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A young friend of mine had a homework problem involving a surface integral over the surface of the equation $x^8+y^8+z^8=64$. The homework problem was solvable using the divergence theorem, but he was curious how to calculate the volume of the bounded complementary component of this surface using triple integrals and I don't remember enough of these tricks to figure it out.

Is there a nice formula for calculating volumes of these 'fat spheres' defined by $x^{2n}+y^{2n}+z^{2n} = a$?

We tried spherical coordinates but the mess was prodigious.

John Samples
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    Have you tried the change of variables theorem? Start with the unit ball and map $f(x,y,z)=(x^4,y^4,z^4)$. Then integrate over the ball in spherical coordinates. Don’t forget to take the absolute value of the Jacobian; integrating over the first octant and multiplying by $8$ is recommended. – Ted Shifrin Dec 08 '23 at 00:37
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    One can use the Gamma function for a general closed form here. See the section "Balls with $L_p$ norms here" They define the $L_p$ ball as satisfying $(x_0^p + x_1^p + ... x_n^p)^{\frac{1}{p}} = r$ we can translate this to $x_0^p + x_1^p ... x_n^p= r^p$ then per the article $\frac{(2\Gamma(1/p + 1))^n}{\Gamma(n/p+1)} R^n$ is the volume. In this case $n=3, p=8, r=2^{\frac{3}{4}}$ which simplifies a bit to $ 32 \sqrt[4]{2} \frac{\Gamma(9/8)^3}{\Gamma(11/8)} $ – Sidharth Ghoshal Dec 08 '23 at 00:50
  • FWIW i have no idea if that formula is even correct as wikipedia doesn't have a citation on that result. So one might have to dig through the citations in wikipedia to check if one of the other sources explains the result OR find another resource altogether. If your reputation depends on it i'd be hesitant about using the formula above and if you can find a derivation then updating the wiki to include a link to the derivation would be quite useful :) – Sidharth Ghoshal Dec 08 '23 at 00:59
  • Related ... Lamé surface: https://mathcurve.com/surfaces.gb/lame/lame.shtml – GEdgar Dec 08 '23 at 02:44
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    @SidharthGhoshal the formula is due to Dirichlet, Uber Eine Neue Methoden zur Bestimmung Vielfacher Integralen; I read it in his collected works. In English, it is presented in Whittaker and Watson, and most complete in the first edition. In later editions more of it was pushed into homework exercises. I have the third edition at home...In the first edition it is pages 191 and 192 and has lots of letters – Will Jagy Dec 08 '23 at 02:53
  • Hint: compute the area of $z^8+y^8=64-z^8$ at fixed $z$ (for $|z|\le2^{3/4}$), then integrate over $z$ to get the volume. You'll need to use this twice. – J.G. Dec 08 '23 at 19:26
  • Cf. this generalization of the analogous 2D problem: https://math.stackexchange.com/questions/4721293/on-bodies-of-revolution-for-y-1-xqp/4721371 – Travis Willse Dec 08 '23 at 21:05

2 Answers2

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The technique is due to Dirichlet. It is spelled out fairly completely in the first edition of Whittaker and Watson, but reduced to a homework exercise by the fifth. Delighted to find that the 1902 edition has been scanned, some pages below. The fifth edition is available for personal use, nice clear typesetting; this material is mostly pages 267-268 in the fifth.

Given $x,y,z > 0,$ the function Whittaker called $f$ is constant $1,$ and $$ \left( \frac{x}{a} \right)^\alpha + \left( \frac{y}{b} \right)^\beta + \left( \frac{z}{c} \right)^\gamma < 1. $$ Next we are taking $p=q=r=1$ we get $$ \int 1 dx dy dz $$ as $$ \frac{ \; a \, b \, c \;\Gamma\left( 1+\frac{1}{\alpha} \right) \Gamma\left( 1+ \frac{1}{\beta} \right) \Gamma\left( 1+\frac{1}{\gamma} \right) }{ \Gamma\left( 1 + \frac{1}{\alpha} + \frac{1}{\beta}+ \frac{1}{\gamma}\right) }$$
and for all $8$ octants we need $$ \frac{ \; 8 \; a \, b \, c \;\Gamma\left( 1+\frac{1}{\alpha} \right) \Gamma\left( 1+ \frac{1}{\beta} \right) \Gamma\left( 1+\frac{1}{\gamma} \right) }{ \Gamma\left( 1 + \frac{1}{\alpha} + \frac{1}{\beta}+ \frac{1}{\gamma}\right) }$$

Next, $a=b=c = 2^{3/4} $ so that $abc = 2^{9/4} = 4 \cdot 2^{1/4}.$ For more than just the first octant we multiply by another $8$ for $ 32 \cdot 2^{1/4}.$ Also $\alpha = \beta = \gamma = 8.$

The final volume is

$$ \frac{ 32 \cdot 2^{1/4} \; \Gamma\left( \frac{9}{8} \right)^3 } { \Gamma\left( \frac{11}{8} \right)} $$

Let's see. The solid is inscribed in an ordinary cube, volume
$ 32 \cdot 2^{1/4} \approx 38.0546 . $ Your solid is this number times $ \Gamma\left( \frac{9}{8} \right)^3 / \Gamma\left( \frac{11}{8} \right) \approx 0.9395875$ resulting in $35.7556542..$

Pages 191-193 of Whittaker 1902.

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Will Jagy
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  • Too telegraphic for me. – Ted Shifrin Dec 08 '23 at 02:24
  • “We get” is totally confusing. Where does this random formula come from? Just google a formula and plug in is not helpful. – Ted Shifrin Dec 08 '23 at 02:36
  • At first, I expected the two values (the measure of the $8$-ball and the measure of its circumscribing $8$-hypercube) to be closer, but it occurs to me that there's a lot of measure in them thar corners. – Brian Tung Dec 08 '23 at 19:53
  • @BrianTung one way to give a better upper bound si to take the tangent plane at each "corner" point $t,t,t$ for $t = 3^{-1/8}$ in a cube with actual corner at $(1,1,1).$ Subtract those eight right tetrahedra out of the cube – Will Jagy Dec 09 '23 at 17:38
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We consider computing the volume of the region $\mathcal{D}$ described by

$$ \mathcal{D} = \left\{ (x, y, z) \in \mathbb{R}^3 : \left|\frac{x}{a}\right|^{\alpha} + \left|\frac{y}{b}\right|^{\beta} + \left|\frac{z}{c}\right|^{\gamma} \leq 1 \right\} $$

for $a, b, c, \alpha, \beta, \gamma > 0$. It is clear that $\operatorname{Vol}(\mathcal{D}) = 8 \operatorname{Vol}(\mathcal{D} \cap \mathbb{R}_{\geq 0}^3)$, so we only consider the volume of part of $\mathcal{D}$ lying in the first octant. Then by substituting

$$ \left\{ \begin{aligned} x &= a (r \cos\phi)^{2/\alpha}, \\ y &= b (r \sin\phi \cos\theta)^{2/\beta}, \\ z &= c (r \sin\phi \sin\theta)^{2/\gamma}, \end{aligned} \right. \qquad \text{where} \quad \begin{cases} 0 \leq r \leq 1, \\ 0 \leq \phi \leq \pi/2, \\ 0 \leq \theta \leq \pi/2, \end{cases} $$

we get

$$ \frac{\partial(x,y,z)}{\partial(r,\psi,\theta)} = \frac{8abc}{\alpha\beta \gamma} r^{\frac{2}{\alpha}+\frac{2}{\beta}+\frac{2}{\gamma}-1} (\cos \phi)^{\frac{2}{\alpha}-1} (\sin \phi)^{\frac{2}{\beta}+\frac{2}{\gamma}-1} (\cos\theta)^{\frac{2}{\beta}-1} (\sin\theta)^{\frac{2}{\gamma}-1}. $$

Hence,

\begin{align*} \operatorname{Vol}(\mathcal{D}) &= 8 \iiint_{\mathcal{D} \cap \mathbb{R}_{\geq 0}^3} 1 \, \mathrm{d}x\mathrm{d}y\mathrm{d}z \\ &= \frac{8abc}{\alpha\beta \gamma} \biggl( 2 \int_{0}^{1} r^{\frac{2}{\alpha}+\frac{2}{\beta}+\frac{2}{\gamma}-1} \, \mathrm{d}r \biggr) \\ &\quad \times \biggl( 2 \int_{0}^{\frac{\pi}{2}} (\cos \phi)^{\frac{2}{\alpha}-1} (\sin \phi)^{\frac{2}{\beta}+\frac{2}{\gamma}-1} \, \mathrm{d}\phi \biggr) \biggl( 2 \int_{0}^{\frac{\pi}{2}} (\cos\theta)^{\frac{2}{\beta}-1} (\sin\theta)^{\frac{2}{\gamma}-1} \, \mathrm{d}\theta \biggr). \end{align*}

Invoking the beta function identity

$$ 2 \int_{0}^{\frac{\pi}{2}} (\cos \theta)^{2p-1} (\sin \theta)^{2q-1} \, \mathrm{d}\theta = B(p, q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}, $$

it therefore follows that

\begin{align*} \operatorname{Vol}(\mathcal{D}) &= 8abc\frac{\Gamma(\alpha^{-1}+1)\Gamma(\beta^{-1}+1)\Gamma(\gamma^{-1}+1)}{\Gamma(\alpha^{-1}+\beta^{-1}+\gamma^{-1}+1)}, \end{align*}

which is in accordance with @Will Jagy's answer.

Sangchul Lee
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