<s>[INST] Volume of the solid bounded by $\sqrt[3]{x^2}+\sqrt[3]{y^2} +\sqrt[3]{z^2} =\sqrt[3]{a^2}$ I'm trying to determine the volume of the solid whose boundary is $$\sqrt[3]{x^2}+\sqrt[3]{y^2} +\sqrt[3]{z^2} =\sqrt[3]{a^2}$$ I've tried doing it using cylindrical coordinates and spherical coordinates but haven't been able to determine what the domain should be. Can anyone help me?
I'm also wondering what the name of this shape is?
 [/INST] It seems cleaner to me to use polar coordinates, at least to compute the cross-sectional area.
Parametrization
$$
\begin{align}
x&=a\cos^3(\theta)\cos^3(\phi)\\
y&=a\sin^3(\theta)\cos^3(\phi)\\
z&=a\sin^3(\phi)\\
\end{align}
$$
Cross-Sectional Area
Using $p=(x,y,0)$ with $a=1$ and $\phi=0$,
$$
\begin{align}
&\int_0^{2\pi}\frac12\overbrace{\left(\cos^3(\theta),\sin^3(\theta),0\right)}^p\times\overbrace{\left(-3\cos^2(\theta)\sin(\theta),3\sin^2(\theta)\cos(\theta),0\right)}^{p'}\cdot(0,0,1)\,\mathrm{d}\theta\\
&=\frac32\int_0^{2\pi}\sin^2(\theta)\cos^2(\theta)\,\mathrm{d}\theta\\
&=\frac38\int_0^{2\pi}\sin^2(2\theta)\,\mathrm{d}\theta\\[3pt]
&=\frac{3\pi}8
\end{align}
$$
Scaling both $p$ and $p'$ by $a\cos^3(\phi)$, the cross-sectional area is $\frac{3\pi}8a^2\cos^6(\phi)$.
Volume
Using $z=a\sin^3(\phi)$,
$$
\begin{align}
&\int_{-\pi/2}^{\pi/2}\overbrace{\tfrac{3\pi}8a^2\cos^6(\phi)}^{\substack{\text{cross-sectional}\\\text{area}}}\,\overbrace{3a\sin^2(\phi)\cos(\phi)}^{z'}\,\mathrm{d}\phi\\
&=\frac{9\pi}4a^3\int_0^{\pi/2}\cos^6(\phi)\sin^2(\phi)\,\mathrm{d}\sin(\phi)\\
&=\frac{9\pi}4a^3\int_0^1\left(1-u^2\right)^3u^2\,\mathrm{d}u\\
&=\frac{9\pi}4a^3\left(\frac13-\frac35+\frac37-\frac19\right)\\
&=\frac{9\pi}4a^3\frac{16}{315}\\
&=\frac{4\pi}{35}a^3
\end{align}
$$

</s>