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