Evaluate $\iiint(x^{2}+y^{2}+z^{2})dxdydz$ for the positive octant, $x^{2}+y^{2}+z^{2}=9$

Question

Evaluate $\displaystyle\iiint(x^{2}+y^{2}+z^{2})dxdydz$ for the positive octant, $x^{2}+y^{2}+z^{2}=9$.

This question needs to be solved in spherical coordinates. I tried putting a limit of $r$ from $0$ to $3$, $\theta$ from $0$ to $\pi/2$ and $\varphi$ from $0$ to $\pi/2$.

But What about $dxdydz$?

Answer 1

Use $dxdydz=r^2dr\sin\theta d\theta d\phi$ (proven here) so your integral is $$\frac18\int_0^{2\pi}d\phi\int_0^\pi\sin\theta d\theta\int_0^3r^4dr=\frac182\pi2\frac{3^5}{5}=\frac{243\pi}{10}.$$

Answer 2

The region projects to the quarter-circle $x^2+y^2\le 9$, $x,y\ge 0$, in the $xy$-plane, and so the triple integral in cartesian coordinates, yucky though it be, is $$\int_0^3\int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{9-x^2-y^2}}(x^2+y^2+z^2)dz\,dy\,dx.$$ You can actually evaluate this with some trigonometric substitutions, but, ultimately, spherical coordinates is the way to go.