Any shorter way to compute this integral over the unit ball?

Question

Let $n$ be a non-negative integer, and let $V$ be the unit ball centered at the origin. To compute: \begin{equation*}I = \iiint\limits_{V}(x+y+z)^{2n}{\mathop{}\!\mathrm{d}}V\end{equation*} I substituted for $x$, $y$, and $z$ with spherical coordinates, did the $r$-integral, then used the binomial expansion on the remaining $((\sin\theta\cos\phi+\sin\theta\sin\phi)+\cos\theta)^{2n}$ and integrated the resulting series. Eventually, after a lot of algebra, I got the result \begin{equation*}I = \frac{8 \cdot 3^n \pi(n+1)}{(2n+3)(2n+2)(2n+1)}\end{equation*} but I am wondering if there is any shorter way to do this computation. Any suggestions would be appreciated (or explanations as to why no shorter method is possible, if that is the case).

Answer 1

If you rotate this such that:

$u = \frac {1}{\sqrt 3} x + \frac {1}{\sqrt 3} y + \frac {1}{\sqrt 3} z\\ v = \frac {1}{\sqrt 2} x - \frac {1}{\sqrt 2} y\\ w = \frac {1}{\sqrt 6} x + \frac {1}{\sqrt 6} y - \frac {2}{\sqrt 6} z\\ $

We have an ortho-normal transformation.

It doesn't do anything to the unit ball. i.e. $u^2 + v^2 + w^2 = 1$

But the integrand becomes $(\sqrt 3 u)^{2n}$

$\iiint 3^n u^{2n} \ du\ dv\ dw$

Then we can convert to spherical.

$3^n\iiint \rho^{2n+2}\cos^{2n}\phi\sin\phi \ d\rho\ d\phi\ d\theta$

$2\pi 3^n (\frac{1}{2n+3})(\frac {2}{2n+1})$

Which is equivalent to what you have.