VC5.58

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Using divergence theorm,we have

\iint _{S}(x^{3}dydz+x^{2}ydzdx+x^{2}zdxdy)\,

=\iiint _{V}[{\frac  {\partial x^{3}}{\partial x}}+{\frac  {\partial (x^{2}y)}{\partial y}}+{\frac  {\partial (x^{2}z)}{\partial z}}]dxdydz\,

=\iiint _{V}(3x^{2}+x^{2}+x^{2})dxdydz=5\iiint _{V}x^{2}dxdydz\,

=5\int _{{x=-a}}^{{a}}\int _{{y=-{\sqrt  {(a^{2}-x^{2})}}}}^{{{\sqrt  {(a^{2}-x^{2})}}}}\int _{{z=0}}^{{b}}x^{2}dxdydz\,

=5\times 2\times 2\int _{{x=0}}^{{a}}\int _{{y=0}}^{{{\sqrt  {(a^{2}-x^{2})}}}}\int _{{z=0}}^{{b}}x^{2}dxdydz\,

=20\int _{0}^{a}\int _{{0}}^{{{\sqrt  {a^{2}-x^{2}}}}}x^{2}[z]_{0}^{b}dxdy\,

=20b\int _{0}^{a}x^{2}[y]_{{0}}^{{{\sqrt  {a^{2}-x^{2}}}}}dx=20b\int _{0}^{a}x^{2}{\sqrt  {a^{2}-x^{2}}}dx\,

=20b\int _{{0}}^{{{\frac  {\pi }{2}}}}a^{2}\sin ^{2}\theta (a\cos \theta )(a\cos \theta )d\theta \, by converting to polar coordinates

=20ba^{4}\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta \cos ^{2}\theta d\theta =5ba^{4}\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin ^{2}2\theta d\theta =5ba^{4}\int _{{0}}^{{{\frac  {\pi }{2}}}}{\frac  {1-\cos 4\theta }{2}}d\theta ={\frac  {5ba^{4}}{2}}[\theta -{\frac  {\sin 4\theta }{4}}]_{{0}}^{{{\frac  {\pi }{2}}}}\,

={\frac  {5ba^{4}\pi }{4}}\,

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