VC5.57

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Given surface is \phi (x,y,z)=x^{2}+y^{2}+z^{2}=a^{2}\, --(1)

From (1),\nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](x^{2}+y^{2}+z^{2})=2(xi+yj+zk)\,

Therefore,n=unit vector to S={\frac  {2(xi+yj+zk)}{{\sqrt  {4(x^{2}+y^{2}+z^{2})}}}}={\frac  {xi+yj+zk}{a}}\,,using (1)

We have xyz=(xyzn)\cdot n={\frac  {xyz(xi+yj+zk)}{a}}\cdot n\,

Now,\iint _{S}xyzdS={\frac  {1}{a}}\iint _{S}(x^{2}yzi+xy^{2}zj+xyz^{2}k)\cdot ndS\,

={\frac  {1}{a}}\iiint _{V}{\mathrm  {div}}(x^{2}yzi+xy^{2}zj+xyz^{2}k)dV\, by using divergence theorm.

={\frac  {1}{a}}\iiint _{V}[{\frac  {\partial (x^{2}yz)}{\partial x}}+{\frac  {\partial (xy^{2}z)}{\partial y}}+{\frac  {\partial (xyz^{2})}{\partial z}}]dV\,

={\frac  {6}{a}}\iiint _{V}xyzdV\,

={\frac  {6}{a}}\int _{{r=0}}^{{a}}\int _{{\theta =0}}^{{{\frac  {\pi }{2}}}}\int _{{\phi =0}}^{{{\frac  {\pi }{2}}}}(r\sin \theta \cos \phi )(r\sin \theta \sin \phi ))r\cos \theta )r^{2}\sin \theta drd\theta d\phi \, [Changing to spherical polar coordinates]

={\frac  {6}{a}}[\int _{0}^{a}r^{5}dr][\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin ^{3}\theta \cos \theta d\theta ][\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin \phi \cos \phi d\phi ]\,

={\frac  {6}{a}}[{\frac  {r^{6}}{6}}]_{0}^{a}\times [{\frac  {\sin ^{4}\theta }{4}}]_{{0}}^{{{\frac  {\pi }{2}}}}\times [{\frac  {\sin ^{2}\phi }{2}}]_{{0}}^{{{\frac  {\pi }{2}}}}={\frac  {2\pi a^{5}}{5}}\,

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