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.
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.