Volume of the solid bounded by $\sqrt[3]{x^2}+\sqrt[3]{y^2} +\sqrt[3]{z^2} =\sqrt[3]{a^2}$

Question

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?

Answer 1

I'll assume $a>0$; otherwise, replace $a$ with $|a|$. The boundary is a generalization of a curve called an astroid into three dimensions, otherwise known as an 'astroidal ellipsoid' or 'hyperbolic octahedron'. In the two-dimensional case we have the equation $x^{2/3}+y^{2/3}=r^{2/3}$; the area of this shape is $$ 4\int _{0}^{r} (r^{2/3}-x^{2/3})^{3/2}\,dx $$ $$ =4r\int _{0}^{1} (1-u^{2/3})^{3/2}\,du $$ $$ =6r^2\int _{0}^{1} t^{1/2}(1-t)^{3/2}\,dt $$ $$ =6r^2\cdot \frac{\Gamma(3/2)\Gamma(5/2)}{\Gamma(3/2+5/2)} = \frac{3\pi r^2}{8} $$

We can find the volume of this shape by using Cavalieri's Principle. Let $A(z)$ denote the volume of the cross-section at height $z$. Then the volume $V$ is $$ V= \int _{-a}^{a} A(z)\,dz = 2\int _0^a A(z)\,dz $$We know the area is proportional to the radius squared; now express the radius in terms of $z$ and integrate. $$ V=\frac{3\pi}{4} \int_0^a r^2\,dz $$ $$ =\frac{3\pi}{4} \int_0^a \left(\left(a^{2/3}-z^{2/3}\right)^{3/2}\right)^2\,dz = \frac{4\pi a^3}{35} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\iiint_{\large\mathbb{R}^{3}} \bracks{\root[\large 3]{x^{2}} + \root[\large 3]{y^{2}} + \root[\large 3]{z^{2}} < \root[\large 3]{a^{2}}} \dd x\,\dd y\,\dd z} \\[5mm] = &\ 8\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \bracks{\root[\large 3]{x^{2}} + \root[\large 3]{y^{2}} + \root[\large 3]{z^{2}} < \root[\large 3]{a^{2}}} \dd x\,\dd y\,\dd z \end{align} With the substitutions $\ds{\pars{~x \mapsto\ x^{3},y \mapsto\ y^{3},z \mapsto\ z^{3}~}}$: \begin{align} &\bbox[5px,#ffd]{\iiint_{\large\mathbb{R}^{3}} \bracks{\root[\large 3]{x^{2}} + \root[\large 3]{y^{2}} + \root[\large 3]{z^{2}} < \root[\large 3]{a^{2}}} \dd x\,\dd y\,\dd z} \\[5mm] = &\ 216\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \bracks{x^{2} + y^{2} + z^{2} < \pars{\verts{a}^{1/3}}^{2}} x^{2}y^{2}z^{2}\,\dd x\,\dd y\,\dd z \\[5mm] = &\ 216\int_{0}^{\pi/2}\int_{0}^{\pi/2} \int_{0}^{\large\verts{a}^{1/3}} \\[2mm] &\ \phantom{216\,\,\,} \bracks{r\sin\pars{\theta}\cos\pars{\phi}}^{\, 2}\, \bracks{r\sin\pars{\theta}\sin\pars{\phi}}^{\, 2}\, \bracks{r\cos\pars{\theta}}^{\, 2}\ \times \\[2mm] &\ \phantom{216\,\,\,} r^{2}\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd\phi \\[5mm] = &\ 216\int_{0}^{\pi/2}\int_{0}^{\pi/2} \int_{0}^{\large\verts{a}^{1/3}} \!\!\!\!\! r^{8}\sin^{5}\pars{\theta}\cos^{2}\pars{\theta} \cos^{2}\pars{\phi}\sin^{2}\pars{\phi}\dd r\,\dd\theta\,\dd\phi \\[5mm] = &\ 216\ \overbrace{\bracks{\int_{0}^{\pi/2}{\sin^{2}\pars{2\phi} \over 4}\, \dd\phi}}^{\ds{\pi \over 16}}\ \overbrace{\bracks{\int_{0}^{\pi/2}\sin^{5}\pars{\theta} \cos^{2}\pars{\theta}\,\dd\theta}}^{\ds{8 \over 105}}\ \times \\[3mm] &\ \phantom{216\,\,\,\,\,} \underbrace{\int_{0}^{\large\verts{a}^{1/3}}r^{8}\dd r} _{\ds{\verts{a}^{3} \over 9}}\ =\ \bbx{{4 \over 35}\,\pi\,\verts{a}^{3}} \\ & \end{align}

Answer 3

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 $z=a\sin^3(\phi)$, $$ \begin{align} &\int_{-\pi/2}^{\pi/2}\overbrace{\tfrac{3\pi}8a^2\cos^6(\phi)}^{\substack{\text{cross-sectional}\\\text{area}}}\,\overbrace{3a\sin^2(\phi)\cos(\phi)}^{z'}\,\mathrm{d}\phi\\ &=\frac{9\pi}4a^3\int_0^{\pi/2}\cos^6(\phi)\sin^2(\phi)\,\mathrm{d}\sin(\phi)\\ &=\frac{9\pi}4a^3\int_0^1\left(1-u^2\right)^3u^2\,\mathrm{d}u\\ &=\frac{9\pi}4a^3\left(\frac13-\frac35+\frac37-\frac19\right)\\ &=\frac{9\pi}4a^3\frac{16}{315}\\ &=\frac{4\pi}{35}a^3 \end{align} $$