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Project the given surface S on xy-plane.Then the projection on the xy-plane is the region R bounded by the circle x^{2}+y^{2}=1,z=0\, Let F=y^{2}z^{2}i+z^{2}x^{2}j+x^{2}y^{2}k\,

Therefore the given surface integral=\iint _{S}F\cdot dS=\iint _{S}F\cdot ndS=\iint _{R}F\cdot n{\frac  {dxdy}{|n\cdot k|}}\,

If the given surface S can be written as \phi (x,y,z)=x^{2}+y^{2}+z^{2}=1\, --(2),then the normal vector to S is \nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](x^{2}+y^{2}+z^{2})=2xi+2yj+2zk\,

Let n denote the unit vector,then n={\frac  {\nabla \phi }{|\nabla \phi |}}={\frac  {2(xi+yj+zk)}{{\sqrt  {4x^{2}+4y^{2}+4z^{2}}}}}=xi+yj+zk\,,by using (2)

Therefore,n\cdot k=(xi+yj+zk)\cdot k=z\, F\cdot n=(y^{2}z^{2}i+z^{2}x^{2}j+x^{2}y^{2}k)\cdot (xi+yj+zk)=xy^{2}z^{2}+yz^{2}x^{2}+zx^{2}y^{2}\,

The required integral is \iint _{R}(xy^{2}z^{2}+yz^{2}x^{2}+zx^{2}y^{2}){\frac  {dxdy}{z}}=\iint _{R}[z(xy^{2}+x^{2}y)+x^{2}y^{2}]dxdy=\iint _{R}[(xy^{2}+x^{2}y){\sqrt  {(1-x^{2}-y^{2})}}+x^{2}y^{2}]dxdy\, using (2)

Now the integration can be evaluated as below.

=\int _{{x=-1}}^{{1}}\int _{{y=-{\sqrt  {1-x^{2}}}}}^{{{\sqrt  {1-x^{2}}}}}[(y^{2}+x^{2}y)({\sqrt  {1-x^{2}-y^{2}}})+x^{2}y^{2}]dxdy=2\times 2\int _{{x=0}}^{{1}}\int _{{y=0}}^{{{\sqrt  {1-x^{2}}}}}x^{2}y^{2}dxdy\,,other integral vanish because their integrands are odd function of x and y respectively.

=4\int _{{0}}^{{1}}x^{2}[{\frac  {y^{3}}{3}}]_{{0}}^{{{\sqrt  {1-x^{2}}}}}dx={\frac  {4}{3}}\int _{0}^{1}x^{2}(1-x^{2})^{{{\frac  {3}{2}}}}dx\,

={\frac  {4}{3}}\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta \cos ^{4}\theta \,d\theta \,, on putting x=\sin \theta ,dx=\cos \theta d\theta \,

={\frac  {4}{3}}{\frac  {\tau ({\frac  {3}{2}})\tau ({\frac  {5}{2}})}{2\tau (4)}}={\frac  {4}{3}}{\frac  {{\frac  {{\sqrt  {\pi }}}{2}}\cdot {\frac  {3}{2}}\cdot {\frac  {1}{2}}{\sqrt  {\pi }}}{2(3\cdot 2\cdot 1)}}={\frac  {\pi }{24}}\,

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