Volume integral of $(x+y+z)^{2n}$ over unit sphere $\iiint_{V}(x+y+z)^{2n}\,dV$ where $V$ is the unit sphere
I tried converting to spherical polars but ended up needing to solve $$\int_0^\pi\int_0^{2\pi}(\sin\theta\cos\phi+\sin\theta\sin\phi + \cos\theta)^{2n}\sin\theta\,d\phi\,d\theta$$ which I cannot see a way of doing, anyone have any ideas?
 A: As @NinadMunshi comments, it is fairly easy to rearrange this particular integral to an essentially one-dimensional integral... and a generalization of this separation-of-variables idea can evaluate $\int_{S^{n-1}}x_1^{2e_1}\ldots x_n^{2e_n}\;dx_1\ldots dx_n$ for all $n$-tuples of non-negative integers $e_i$. (And, equivalently, the corresponding integral over the solid ball.)
The trick (very good one, because it still does admit further generalizations) is to evaluate $\int_{\mathbb R^n} x_1^{2e_1}\ldots x_n^{2e_n}\,e^{-(x_1^2+\ldots+x_n^2)}\;dx$ in two different ways.
The first is to evaluate it as a product of one-dimensional integrals
$$
\int_{\mathbb R} x_j^{2e_j}e^{-x_j^2}\,dx_j
\;=\; 2\int_0^\infty x_j^{2e_j+1} e^{-x_j^2}\;{dx\over x}
\;=\; \int_0^\infty x_j^{e_j+{1\over 2}} e^{x_j}\;{dx\over x}
\;=\; \Gamma(e_j+{1\over 2})
$$
From $\Gamma(1/2)=\sqrt{\pi}$ and $s\Gamma(s)=\Gamma(s+1)$, these special values are explicit and elementary.
On the other hand, integrating over $(0,\infty)\times S^{n-1}$ gives
$$
\int_0^\infty t^{2e_1+\ldots+2e_n} e^{-t^2}\,dt
\times \int_{S^{n-1}} x_1^{2e_1}\ldots x_n^{2e_n}\;dx
$$
As in the previous paragraph, the first integral is ${1\over 2}\Gamma(e_1+\ldots e_n+{1\over 2})$.
A: As @NinadMunshi commented, the trick here is through a rotation of coordinates, so that the original coordinate vector $r$ and the new coordinate vector $r'$ are related by
$ r = R r' $
were $R = [u_1, u_2, u_3] $.  Taking the third column vector of $R$ to be
$u_3 = [1, 1, 1]/\sqrt{3} $
Then $ x + y + z = [1, 1, 1] r = [1, 1, 1] R r' = \sqrt{3} z' $
And using spherical coordiantes for $r'$ we have
$ x' = \rho \sin(t) \cos(s) , y' = \rho \sin(t) \sin(s) , z' = \rho \cos(t) $
Then the integral becomes
$I = \displaystyle \int_{\rho = 0}^{1} \int_{s=0}^{2\pi} \int_{t=0}^{\pi} (\sqrt{3} \rho \cos(t) )^{2n} \rho^2 \sin(t) dt ds d\rho $
Integrating with respect to $\rho$ and $\phi$, reduce the integral to
$I =(2 \pi) \dfrac{(3)^n}{2n+3} \displaystyle \int_0^\pi \cos^{2n}(t) \sin(t) dt $
And this evaluates to
$I = \dfrac{(4 \pi) (3)^n}{(2n+3)(2n + 1)} $
