# VC4.22

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 $x=\sin \theta ,dx=\cos \theta d\theta \,$

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