Q:

Evaluate e y2z2 dv, where e lies above the cone ϕ = π/3 and below the sphere ρ = 1.

Accepted Solution

A:
In spherical coordinates, we set[tex]x=\rho\cos\theta\sin\varphi[/tex][tex]y=\rho\sin\theta\sin\varphi[/tex][tex]z=\rho\cos\varphi[/tex]so that the volume element under this transformation becomes[tex]\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=|\det\mathbf J|\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]The region [tex]E[/tex] is given by the set[tex]\left\{(\rho,\theta,\varphi)\mid0\le\rho\le1,0\le\theta\le2\pi,0\le\varphi\le\dfrac\pi3\right\}[/tex]so that the integral is[tex]\displaystyle\iiint_Ey^2z^2\,\mathrm dV=\int_{\varphi=0}^{\varphi=\pi/3}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=1}\rho^6\sin^2\theta\sin^3\varphi\cos^2\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex][tex]\displaystyle=\left(\int_0^{\pi/3}\sin^3\varphi\cos^2\varphi\,\mathrm d\varphi\right)\left(\int_0^{2\pi}\sin^2\theta\,\mathrm d\theta\right)\left(\int_0^1\rho^6\,\mathrm d\rho\right)[/tex][tex]=\dfrac{47}{480}\cdot\pi\cdot\dfrac17=\dfrac{47\pi}{3360}[/tex]