Triple Integral $\iiint x^{2n}+y^{2n}+z^{2n}dV$

Question

Evaluate: $$\iiint_{x^2+y^2+z^2 \leqslant 1} x^{2n}+y^{2n}+z^{2n} dV $$

I have tried to convert to spherical polars and then compute the integral, but it gets really messy because of the 2n power. Any tips?

Answer 1

First observation: it is symmetric in $x,y,z$, so by linearity we have $$\iiint_{x^2+y^2+z^2 \leqslant 1} x^{2n}+y^{2n}+z^{2n} dV =3\iiint_{x^2+y^2+z^2 \leqslant 1} z^{2n} dV.$$

Choosing spherical coordinates it becomes $$3\iiint_{x^2+y^2+z^2 \leqslant 1} (r\cos \theta)^{2n} dV$$ where $dV= r^2 \sin \theta \ \text{d}r \ \text{d}\theta \ \text{d}\phi$. Thus the integral simplifies to $$3 \int_0^{2\pi}\int_0^{\pi}\int_0^1 r^{2(n+1)} (\cos \theta)^{2n} \sin \theta \ \text{d}r \ \text{d}\theta \ \text{d}\phi = \frac{3}{2n+3}2 \pi \int_0^{\pi}(\cos \theta)^{2n} \sin \theta \ \text{d}\theta. $$ Using that $$\int_0^{\pi}(\cos \theta)^{2n} \sin \theta \ \text{d}\theta = \frac{2}{2n+1} $$ we have $$\iiint_{x^2+y^2+z^2 \leqslant 1} x^{2n}+y^{2n}+z^{2n} dV= \frac{3}{2n+3}2 \pi \frac{2}{2n+1}.$$