Triple Integral Over a Ball How do I compute the following:
$\iiint _V (x^{2n} + y^{2n} +z^{2n}) dxdydz$ where $V = \{x^2 +y^2 +z^2 \leq 1\}$?
Due to symmetry
$$\iiint _V (x^{2n} + y^{2n} +z^{2n}) dxdydz
= 3\iiint _V z^{2n} dxdydz\\
= 3\int_0^{2\pi}\int_0^\pi \int_0^1 r^{2n}\cos^{2n}\theta \&gt;r^2\sin\theta drd\theta d\phi
 = \frac{12\pi}{(2n+3)(2n+1)}
$$
Your question as it stands is likely incorrect.  Are you really interested in the cases where $n&lt;0$, for instance??
Assuming $n \geq 1/2$, and using symmetry:
$$8 \int\limits_{x=0}^1 \int\limits_{y=0}^{\sqrt{1-x^2}} \int\limits_{z=0}^{\sqrt{1 - x^2 - y^2}} (x^{2 n} + y^{2 n} + z^{2 n})\ dx\ dy\ dz = \frac{12 \pi }{4 n (n+2)+3}$$,
which is the same as @Quanto's later answer.  (Though somehow mine is downvoted.)