac  {\sin 2\theta }{2}}]_{{0}}^{{2\pi }}=-\pi \, --(3)

Now,{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\y&z&x\end{vmatrix}}=-i-j-k\, --(4)

Let R be the plane circular region PQK bounded by C.Let S1 be the surface consisting of the surfaces S and R. Then S1 is a closed surface enclosing volume V(say). By Gauss Divergence theorm,we have

\iint _{{S_{1}}}{\mathrm  {curl}}F\cdot ndS=\iiint _{V}{\mathrm  {div}}({\mathrm  {curl}}F\cdot ndS+)dV=0\, as {\mathrm  {div}}({\mathrm  {curlF}})=0\, --(5)

Since S1 consists of S and R,we have

\iint _{{S_{1}}}{\mathrm  {curl}}F\cdot ndS=\iint _{S}{\mathrm  {curl}}F\cdot ndS+\iint _{R}{\mathrm  {curl}}F+\iint _{S}{\mathrm  {curl}}F+\iint _{R}{\mathrm  {curl}}F\cdot (-k)dS\, --(6),noting that n denotes unit vector to S1.

From (5) and (6),we have \iint _{S}{\mathrm  {curl}}F\cdot ndS-\iint _{R}{\mathrm  {curl}}F\cdot kdS=0\, --(7)

Therefore,RHS of (2)=\iint _{R}{\mathrm  {curl}}F\cdot kdS\, by using (7)

=\iint _{R}(-i-j-k)\cdot kdS\, using(4)

=-\iint _{R}dS=-R=-(areaofthecircle)=-\pi (1)^{2}=-\pi \,

From (3) and (8),we obtain (2) and this verifies Stokes'theorm.

Main Page