Finding the volume of a solid denoted by a set Say I would like to find the volume of the set
$$S = \left\{\left\vert x\right\vert^c +
\left\vert y\right\vert^c + \left\vert z\right\vert^{c}
\leq 1\right\}\ \mbox{with}\ c\ \mbox{being positive.}
$$
I am thinking about using fubini theorem here, and expanding into a series of three iterated integrals like so:
$$(\int_{-1}^1(\int_{-(1-|x|^c)}^{1-|x|^c}(\int_{-(1-|x|^c-|y|^c)}^{1-|x|^c-|y|^c}1dz)dy)dx)$$. I.e as I would for a general sphere, where $c=2$.
Does this seem correct? How might I simplify this into an expression given the absolute values then? help appreciated
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{{\displaystyle #1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{\bracks{\cdots}}$ is an $\ds{Iverson Bracket}$,
$\ds{\on{H}}$ is the $\ds{Heaviside\ Step\ Function}$ and
$\ds{\cal H}$ is a $\ds{Hankel\ Contour}$.
\begin{align}
& \color{#44f}{\left.\iiint_{\mathbb{R}^{3}}\bracks{\verts{x}^{c} + \verts{y}^{c} + \verts{z}^{c} < 1}\dd^{3}\vec{r}\,
\right\vert_{\, c\ >\ 0}}
\\[5mm] = & \
\iiint_{\mathbb{R}^{3}}
\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}
{\exp\pars{\braces{1 - \verts{x}^{c} - \verts{y}^{c} - \verts{z}^{c}}s} \over s}{\dd s \over 2\pi\ic}
\dd^{3}\vec{r}
\\[5mm] = & \
\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}{\expo{s} \over s}
\pars{\int_{-\infty}^{\infty}\expo{-s\verts{x}^{\,c}}\dd x}^{3}
{\dd s \over 2\pi\ic}
\\[5mm] = & \
\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}{\expo{s} \over s}
\braces{2s^{-1/c}\,\,\Gamma\pars{1 + {1 \over c}}}^{3}
{\dd s \over 2\pi\ic}
\\[5mm] = & \
8\,\Gamma^{\,3}\pars{1 + {1 \over c}}\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}{\expo{s} \over s^{1 + 3/c}\,}\,\,
{\dd s \over 2\pi\ic}
\\[5mm] = & \
8\,\Gamma^{\,3}\pars{1 + {1 \over c}}\oint_{\cal H}{\expo{s} \over s^{1 + 3/c}\,}\,\,
{\dd s \over 2\pi\ic} =
\bbx{\color{#44f}{{8\,\Gamma^{\,3}\pars{1 + 1/c} \over
\Gamma\pars{1 + 3/c}}}} \\ &
\end{align}
$\ds{s}$-integral in the second line is
$\ds{\on{H}\pars{1 - \verts{x}^{c} - \verts{y}^{c} - \verts{z}^{c}}}$.
A: From the paper  "Volumes of Generalized Unit Balls", we know that the solution is
$$\frac{8(\Gamma(1+\frac1c))^3}{\Gamma(1+\frac3{c})}.$$
The question is how to express it without the Euler Gamma function.
\begin{align}
\frac{8(\Gamma(1+\frac1c))^3}{\Gamma(1+\frac3{c})}&=\frac{8(\Gamma(1+\frac1c))^3}{\Gamma(1+\frac3{c})}\cdot \frac{\Gamma(3+\frac3c)}{\Gamma(3+\frac3c)}\\
&=8B\left(1+\frac1c, 1+\frac1c, 1+\frac1c\right)\cdot \frac{\Gamma(3+\frac3c)}{\Gamma(1+\frac3c)} \\
&=8B\left(1+\frac1c, 1+\frac1c, 1+\frac1c\right)\cdot \left(2+\frac3c\right)\left(1+\frac3c\right)  \\
&=8 \left(2+\frac3c\right)\left(1+\frac3c\right)\int x^{\frac1c}y^\frac1{c}z^\frac1c dS
\end{align}
Here $S$ is the unit simplex ($x+y+z=1$,$x,y,z \ge 0$) and $B$ is the multivariable beta function. The last line is due to definition of Dirichlet distribution
A: $(x,y,z)\rightarrow (u^{2/c},v^{2/c},w^{2/c})$
$V=8\int\int\int_{u^2+v^2+w^2\leq 1}^{+++}\frac{8}{c^3}u^{2/c-1}v^{2/c-1}w^{2/c-1}dudvdw$
$V=8\int_0^{\pi/2}\int_0^{\pi/2}\int_0^1\frac{8}{c^3}(\rho^3\sin\theta\cos\theta\sin^2\phi\cos\phi)^{2/c-1}\rho^2\sin\phi d\rho d\theta d\phi$
$V=\frac{64}{c^3}(\int_0^1\rho^{6/c-1}d\rho)(\int_0^{\pi/2} (\sin\theta)^{2/c-1}(\cos\theta)^{2/c-1} d\theta)(\int_0^{\pi/2} (\sin\phi)^{4/c-1}(\cos\phi)^{2/c-1} d\phi)$
$V=\frac{8}{3c^2}B(\frac{1}{c},\frac{1}{c})B(\frac{2}{c},\frac{1}{c})$
A: Since this is symmetric, you can consider the volume of $S$ in, say the first octant ($x,y,z\geq0$ and obtain the whole volume by multiply this partial volume by 8.
Another way to do this is to use polar coordinates ($0\leq r\leq 1$: magnitude of the point; $0\leq\phi\leq\frac{\pi}{2}$: the angle of the point vector and the positive $z$-axis; $0\leq\theta\leq2\pi$: the angle of the vector projection of the point vector onto $xy$-plane and the positive $x$-axis.
