Question

Use spherical coordinates. evaluate xyz dv e , where e lies between the spheres ρ = 4 and ρ = 6 and above the cone ϕ = π/3.

Answer 1

Set

$x=\rho\cos\theta\sin\varphi$

$y=\rho\sin\theta\sin\varphi$

$z=\rho\cos\varphi$

so that the volume element is

$\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

The integral is then

$\displaystyle\iiint_Exyz\,\mathrm dV=\int_{\varphi=0}^{\varphi=\pi/3}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=4}^{\rho=6}\rho^5\cos\theta\sin\theta\sin^3\varphi\cos\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\rho$

$=\displaystyle\left(\int_0^{\pi/3}\sin^3\varphi\cos\varphi\,\mathrm d\varphi\right)\underbrace{\left(\int_0^{2\pi}\cos\theta\sin\theta\,\mathrm d\theta\right)}_0\left(\int_4^6\rho^5\,\mathrm d\rho\right)$

and so evaluates to 0.