Triple Integral over a Superquadric

Question

How do I solve the following:

$\iiint _V (x^2+y^2+z^2) dxdydz $, where $V=\{x^{2n}+y^{2n}+z^{2n} \leq 1\}$ is a superquadric?

Answer 1

By symmetry, $$ \iiint _V (x^2+y^2+z^2) dxdydz=24\iiint_{V^+} x^2 dxdydz $$ where $$ V^+=\{x^{2n}+y^{2n}+z^{2n} \leq 1, x\ge0,y\ge0,z\ge0\}. $$ Let $$ x=r^{\frac1n}\sin^{\frac1n}(\theta)\cos^{\frac1n}(\phi), x=r^{\frac1n}\sin^{\frac1n}(\theta)\cos^{\frac1n}(\phi),z=r^{\frac1n}\cos^{\frac1n}(\theta). $$ Then $V^+$ becomes $$ \bar{V}^+=\{(r,\theta,\pi): 0\le r\le1, 0\le\theta,\phi\le\frac{\pi}{2}\}. $$ Now $$ \det\frac{\partial(x,y,z)}{\partial r\partial \theta\partial\psi}=\frac{1}{n^3}r^{\frac{3}{n}-1} \sin ^{\frac{2}{n}-1}(\theta ) \cos ^{\frac{1}{n}-1}(\theta ) \sin ^{\frac{1}{n}-1}(\phi ) \cos ^{\frac{1}{n}-1}(\phi ) $$ and then \begin{eqnarray} &&\iiint _V (x^2+y^2+z^2) dxdydz\\ &=&24\iiint_{V^+} x^2 dxdydz\\ &=&24\int_0^1\int_0^{\pi/2}\int_0^{\pi/2} r^{\frac2n}\sin^{\frac2n}\theta\cos^{\frac2n}\phi \det\frac{\partial(x,y,z)}{\partial r\partial \theta\partial\phi}drd\theta d\phi\\ &=&\frac{24}{n^3}\int_0^1\int_0^{\pi/2}\int_0^{\pi/2}r^{\frac{5}{n}-1} \sin ^{\frac{4}{n}-1}(\theta ) \cos ^{\frac{1}{n}-1}(\theta ) \sin ^{\frac{1}{n}-1}(\phi ) \cos ^{\frac{3}{n}-1}(\phi )drd\theta d\phi\\ &=&\frac{24}{n^3}\int_0^1r^{\frac{5}{n}-1}dr\int_0^{\pi/2}\sin ^{\frac{4}{n}-1}(\theta ) \cos ^{\frac{1}{n}-1}(\theta )d\theta \int_0^{\pi/2} \sin ^{\frac{1}{n}-1}(\phi ) \cos ^{\frac{3}{n}-1}(\phi )d\phi\\ &=&\frac{24}{n^3}\cdot\frac{n}{5}\cdot \frac{\Gamma \left(\frac{2}{n}\right) \Gamma \left(\frac{1}{2n}\right)}{2 \Gamma \left(\frac{5}{2n}\right)}\cdot\frac{\Gamma \left(\frac{1}{2n}\right) \Gamma \left(\frac{3}{2n}\right)}{2 \Gamma \left(\frac{2}{n}\right)}\\ &=&\frac{6}{5n^2}\frac{ \Gamma^2 \left(\frac{1}{2n}\right)\Gamma \left(\frac{3}{2n}\right)}{\Gamma \left(\frac{5}{2n}\right)}. \end{eqnarray} Here $$ \int_0^{\pi/2}\sin^{-1+p}(t)\cos^{-1+q}(t)dt=\frac{\Gamma \left(\frac{p}{2}\right) \Gamma \left(\frac{q}{2}\right)}{2 \Gamma \left(\frac{p+q}{2}\right)} $$ is used.

Answer 2

I found an online scan of the first or second edition of Whittaker and Watson, A Course of Modern Analysis. It includes Dirichlet's method, pages 191-193 of See if this works, links to the individual page 191, page 192, page 193.

There was a way to magnify the image, here are the beginning set-up and the final calculation. For the current problem $f=1.$ To take the integral of $x^2$ (on the positive octant) I took $$ f=1 \; , \; \; \alpha = \beta = \gamma = 2n \; , \; \; a=b=c=1 \; , \; \; p=3, q=1,r=1 $$ Then $$ \frac{p}{\alpha} + \frac{q}{\beta} + \frac{r}{\gamma} = \frac{5}{2n} $$ so that the one-variable integral on the right is $ \frac{2n}{5} $

For the integral over the whole body of $x^2 + y^2 + z^2$ we get $$ 8 \cdot \frac{\Gamma \left( 1 + \frac{3}{2n} \right)\Gamma \left( 1 + \frac{1}{2n} \right)^2}{\Gamma \left( 1 + \frac{5}{2n} \right)} $$ Here I have used the recurrence relation $$ z \; \Gamma(z) = \Gamma(1+z) $$ which is 6.1.15 in Abramowitz and Stegun. Also found it in the Wikipedia page.

This gives the correct figure for the sphere ($n=1$) which is $4 \pi / 5.$ It also gives the correct limit as $n$ goes to positive infinity, namely $8$