MvCalc64

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Let \rho =kr\,,then M=\iiint \rho dV=8\int _{{0}}^{{{\frac  {\pi }{2}}}}\int _{{0}}^{{{\frac  {\pi }{2}}}}\int _{0}^{a}(kr)(r^{2}\sin \phi )drd\theta d\phi \,

=8k{\frac  {1}{4}}a^{4}[-\cos \phi ]_{{0}}^{{{\frac  {\pi }{2}}}}{\frac  {\pi }{2}}\,

=k\pi a^{4}\,

We have I_{x}+I_{y}+I_{z}=2I_{0}\,

But,by symmetry,I_{x}=I_{y}=I_{z}\,

Hence I_{x}=I_{y}=I_{z}={\frac  {2}{3}}I_{0}\,

I_{0}=\iiint \rho (x^{2}+y^{2}+z^{2})dV\,

We have I_{z}={\frac  {2}{3}}I_{0}={\frac  {2}{3}}8\int _{{0}}^{{{\frac  {\pi }{2}}}}\int _{{0}}^{{{\frac  {\pi }{2}}}}\int _{0}^{a}(kr)(r^{2})r^{2}\sin \phi drd\theta d\phi \,

={\frac  {16}{3}}k{\frac  {1}{6}}a^{6}[-\cos \phi ]_{{0}}^{{{\frac  {\pi }{2}}}}{\frac  {\pi }{2}}\,

={\frac  {4}{9}}k\pi a^{6}\,

={\frac  {4}{9}}Ma^{2}\,

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