Solve integral $\iiint_Ax^pdxdydz$ on $A=\{(x,y,z):x^2+y^2+z^2We have following integral to count:
$$
\iiint_Ax^pdxdydz
$$
where $p$ is constant real number and $A=\{(x,y,z):x^2+y^2+z^2<x^{\frac{1}{3}}\}$
I tried spherical substitution and cylindrical.
 A: By setting $x=w^3$, the triple integral turns into
$$3\iiint_{w^6+y^2+z^2<w}w^{3p+2}\,dw\,dy\,dz \tag{1}$$
that equals:
$$ 3 \int_{0}^{1}\iint_{y^2+z^2<w-w^6}w^{3p+2}\,dy\,dz   \,dw\tag{2}$$
or:
$$ 3\pi \int_{0}^{1}w^{3p+2}(w-w^6)\,dw = \color{red}{\frac{5\pi}{(p+3)(3p+4)}}.\tag{3}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\int_{0}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
x^{p}\bracks{x^{2} + y^{2} + z^{2} < x^{1/3}}\dd y\,\dd z\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}x^{p}\int_{0}^{2\pi}\int_{0}^{\infty}
\bracks{x^{2} + \rho^{2} < x^{1/3}}\rho\,\dd\rho\,\dd\phi\,\dd x
\\[5mm] = &\
\pi\int_{0}^{\infty}x^{p}\int_{0}^{\infty}
\bracks{\rho < x^{1/3} - x^{2}}\,\dd\rho\,\dd x =
\pi\int_{0}^{\infty}x^{p}\bracks{x^{1/3} - x^{2} > 0}
\int_{0}^{x^{1/3}\ -\ x^{2}}\,\dd\rho\,\dd x
\\[5mm] = &\
\pi\int_{0}^{\infty}\pars{x^{p + 1/3} - x^{p + 2}}
\bracks{x < 1}\,\dd x =
\bbx{\ds{{5\pi \over \pars{3p + 4}\pars{p + 3}}\,,\qquad
\Re\pars{p} > -\,{4 \over 3}}}
\end{align}
