MvCalc21

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The region of integration is the upper half of the circle x^{2}+y^{2}=2ax\,.If we change to polars,this region becomes r=2a\cos \theta \, from \theta =0\, to \theta ={\frac  {\pi }{2}}\,

Hence the integral becomes \int _{{0}}^{{{\frac  {\pi }{2}}}}\int _{{0}}^{{2a\cos \theta }}r^{2}(rdr)d\theta =\int _{{0}}^{{{\frac  {\pi }{2}}}}\int _{{0}}^{{2a\cos \theta }}r^{3}drd\theta \,

=\int _{{0}}^{{{\frac  {\pi }{2}}}}[{\frac  {r^{4}}{4}}]_{{0}}^{{2a\cos \theta }}d\theta \,

={\frac  {1}{4}}\int _{{0}}^{{{\frac  {\pi }{2}}}}16a^{4}\cos ^{4}\theta d\theta \,

=4a^{4}\int _{{0}}^{{{\frac  {\pi }{2}}}}\cos ^{4}\theta d\theta \,

=4a^{4}[{\frac  {3}{16}}\pi ={\frac  {3}{4}}a^{4}\pi \,

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