# Triple integral depending on a parameter

1. Feb 4, 2013

### Draconian28

1. The problem statement, all variables and given/known data

Find $$\iiint (x^{2n} + y^{2n} + z^{2n})\,dV$$ where the integral is taken over the region of 3D space where $$x^{2} + y^{2} + z^{2} \leq 1$$

2. Relevant equations

3. The attempt at a solution

I tried doing this in Cartesian coordinates, but the limits of integration got very messy and I got stuck after doing the first integral. I also tried using spherical polar coordinates, and then the limits of integration are quite simple, but the integrand gets complicated, unless $n = 1$, in which case the integral is quite easy to do.

I then thought that, since the only case where this looks simple enough to do directly is $n = 1$, I could try to make a conjecture as to what the value of the integral is for general n and then try to prove it by induction. The problem with that, though, is that I don't see how to go from the case $n = k + 1$ to the case $n = k$.

Any ideas?

2. Feb 4, 2013

### LCKurtz

I haven't tried it myself yet, and the wife wants the computer just now, but have you tried spherical coordinates? That would be the natural first choice. May or may not work...

[Edit later] I think you can work it with spherical coordinates using symmetry and the formula$$\int_0^\frac \pi 2 \sin^n x\, dx =\int_0^\frac \pi 2 \cos^n x\, dx = \frac{1\cdot 3\cdot 5\cdot\cdot\cdot(n-1)}{2\cdot 4\cdot 6\cdot\cdot\cdot n}\frac \pi 2$$which is valid for $n$ even and $n \ge 2$.

Last edited: Feb 4, 2013
3. Feb 4, 2013

### SteamKing

Staff Emeritus
This integral covers the volume enclosed by a sphere with radius = 1. It can be broken up into the sum of its parts, since the integrand is a sum.

For n = 1, the integral will evaluate to the sum of the inertias of a sphere about the three coordinate axes, assuming a unit density for the sphere.