$\iiint_V \ x^{2n} + y^{2n} + z^{2n} \,dx\,dy\,dz$

You should use the symmetry to reduce it to $$3\iiint_V \ z^{2n} \,dx\,dy\,dz,$$ then use spherical coordinates to integrate.


In spherical coordinates : $$\begin{cases} x=r\cos(\theta)\sin(\phi) \\ y=r\sin(\theta)\sin(\phi) \\ z=r\cos(\phi) \\ \end{cases}$$ $$\iiint_V \ x^{2n} + y^{2n} + z^{2n} \,dx\,dy\,dz = \iiint_V \ \left( \cos^{2n}(\theta)\sin^{2n}(\phi) + \sin^{2n}(\theta)\sin^{2n}(\phi) +\cos^{2n}(\phi) \right) r^{2n+2}\sin(\phi)\,dr\,d\theta\,d\phi $$ in : $0<r<1\quad;\quad 0<\theta<2\pi \quad;\quad 0<\phi<\pi$

With the runaroud's remark : $$ = 3\iiint_V \ r^{2n+2}\cos^{2n}(\phi)\sin(\phi)\,dr\,d\theta\,d\phi = 3 \int_0^{1}r^{2n+2}dr \int_0^{2\pi}d\theta \int_0^{\pi}\cos^{2n}(\phi)\sin(\phi)d\phi = 3 \frac{1}{2n+3}2\pi\frac{2}{2n+1} = \frac{12\pi}{(2n+3)(2n+1)}$$


Following runaround's excellent hint, the required integral is $$3\int_{-1}^1 z^{2n}\hskip-1.4cm \underbrace{\int\int dx\,dy}_{\mbox{area of disk of radius }\sqrt{1-z^2}} \hskip-1.3cm dz=3\pi \int_{-1}^1 z^{2n}(1-z^2)\,dz={12\pi\over(2n+3)(2n+1)}.$$

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Integration

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