Triple integral over a sphere with parameter $2n$?

Question

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.

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\large\mbox{If}\quad {\tt n = 0,1,2,\ldots}$: \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}