VC5.20

From Example Problems
Jump to: navigation, search

The given surface z=4-(x^{2}+y^{2})\, meets the xy-plane in the circle C whose equations are x^{2}+y^{2}=4,z=0\,. Hence the boundary of the surface S is the circle C. The parametric equations of the curve C are

x=2\cos \theta ,y=2\sin \theta ,z=0,0\leq \theta \leq 2\pi \, --(1)

Hence,by Stokes'theorm,we have

\iint _{S}{\mathrm  {curl}}F\cdot ndS=\int _{C}F\cdot dr\,

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

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

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

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

=\int _{{0}}^{{2\pi }}[[4\cos ^{2}\theta +2\sin \theta -4](-2\sin \theta )+(12\cos \theta \sin \theta )(2\cos \theta ))]d\theta \,

=\int _{{0}}^{{2\pi }}(16\cos ^{2}\theta \sin \theta -4\sin ^{2}\theta +8\sin \theta )d\theta \,

=16\int _{{0}}^{{\theta }}\cos ^{2}\theta \sin \theta d\theta -2\int _{{0}}^{{2\pi }}(1-\cos 2\theta )d\theta +8\int _{{0}}^{{2\pi }}\sin \theta d\theta \,

=16[{\frac  {\cos ^{3}\theta }{3}}]_{{0}}^{{2\pi }}-2[\theta -{\frac  {\sin 2\theta }{2}}]_{{0}}^{{2\pi }}+8[-\cos \theta ]_{{0}}^{{2\pi }}=-4\pi \,,on simplification.

Main Page