Solve integral $\iiint_Ax^pdxdydz$ on $A=\{(x,y,z):x^2+y^2+z^2<x^{\frac{1}{3}}\}$

Question

We 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.

Answer 1

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: $$ 3\pi \int_{0}^{1}w^{3p+2}(w-w^6)\,dw = \color{red}{\frac{5\pi}{(p+3)(3p+4)}}.\tag{3}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\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}