Volume bounded by the surface $x^n + y^n + z^n = a^n$

Question

Calculate the volume bounded by the surface $x^n + y^n + z^n = a^n$ $(x>0,y>0,z>0)$.

$$\iiint\limits_{x^n+y^n+z^n \le a^n \\ \ \ \ \ \ \ x,y,z > 0}\mathrm dx~ \mathrm dy ~\mathrm dz = \begin{bmatrix}x = r\cos\varphi\sin\psi \\ y = r\sin\varphi \sin\psi \\ z = r\cos\psi\end{bmatrix} = \iiint\limits_{r^n \le a^n} \underbrace{r^2 \sin \psi}_{J} ~\mathrm d\varphi ~\mathrm d\psi ~\mathrm dr =\\= \int_0^a r^2\mathrm dr \int_0^{\pi/2}\mathrm d\varphi \int_0^{\pi/2}\sin\psi~ \mathrm d\psi$$ Am I going right? I'm not sure about the bounds of the last three integrals.

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For even $n$ the graph looks like the following:

For odd $n$ the first quadrant is alike.

Answer 1

$x^n+y^n+z^n$ is not the same as $r^n$, and I suspect that transforming to spherical or cylindrical coordinates will just make this problem more gross.

These shapes are a specific type of superellipsoid (https://en.wikipedia.org/wiki/Superellipsoid) You can find an answer about the formula for their volumes here: https://math.stackexchange.com/a/261652/908546

Answer 2

Considering the $d$-dimensional generalisation $\{x\in\mathbb{R}_+^d : \lVert x\lVert_n\leqslant a\}$, where $a>0$ and $$x=(x_1,\dots,x_d)\in\mathbb{R}^d\implies\lVert x\lVert_n:=(x_1^n+\dots+x_d^n)^{1/n},$$ we can show (using the approach from here) that for any "good enough" function $f$ $$\int_{\mathbb{R}_+^d}f(\lVert x\lVert_n)\,dx=dK_{n,d}\int_0^\infty t^{d-1}f(t)\,dt,\qquad K_{n,d}=\frac{\Gamma^d(1+1/n)}{\Gamma(1+d/n)}.$$ Our volume is obtained with $f$ the indicator function of $[0,a]$: $$\int_{\substack{x\in\mathbb{R}_+^d\\\lVert x\lVert_n\leqslant a}}dx=dK_{n,d}\int_0^a t^{d-1}\,dt=K_{n,d}\,a^d.$$