Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$
when $E$ is bounded by $x^2+y^2+z^2-x=0$?
I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\sin\theta\sin\varphi,\quad z=r\cos\theta$$
and this function in spherical coordinates is
\begin{align*}
&\iiint\limits_E\frac{yzdxdydz}{x^2+y^2+z^2} = \iiint\limits_E\frac{r^2\sin\theta}{r^2\sin^2\theta\cos^2\varphi + r^2\sin^2\theta\sin^2\varphi + r^2\cos^2\theta}drd\theta d\varphi = \\ &\iiint\limits_E\frac{r^2\sin\theta}{r^2(\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta)}drd\theta d\varphi = \iiint\limits_E\frac{\sin\theta}{\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta}drd\theta d\varphi
\end{align*}
but I don't know how to write $E$ as set and convert it to spherical coordinates, and also what happens with this function after conversion. Triple integrals is now topic for me and I have never used spherical coordinates before, so I would be grateful if anyone can help me with this.

$x^2+y^2+z^2-x=0 \implies (x-\frac{1}{2})^2 + y^2 + z^2 = (\frac{1}{2})^2$
So it is a sphere with radius $\frac{1}{2}$ centered at $(\frac{1}{2},0,0)$.
In spherical coordinates, $x = \rho \cos \theta \sin \phi, \, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi \,$ where $\theta$ is the azimuthal angle and $\phi$ is the polar angle (just opposite of your convention).
$x^2 + y^2 + z^2 = \rho^2$ so substituting in the equation of our sphere,
$\rho^2 = \rho \cos \theta \sin \phi \implies \rho = \cos \theta \sin \phi$.
Now the part that you have to be careful about is the bounds of azimuthal angle. Please consider a circle in XY plane with center at $(\frac{1}{2},0)$ and radius of $\frac{1}{2}$ and equation of $r = \cos \theta$. $\theta$ varies betweeen  -$\frac{\pi}{2}$ and $\frac{\pi}{2}$ (range of $\pi$ instead of $2 \pi$ for a sphere centered at the origin). It is same for this sphere.
So the integral becomes
$I = \displaystyle \int_{-\pi/2}^{\pi/2} \int_{0}^{\pi} \int_{0}^{\cos\theta \sin\phi} \rho^2 \sin \theta \sin^2 \phi \cos \phi  \, d\rho \, d\phi \, d\theta$
And based on the given function, its integral above $XY$ plane and below will cancel each other.
If I was integrating over the part of the sphere only in the first octant, the integral will be
$I = \displaystyle \int_{0}^{\pi/2} \int_{0}^{\pi/2} \int_{0}^{\cos\theta \sin\phi} \rho^2 \sin \theta \sin^2 \phi \cos \phi  \, d\rho \, d\phi \, d\theta = \frac{1}{72}$
