Triple integral over a sphere with parameter $2n$? I need to integrate $x^{2n}+y^{2n}+z^{2n}$ over a sphere of equation $x²+y²+z²=1$.
I have thought of changing the coordinates from cartesian to spherical but I don't know how to deal with the integrant in this case.
Or maybe should I divide the integral in three parts? Or even use the vector field $r=(x,y,z)$, but then how?
Any help would be amazing there.
Thank you.
 A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\large\mbox{If}\quad {\tt n = 0,1,2,\ldots}$:
\begin{align}
&\color{#00f}{\large\iiint_{x^{2} + y^{2} + z^{2} < 1}
\pars{x^{2n} + y^{2n} + z^{2n}}\,\dd x\,\dd y\,\dd z}
=3\iiint_{x^{2} + y^{2} + z^{2} < 1}
z^{2n}\,\dd x\,\dd y\,\dd z
\\[3mm]&=3\int_{0}^{1}r^{2n + 2}\,\dd r\int_{0}^{2\pi}\dd\phi
\int_{0}^{\pi}\cos^{2n}\pars{\theta}\sin\pars{\theta}\,\dd\theta
={6\pi \over 2n + 3}\int_{-1}^{1}\xi^{2n}\,\dd\xi
\\[3mm]&=\color{#00f}{\large{12\pi \over \pars{n + 1}\pars{2n + 3}}}
\end{align}
A: Let $f(x,y,z) = x^{2n} + y^{2n} + z^{2n}$
See $\Delta . f(x,y,z) = i 2n x^{2n -1} + j 2ny^{2n-1} + k2nz^{2n-1} = A$ (Let us write down)
Your $S(x,y,z) = x^{2} + y^{2} + z^{2}$
So $\Delta S = 2xi + 2yj + 2zk$
Unite normal vector to the given surface is $n = \frac{2xi + 2yj+ 2zk}{\sqrt{4x^2 + 4y^2 + 4z^2}} = \frac{xi + yj+ zk}{r}$
Now see $A.n = \frac{2n}{r}f(x,y,z)$. Calculate, very easy.
Thus $f = \frac{r}{2n} A.n$.
Now intrgrate.
$$\int \int_S f.dS = \int \int \int_V \Delta. A dV = 0$$
as divergence (Gradient $\phi$) = 0, for a scaler valued function and here $\phi = f$
Please edit typesetting of the answer.
