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\iint _{S}(x^{3}dydz+y^{3}dzdx+z^{3}dxdy)\,

=\iiint _{V}[{\frac  {\partial }{\partial x}}(x^{3})+{\frac  {\partial }{\partial y}}(y^{3})+{\frac  {\partial }{\partial z}}(z^{3})]dxdydz\, by divergence theorm.

=3\iiint _{V}(x^{2}+y^{2}+z^{2})dxdydz\, where V is the volume bounded by the given sphere.

We use the spherical polar coordinats (r,\theta ,\phi )\, so that

x=r\sin \theta \cos \phi ,y=r\sin \theta \sin \phi ,z=r\cos \theta \,

and dxdydz=dV=(dr)(rd\theta )(r\sin \theta d\theta )=r^{2}\sin \theta drd\theta d\phi \,


The limits are r=0 to r=1,\theta =0\,to \pi \, and those of \phi \, are from 0 to 2\pi \,

Therefore,\iiint _{V}(x^{2}+y^{2}+z^{2})dxdydz\,

=\int _{0}^{1}\int _{{\theta =0}}^{{\pi }}\int _{{\phi =0}}^{{2\pi }}r^{2}(r^{2}\sin \theta )drd\theta d\phi \,

=\int _{0}^{1}\int _{{\theta =0}}^{{\pi }}r^{4}\sin \theta [\phi ]_{{0}}^{{2\pi }}drd\theta \,

=2\pi \int _{0}^{1}\int _{{0}}^{{\pi }}r^{4}\sin \theta drd\theta \,

=2\pi \int _{0}^{1}r^{4}[-\cos \theta ]_{{0}}^{{\pi }}dr=4\pi \int _{0}^{1}r^{4}dr=4\pi [{\frac  {r^{5}}{5}}]_{0}^{1}={\frac  {4\pi }{5}}\,

From (1),\iint _{S}(x^{3}dydz+y^{3}dzdx+z^{3}dxdy)=3[{\frac  {4\pi }{5}}]={\frac  {12\pi }{5}}\,

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