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The equation of circle is x^{2}+y^{2}=16\, and the parametric equations of C are x=4\cos \theta ,y=4\sin \theta ,z=0,0\leq \theta \leq 2\pi \, --(1)

We wish to verify Stokes' theorm that is

\oint _{C}F\cdot dr=\iint _{S}{\mathrm  {curl}}F\cdot ndS\, --(2)

LHS of (2)=\oint _{C}[(x^{2}+y-4)i+3xyj+(2xy+z^{2})k]\cdot (dxi+dyj+dzk)\,

=\oint _{C}[(x^{2}+y-4)dx+3xydy+(2xz+z^{2})dz]\,

=\oint _{C}[(x^{2}+y-4)dx+3xydy]\, as on circle(1),z=0,dz=0

=\int _{{0}}^{{2\pi }}[(x^{2}+y-4){\frac  {dx}{d\theta }}+3xy{\frac  {dy}{d\theta }}]d\theta \,

=\int _{{0}}^{{2\pi }}[(16\cos ^{2}\theta +4\sin \theta -4)(-4\sin \theta )+(48\sin \theta \cos \theta )(4\cos \theta )]d\theta \,

=128\int _{{0}}^{{2\pi }}\cos ^{2}\theta \sin \theta d\theta -16\int _{{0}}^{{2\pi }}\sin ^{2}\theta d\theta +16\int _{{0}}^{{2\pi }}\sin \theta d\theta \,

=-\int _{{0}}^{{2\pi }}\cos ^{2}\theta (-\sin \theta )d\theta -16\int _{{0}}^{{2\pi }}{\frac  {1-\cos 2\theta }{2}}d\theta +16[-\cos \theta ]_{{0}}^{{2\pi }}\,

=-[{\frac  {\cos ^{3}\theta }{3}}]_{{0}}^{{2\pi }}-8[\theta -{\frac  {\sin 2\theta }{2}}]_{{0}}^{{2\pi }}+16(-1+1)\,

=-{\frac  {1}{3}}(0-0)-8(2\pi -0)+0=-16\pi \, --(3)

Now,{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\x^{2}+y-4&3xy&2xz+z^{2}\end{vmatrix}}\,

=[{\frac  {\partial }{\partial y}}(2xz+z^{2})-{\frac  {\partial }{\partial z}}(3xy)]i-[{\frac  {\partial }{\partial x}}(2xz+z^{2})-{\frac  {\partial }{\partial z}}(x^{2}+y^{2}-4)]j+[{\frac  {\partial }{\partial x}}(3xy)-{\frac  {\partial }{\partial y}}(x^{2}+y-4)]k=-2zj+(3y-1)k\, --(4)

Let R be the plane circular region PQK bounded by C.Let S1 be the surface consisting of 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)dV=0\, --(5)

Since S1 consists of the two surfaces,we have

\iint _{{S_{1}}}{\mathrm  {curl}}F\cdot ndS=\iint _{S}{\mathrm  {curl}}F\cdot ndS+\iint _{R}{\mathrm  {curl}}F\cdot ndS+\iint _{S}{\mathrm  {curl}}F\cdot ndS+\iint _{S}{\mathrm  {curl}}F\cdot (-k)dS\, --(6)

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

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

=\iint _{R}(-2zi+(3y-1)k)\cdot kdS=\iint _{R}(3y-1)dS\,

=\int _{{\theta =0}}^{{2\pi }}\int _{{r=0}}^{{4}}(3r\sin \theta -1)(rd\theta dr)\,

=\int {0}^{{2\pi }}[3\sin \theta {\frac  {r^{3}}{3}}-{\frac  {r^{2}}{2}}]_{0}^{4}d\theta \,

=\int _{{0}}^{{2\pi }}(64\sin \theta -8)d\theta =[64(-\cos \theta )-8\theta ]_{{0}}^{{2\pi }}\,

=-64(1-1)-16\pi =-16\pi \, --(8)

Hence from (3) and (8),the Stokes'theorm is verified.

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