Question

2) Use spherical coordinates to find the integral of f(x,y,z) = z over x^2 + y^2 + z^2 < 4 and x,y,z < 0

Answer 1

1. Denote by $\mathcal D$ the region bounded by the cylinder $x^2+y^2=9$ and the planes $z=0$ and $z=6$. Converting to cylindrical coordinates involves setting


$\begin{cases}x=r\cos u\\y=r\sin u\\z=v\end{cases}$


and $\mathcal D$ is obtained by varying the parameters over $0\le r\le3$, $0\le u\le2\pi$, and $0\le v\le6$. The volume element is


$\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm du\,\mathrm dv$

so the integral becomes

$\displaystyle\iiint_{\mathcal D}xz\,\mathrm dV=\int_{v=0}^{v=6}\int_{u=0}^{u=2\pi}\int_{r=0}^{r=3}r^2v\cos u\,\mathrm dr\,\mathrm du\,\mathrm dv=0$

2. Now let $\mathcal D$ denote the region bounded by the sphere $x^2+y^2+z^2=4$ and the coordinate planes in the octant in which each coordinate of the point $(x,y,z)$ is negative.

Converting to spherical coordinates, we set

$\begin{cases}x=\rho\cos u\sin v\\y=\rho\sin u\sin v\\z=\rho\cos v\end{cases}$

where we obtain $\mathcal D$ by varying over $0\le\rho\le2$, $\pi\le u\le\dfrac{3\pi}2$, and $\dfrac\pi2\le v\le\pi$. The volume element is now

$\mathrm dV=\rho^2\sin v\,\mathrm d\rho\,\mathrm du\,\mathrm dv$

so the integral becomes

$\displaystyle\iiint_{\mathcal D}z\,\mathrm dV=\int_{v=\pi/2}^{v=\pi}\int_{u=\pi}^{u=3\pi/2}\int_{\rho=0}^{\rho=2}\rho^3\cos v\sin v\,\mathrm d\rho\,\mathrm du\,\mathrm dv=-\pi$