VC5.50

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Evaluate with the help of divergence theorm the integral \iint _{S}[xz^{2}dydz+(x^{2}y-z^{3})dzdx+(2xy+y^{2}z)dxdy]\,, where S is the entire surface of the hemispherical region bounded by z={\sqrt  {(a^{2}-x^{2}-y^{2})}},z=0\,.


Let V be the volume enclosed by S.Hence S is the point of the sphere given.

By divergence theorem,\iint _{S}[xz^{2}dydz+(x^{2}y-z^{3})dzdx+(2xy+y^{2}z)dxdy]\,

=\iiint _{V}[{\frac  {\partial }{\partial x}}(xz^{2})+{\frac  {\partial }{\partial y}}(x^{2}y-z^{3})+{\frac  {\partial }{\partial z}}(2xy+y^{2}z)]dV\,

=\iiint _{V}(z^{2}+x^{2}+y^{2})dV\, --(1)

To simplify the triple integral,we use the spherical polar coordinates (r,\theta ,\phi )\, so that x=r\sin \theta \cos \phi ,y=r\sin \theta \sin \phi ,z=r\cos \theta ,dV=(dr)(rd\theta )(r\sin \theta d\phi )=r^{2}\sin \theta drd\theta d\phi \, --(2)

In order to cover the volume,the limits are r=0 to a \theta =0,{\frac  {\pi }{2}},\phi =0,2\pi \,

Here \iiint _{V}(x^{2}+y^{2}+z^{2})dV\,

=\int _{{r=0}}^{{a}}\int _{{\theta =0}}^{{{\frac  {\pi }{2}}}}\int _{{\phi =0}}^{{2\pi }}r^{2}\cdot r^{2}\sin \theta drd\theta d\phi \,

=\int _{{\theta =0}}^{{{\frac  {\pi }{2}}}}\int _{{\phi =0}}^{{2\pi }}[{\frac  {r^{5}}{5}}]_{0}^{a}\sin \theta d\theta d\phi \,

={\frac  {a^{5}}{5}}\int _{{\theta =0}}^{{{\frac  {\pi }{2}}}}\int _{{\phi =0}}^{{2\pi }}\sin \theta d\theta d\phi \,

={\frac  {a^{5}}{5}}\int _{{\theta =0}}^{{{\frac  {\pi }{2}}}}\sin \theta [\phi ]_{{0}}^{{2\pi }}d\theta \,

={\frac  {2\pi a^{5}}{5}}\int _{{\theta =0}}^{{{\frac  {\pi }{2}}}}\sin \theta d\theta \,

={\frac  {2\pi a^{5}}{5}}[-\cos \theta ]_{{0}}^{{{\frac  {\pi }{2}}}}={\frac  {2\pi a^{5}}{5}}\, --(3)

Using (3) in (1),we see that the value of the given surface integral is {\frac  {2\pi a^{5}}{5}}\,

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