VC5.55

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Let V be the volume of the cone enclosed by the given closed surface S.O is the vertex and base ABC in the circle x^{2}+y^{2}=1,z=1\,

Now,by divergence theorm,we have

\iint _{S}(xi+yj+z^{2}k)\cdot ndS\,

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

=2\int _{{z=0}}^{{1}}\int _{{y=-z}}^{{z}}\int _{{x=-{\sqrt  {(z^{2}-y^{2})}}}}^{{{\sqrt  {(z^{2}-y^{2})}}}}(1+z)dxdydz\,

=2\int _{0}^{1}\int _{{-z}}^{{z}}(1+z)[x]_{{x=-{\sqrt  {(z^{2}-y^{2})}}}}^{{{\sqrt  {(z^{2}-y^{2})}}}}dydz\,

=8\int _{0}^{1}\int _{0}^{z}(1+z){\sqrt  {(z^{2}-y^{2})}}dydz\,

=8\int _{0}^{1}(1+z)[{\frac  {y}{2}}{\sqrt  {(z^{2}-y^{2})}}+{\frac  {z^{2}}{2}}\arcsin {\frac  {y}{z}}]_{0}^{z}dz\,

=8\int _{0}^{1}(1+z)[{\frac  {z^{2}}{2}}\times {\frac  {\pi }{2}}]dz=2\pi \int _{0}^{1}(z^{2}+z^{2})dz\,

=2\pi [{\frac  {z^{3}}{3}}+{\frac  {z^{4}}{4}}]_{0}^{1}=2\pi [{\frac  {1}{3}}+{\frac  {1}{4}}]={\frac  {7\pi }{16}}\,


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