# VC5.22

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.