I need to evaluate $\iiint_V \frac{1}{x}\: dV$ where $V$ is the inside of a sphere given by $x^2+y^2+z^2=x$.
I write the equation as a sphere with centre in $\left(\frac{1}{2},0,0\right)$ and radius of $\frac{1}{2}$. Now I know that my sphere and sphere in $(0,0,0)$ with radius $\frac{1}{2}$ would have the same volume, but I'm confused about what do do with $\frac{1}{x}$.
I tried something like this: $$x=\frac{1}{2}+\frac{1}{2} R\cos\phi$$ $$y=\frac{1}{2} R\cos\phi$$ $$z=\frac{1}{2z}$$ and using Jacobian of $|J|= \frac{1}{2R}$
Then I tried writing my coordinates as $$x=x$$ $$y=R\cos\phi$$ $$z=R\sin\phi$$
I didn't mention my limits because I think the problem is with my choice of coordinates. Any hints on what the easiest substitution would be?
Using the following spherical coordinates
$$\begin{cases} x = \rho \cos \phi \\ y = \rho\sin\phi\sin\theta \\ z = \rho \sin\phi\cos\theta \\ \end{cases} \implies J = \rho^2\sin\phi$$
we get that the boundary is given by
$$x^2+y^2+z^2 = x \implies \rho^2 = \rho \cos\phi \implies \rho = \cos\phi$$
which makes the integral
$$\iiint_E\frac{1}{x}\:dV = \int_0^{2\pi}\int_0^{\frac{\pi}{2}}\int_0^{\cos\phi}\rho \tan\phi \:d\rho \:d\phi \:d\theta$$
$$ = \pi \int_0^{\frac{\pi}{2}} \sin\phi\cos\phi\:d\phi = \frac{\pi}{2}$$