MvCalc66

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\int _{{-2}}^{{2}}\int _{{0}}^{{{\sqrt  {4-y^{2}}}}}\int _{{-{\sqrt  {4-x^{2}-y^{2}}}}}^{{{\sqrt  {4-x^{2}-y^{2}}}}}y^{2}{\sqrt  {x^{2}+y^{2}+z^{2}}}dzdxdy\,

The region of integration is a hemisphere along the positive x axis intersected by the y-z plane


In spherical co-ordinates the region is given by:


R={(\rho ,\theta ,\phi ):0\leq \rho \leq 2,-{\frac  {\pi }{2}}\leq \theta \leq {\frac  {\pi }{2}},0\leq \phi \leq \pi }\,


x=\rho \sin \phi \cos \theta \,

y=\rho \sin \phi \sin \theta \,

z=\rho \cos \phi \,

\rho ^{2}=x^{2}+y^{2}+z^{2}\,


So.. \int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\int _{{0}}^{{\pi }}\int _{{0}}^{{2}}(\rho \sin \phi \sin \theta )^{2}(\rho )\rho ^{2}\sin \phi d\rho d\phi d\theta \,


\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\int _{{0}}^{{\pi }}\int _{{0}}^{{2}}\rho ^{2}\sin ^{2}\phi \sin ^{2}\theta (\rho )\rho ^{2}\sin \phi d\rho d\phi d\theta \,


\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta d\theta \int _{{0}}^{{\pi }}\sin ^{3}\phi d\phi \int _{{0}}^{{2}}\rho ^{5}d\rho \,


\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta d\theta \int _{{0}}^{{\pi }}\sin ^{3}\phi d\phi [{\frac  {1}{6}}\rho ^{6}]_{{0}}^{{2}}\,


{\frac  {32}{3}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta d\theta \int _{{0}}^{{\pi }}\sin ^{3}\phi d\phi \,


{\frac  {32}{3}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta d\theta \int _{{0}}^{{\pi }}\sin \phi (1-\cos ^{2}\phi )d\phi \,


{\frac  {32}{3}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta d\theta \int _{{0}}^{{\pi }}(\sin \phi -\sin \phi \cos ^{2}\phi )d\phi \,


{\frac  {32}{3}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta d\theta [\int _{{0}}^{{\pi }}\sin \phi d\phi -\int _{{0}}^{{\pi }}\sin \phi \cos ^{2}\phi d\phi ]\,

By u substitution let u = cosφ so then du = -sinφ dφ and when φ = π, u = -1 and when φ = 0, u = 1

{\frac  {32}{3}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta d\theta [\int _{{0}}^{{\pi }}\sin \phi d\phi -\int _{{1}}^{{-1}}-u^{2}du]\,


{\frac  {32}{3}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta d\theta [[-\cos \phi ]_{{0}}^{{\pi }}+[{\frac  {1}{3}}u^{3}]_{{1}}^{{-1}}]\,


{\frac  {32}{3}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta d\theta [2-{\frac  {2}{3}}]\,


{\frac  {128}{9}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta d\theta \,

By half angle formula..

\sin ^{2}\theta =({\frac  {1}{2}}-{\frac  {1}{2}}\cos(2\theta ))\,

So..


{\frac  {128}{9}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}({\frac  {1}{2}}-{\frac  {1}{2}}\cos(2\theta ))d\theta \,


{\frac  {64}{9}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}d\theta -{\frac  {64}{9}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}\cos(2\theta ))d\theta \,

By u substitution let u = 2θ so then 1/2du = dθ and when θ = π/2, u = π and when θ = -π/2, u = -π


{\frac  {64}{9}}\int _{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}d\theta -{\frac  {64}{9}}\int _{{-\pi }}^{{\pi }}\cos(u))du\,


{\frac  {64}{9}}[\theta ]_{{-{\frac  {\pi }{2}}}}^{{{\frac  {\pi }{2}}}}-{\frac  {64}{9}}[sin(u)]_{{-\pi }}^{{\pi }}\,

{\frac  {64\pi }{9}}-{\frac  {64}{9}}[0]\,


Hence \int _{{-2}}^{{2}}\int _{{0}}^{{{\sqrt  {4-y^{2}}}}}\int _{{-{\sqrt  {4-x^{2}-y^{2}}}}}^{{{\sqrt  {4-x^{2}-y^{2}}}}}y^{2}{\sqrt  {x^{2}+y^{2}+z^{2}}}dzdxdy={\frac  {64\pi }{9}}\,