# Triple Integrate $x^{2n}+y^{2n}+z^{2n}$

Evaluate: $$\iiint x^{2n}+y^{2n}+z^{2n} \; \mathrm dV$$

over the region $x^2+y^2+z^2 \le 1$

I think I know how to find the limits, but I don't know how to do the integration. I see that I have to do a change of variable, but I don't know what coordinates to change into, since the $n$ in the power $2n$ is making it horrible.

I thought of changing $x^{2n}+y^{2n}+z^{2n}$ into a dot product between $(x^{n},y^{n},z^{n})$ with itself, but that didn't get me anywhere at all (or I just don't know how to proceed...?)

Any suggestions anyone? Thanks a lot!

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By symmetry, you only need to figure out how to calculate $\int x^{2n} dx dy dz$. Now if you fix $x$ and integrate over $dy dz$ first, you will get something very simple... –  achille hui Feb 10 '13 at 18:48

Let $x = r\sin(\theta) \cos(\phi)$, $y = r\sin(\theta) \sin(\phi)$ and $z = r \cos(\theta)$, where $\theta \in [0, \pi]$ and $\phi[0, 2 \pi)$. Let $$I = \int_{S^3} \left(x^{2n} + y^{2n} + z^{2n}\right) dx dy dz$$
Hence, \begin{align} I & = \int_{r=0}^1 \int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi} r^{2n} \left(\sin^{2n}(\theta) \cos^{2n}(\phi) + \sin^{2n}(\theta) \sin^{2n}(\phi) + \cos^{2n}(\theta)\right) r^2 \sin(\theta) d \theta d \phi dr\\ & = \dfrac1{2n+3}\int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi} \left(\sin^{2n}(\theta) \cos^{2n}(\phi) + \sin^{2n}(\theta) \sin^{2n}(\phi) + \cos^{2n}(\theta)\right)\sin(\theta) d \theta d \phi \end{align} We have $$\int_{\phi=0}^{2 \pi} \int_{\theta=0}^{\pi} \sin^{2n+1}(\theta) \cos^{2n}(\phi) d\theta d \phi = \left(\int_{\theta=0}^{\pi} \sin^{2n+1}(\theta) d \theta\right)\left(\int_{\phi=0}^{2\pi} \cos^{2n}(\phi) d \phi\right) \,\,\,\, (\spadesuit)$$ $$\int_{\phi=0}^{2 \pi} \int_{\theta=0}^{\pi} \sin^{2n+1}(\theta) \sin^{2n}(\phi) d\theta d \phi = \left(\int_{\theta=0}^{\pi} \sin^{2n+1}(\theta) d \theta\right)\left(\int_{\phi=0}^{2\pi} \sin^{2n}(\phi) d \phi\right) \,\,\,\, (\diamond)$$
$$\int_{\phi=0}^{2 \pi} \int_{\theta=0}^{\pi} \cos^{2n+1}(\theta) \sin(\theta) d\theta d \phi = 2 \pi \int_{\theta=0}^{\pi} \cos^{2n+1}(\theta) \sin(\theta) d\theta = 2 \pi \int_{t=-1}^{t=1} t^{2n+1} dt\\ = 2 \pi \left(\dfrac1{2n+2} + \dfrac1{2n+2} \right) = \dfrac{2 \pi}{n+1}$$ Now $$\int_{\phi=0}^{2\pi} \sin^{2n}(\phi) d \phi = \int_{\phi=0}^{2\pi} \cos^{2n}(\phi) d \phi = 4 \left(\dfrac{2n-1}{2n} \cdot \dfrac{2n-3}{2n-2} \cdot \dfrac{2n-5}{2n-4} \cdots \dfrac34 \cdot \dfrac12 \cdot \dfrac{\pi}2\right)$$ and $$\int_{\theta=0}^{\pi} \sin^{2n+1}(\theta) d \theta = 2 \left(\dfrac{2n}{2n+1} \cdot \dfrac{2n-2}{2n-1} \cdots \dfrac45 \cdot \dfrac23 \right)$$ where the last couple of integrals are computed here.