# MvCalc66

$\displaystyle \int_{-2}^{2}\int_{0}^{\sqrt{4-y^2}}\int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}} y^2\sqrt{x^2+y^2+z^2} dzdxdy \,$

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:

$\displaystyle R = {(\rho,\theta,\phi):0\le\rho\le2, -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}, 0\le\phi\le\pi} \,$

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

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

$\displaystyle z = \rho\cos\phi \,$

$\displaystyle \rho^2 = x^2+y^2+z^2 \,$

So.. $\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{\pi}\int_{0}^{2} (\rho\sin\phi\sin\theta)^2(\rho)\rho^2\sin\phi d\rho d\phi d\theta \,$

$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{\pi}\int_{0}^{2} \rho^2\sin^2\phi\sin^2\theta(\rho)\rho^2\sin\phi d\rho d\phi d\theta \,$

$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2\theta d\theta \int_{0}^{\pi}\sin^3\phi d\phi \int_{0}^{2} \rho^5 d\rho \,$

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

$\displaystyle \frac{32}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2\theta d\theta \int_{0}^{\pi}\sin^3\phi d\phi\,$

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

$\displaystyle \frac{32}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2\theta d\theta \int_{0}^{\pi}(\sin\phi-\sin\phi\cos^2\phi) d\phi\,$

$\displaystyle \frac{32}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2\theta d\theta [\int_{0}^{\pi}\sin\phi d\phi-\int_{0}^{\pi}\sin\phi\cos^2\phi d\phi]\,$

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

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

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

$\displaystyle \frac{32}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2\theta d\theta [2-\frac{2}{3}]\,$

$\displaystyle \frac{128}{9}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2\theta d\theta \,$

By half angle formula..

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

So..

$\displaystyle \frac{128}{9}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\frac{1}{2}-\frac{1}{2}\cos(2\theta))d\theta \,$

$\displaystyle \frac{64}{9}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta-\frac{64}{9}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(2\theta))d\theta \,$

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

$\displaystyle \frac{64}{9}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta-\frac{64}{9}\int_{-\pi}^{\pi}\cos(u))du \,$

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

$\displaystyle \frac{64\pi}{9} -\frac{64}{9} \,$

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