Triple integral, how to convert to spherical coordinates for $x^2+y^2+z^2\le x$

Question

$$\iiint_\Omega{\sqrt{x^2+y^2+z^2}dxdydz}$$ where $\Omega=\{x^2+y^2+z^2\le x\}$.

I've only found answers for $x^2+y^2+z^2=R^2$, and I don't know how to proceed - I believe I need to convert to spherical coordinates to solve the integral but I'm not sure how.

Answer 1

Note that by switching $x$ and $z$ we have that $$I:=\iiint_\Omega{\sqrt{x^2+y^2+z^2}dxdydz}=\iiint_{\Omega'}{\sqrt{x^2+y^2+z^2}dxdydz}$$ where $\Omega'=\{x^2+y^2+z^2\le z\}=\{x^2+y^2+(z-1/2)^2\le 1/2^2\}$.

$\Omega'$ of is the ball with center $(0,0,1/2)$ and radius $1/2$. It is entirely in the half-space $z \ge 0$. Therefore, in spherical coordinates, $\phi \in [0,\pi/2]$, $\theta \in [0,2\pi]$ and $$\rho^2\leq z = \rho \cos{\phi} \implies \rho\leq \cos{\phi}.$$ Hence $$I=\int_{0\phi=}^{\pi/2} \int_{\rho=0}^{\cos{\phi}}\int_{\theta=0}^{2 \pi}\rho\cdot (\rho^2\sin{\phi})\ d\theta d\rho d\phi \\=\int_0^{\pi/2} \sin{\phi}d\phi \: \int_0^{\cos{\phi}} \rho^3 d\rho\: \int_0^{2 \pi} d\theta=\frac{\pi}{10}.$$

Answer 2

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$\ds{\Omega \equiv \braces{\pars{x,y,z}\ \mid\ x^{2} + y^{2} + z^{2} \leq x}}$. Since there is a 'symmetry' which involves the $\ds{y}$ and $\ds{z}$ variables, it's convenient to use Cylindrical Coordinates which are defined by: $$ \left.\begin{array}{rcl} \ds{x} & \ds{=} & \ds{x} \\[2mm] \ds{y} & \ds{=} & \ds{\rho\cos\pars{\phi}} \\[2mm] \ds{z} & \ds{=} & \ds{\rho\sin\pars{\phi}} \end{array}\right\}\,,\qquad x \in \mathbb{R}\,,\qquad \rho > 0\,,\quad \phi \in \pars{0,2\pi} $$

\begin{align} &\color{#f00}{\iiint_\Omega\root{x^{2} + y^{2} + z^{2}}\,\dd x\,\dd y\,\dd z} = \left.\int_{-\infty}^{\infty}\,\int_{0}^{2\pi}\int_{0}^{\infty} \root{\rho^{2} + x^{2}}\,\rho\,\dd\rho\,\dd\phi\,\dd x \,\right\vert_{\ x^{2}\,\, +\, \rho^{2}\,\, \leq\,\, x} \\[5mm] = & \left. 2\pi\int_{-\infty}^{\infty}\int_{0}^{\infty} \root{\rho^{2} + x^{2}}\,\rho\,\dd\rho\,\dd x \,\right\vert_{\ x^{2}\,\, +\ \rho^{2}\,\, \leq\,\, x}\,\, = \left.\pi\int_{-\infty}^{\infty}\,\int_{0}^{\infty} \!\!\!\root{\rho + x^{2}}\,\dd\rho\,\dd x\, \right\vert_{\ \rho\ \leq\,\, x\ -\ x^{2}} \\[5mm] = &\ \pi\int_{0}^{1}\int_{0}^{x - x^{2}}\root{\rho + x^{2}}\,\dd\rho\,\dd x \quad\ \pars{~\mbox{because}\ \rho > 0\ \imp\ x - x^{2} > 0\ \imp\ x \in \pars{0,1}~} \\[5mm] = &\ \pi\int_{0}^{1}\left.{2 \over 3}\pars{\rho + x^{2}}^{3/2} \,\,\right\vert_{\ \rho\ =\ 0}^{\ \rho\ =\ x - x^{2}}\,\,\,\,\dd x = {2\pi \over 3}\int_{0}^{1}\pars{x^{3/2} - x^{3}}\,\dd x = {2\pi \over 3}\pars{{2 \over 5} - {1 \over 4}} \\[5mm] = &\ \color{#f00}{\pi \over 10} \approx 0.3142 \end{align}