Integral over shifted unit ball

Question

I am trying to find \begin{equation*}\iiint\limits_{D}\sqrt{x^2+y^2+z^2} \mathop{}\!\mathrm{d}x \mathop{}\!\mathrm{d}y \mathop{}\!\mathrm{d}z\end{equation*} where $D$ is the set of points $(x,y,z)$ satisfying $x^2+y^2+z^2-2x<0$: in other words, the unit ball, but centred at $(1,0,0)$ rather than the origin. If it were centered at the origin, this would be easy, as the integrand in spherical coordinates would just be $r$, but it's not clear how to proceed given the shifted centre. Any assistance would be much appreciated.

Answer 1

First, note that we can shift the translation onto the $z$ direction instead and get the same answer (this is equivalent to basing spherical coordinates off of the $x$ axis instead of the $z$). Then we get the equation of the surface to be

$$x^2+y^2+z^2 = 2z \implies r^2 = 2r\cos\theta$$

From here setting up spherical coordinates is easy

$$\iiint_D \sqrt{x^2+y^2+z^2}\:dV = \int_0^{2\pi}\int_0^{\frac{\pi}{2}}\int_0^{2\cos\theta} r^3\sin\theta ~dr ~d\theta ~d\phi = \frac{8\pi}{5}$$

Answer 2

$x = \rho\cos\phi\\ y = \rho\sin\theta\sin\phi\\ z = \rho\cos\theta\sin\phi$

These are very close to standard spherical coordinates, but we have applied a rotation such that $\phi = 0$ corresponds to $(1,0,0)$

$\rho^2 - 2\rho\cos\phi < 0$

$\rho < 2\cos \phi $

$\int_0^{2\pi}\int_0^{\frac {\pi}{2}}\int_0^{2\cos\phi} \rho^3\sin\phi\ d\rho\ d\phi\ d\theta$