VC4.25

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

On the xy-plane the region R is surrounded by x^{2}+y^{2}=a^{2},z=0\,

The required surface integral is \iint _{S}F\cdot dS=\iint _{S}F\cdot ndS\, --(1)

=\iint _{R}F\cdot n{\frac  {dxdy}{|n\cdot k|}}\,

n={\frac  {\nabla \phi }{|\nabla \phi |}}\,

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

Therefore,n={\frac  {2(xi+yj+zk)}{{\sqrt  {4x^{2}+4y^{2}+4z^{2})}}}}={\frac  {x}{a}}i+{\frac  {y}{a}}j+{\frac  {z}{a}}k\,

Therefore,n\cdot k={\frac  {z}{a}}\,

and F\cdot n=(y^{2}zi+z^{2}xj+x^{2}yk)\cdot ({\frac  {x}{a}}i+{\frac  {y}{a}}j+{\frac  {z}{a}}k={\frac  {1}{a}}(xy^{2}z+xyz^{2}+x^{2}yz)\,

Therefore,from (1),the required surface integral is \iint _{R}{\frac  {1}{a}}(xy^{2}z+xyz^{2}+x^{2}yz)\cdot {\frac  {a}{z}}dxdy\,

=\iint _{R}(xy^{2}+xyz+x^{2}y)dxdy=\iint _{R}[xy^{2}+xy{\sqrt  {(a^{2}-x^{2}-y^{2})}}+x^{2}y]dxdy\,

The integral can be evaluated in the following way.

=\int _{{x=0}}^{{a}}\int _{{y=0}}^{{{\sqrt  {a^{2}-x^{2}}}}}[xy^{2}+xy{\sqrt  {(a^{2}-x^{2}-y^{2})}}+x^{2}y]dxdy\,

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

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

=[-{\frac  {1}{3}}{\frac  {(a^{2}-x^{2})^{{{\frac  {5}{2}}}}}{{\frac  {5}{2}}}}+{\frac  {1}{6}}x^{3}a^{2}-{\frac  {1}{10}}x^{5}]_{{0}}^{{a}}={\frac  {1}{5}}a^{5}\,

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