VC5.24

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Here F=zi+xj+yk\, --(1)

Also,r=xi+yj+zk,dr=dxi+dyj+dzk\,

F\cdot dr=zdx+xdy+ydz\,

Therefore,\oint _{C}F\cdot dr=\oint _{C}(zdx+xdy+ydz)\, --(2)

On the circle C, x^{2}+y^{2}=1,z=0\, on the xy-plane. Hence on C,we have z=0 so that dz=0.Hence,from (2)reduces to \oint _{C}F\cdot dr=\oint _{C}xdy\, --(3)

Now the parametric equations of C are x=\cos \theta ,y=\sin \theta \, --(4)

Using (4),(3) reduces to \oint _{C}F\cdot dr=\int _{{0}}^{{2\pi }}\cos \theta \cos \theta d\theta \,

=\int _{{0}}^{{2\pi }}{\frac  {1+\cos 2\theta }{2}}d\theta ={\frac  {1}{2}}[\theta +{\frac  {\sin 2\theta }{2}}]_{{0}}^{{2\pi }}=\pi \, --(5)

Let P(x,y,z) be any point on the surface of the hemisphere x^{2}+y^{2}+z^{2}=1\,. The spherical polar coordinates of P are given by x=\sin \phi \cos \theta ,y=\sin \phi \sin \theta ,z=\cos \phi \, --(6)

Here x^{2}+y^{2}+z^{2}=1\, hence the direction cosines of OP are in \sin \phi \cos \theta ,\sin \phi \sin \theta ,\cos \theta \,.But OP is also the direction of outward drawn normal at P.

Therefore,n=unit normal vector to hemisphere S=\sin \theta \cos \phi i+\sin \theta \sin \phi j+\cos \theta k\, --(7)

Also,{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\z&x&y\end{vmatrix}}\,

=i+j+k\, on simplification. --(8)

Therefore,{\mathrm  {curl}}F\cdot n=\sin \theta \cos \phi +\sin \theta \sin \phi +\cos \theta \, from (7) and (8) --(9)

Therefore,\iint _{S}{\mathrm  {curl}}F\cdot ndS\,

=\int _{{\phi =0}}^{{{\frac  {\pi }{2}}}}\int _{{\theta =0}}^{{2\pi }}(\sin \theta \cos \phi +\sin \theta \sin \phi +\cos \theta )\sin \theta d\theta d\phi \,

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

=\int _{{\phi =0}}^{{{\frac  {\pi }{2}}}}(0+0+2\pi \sin \phi \cos \phi )d\theta =\pi \int _{{\phi =0}}^{{{\frac  {\pi }{2}}}}\sin 2\phi d\phi =\pi [-{\frac  {\cos 2\phi }{2}}]_{{0}}^{{{\frac  {\pi }{2}}}}=-{\frac  {\pi }{2}}[-1-1]=\pi \, --(10)

From (5) and (10),Stokes'theorm is verified.

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