Finding the charge density in a solid An object occupies the solid region in the first octant bounded by the coordinate planes and two cylinders $x^2 + y^2 = 4$ and $y^2 + z^2 = 4$ If the charge density at any point is $x$, what's the total charge?

So I know $Q = pV$, where $p$ is charge density and $V$ is volume. 
I think that I need to do this in spherical coordinates, but I'm not too sure. I define $f(x, y, z) = x$, then I have $f( \theta, \phi) = ??$ 
I've never used spherical coordinates before. I'm learning multivariable calc through self study, so any help would be appreciated. I think my integral will be something like this
$$\int_{0}^{2\pi}\int_{0}^{\pi} f(\theta, \phi) d\theta d\phi \cdot x,$$ 
but I'm not sure about how to get the function. Any help is appreciated.
 A: Firstly, $Q=pV$ only holds if the charge density is uniform. In this case $p(\vec{x})=x$, so it varies spatially and instead we must evaluate the total charge using an integral. To do so, we integrate the charge density over the charged volume;
$$Q=\int_{V}p(\vec{x})dV$$
Where $V$ is the volume you described.
We now need to pick a coordinate system and describe $V$ in terms of it. I don't think spherical or cylindrical coordinates quite work here (the union of the cylinders is a bit messy to describe), so we'll stick with good old cartesian coordiantes.
Since for any point $(x, y, z)$ we have $x^{2}+y^{2}=4$ and since $x\geq 0$ as we are in the first octant, we must have;
$$0\leq x\leq\sqrt{4-y^{2}}$$
Similarly, $0\leq y\leq\sqrt{4-z^{2}}$. For $z$ we know we are in the first octant and from $z^{2}+y^{2}=4$ we know that $z\leq 2$, so our final inequality is $0\leq z\leq 2$.
We now have our volume $V$ described in terms of three inequalities in cartesian coordinates, so our integral becomes;
$$Q=\int_{0}^{2}\int_{0}^{\sqrt{4-z^{2}}}\int_{0}^{\sqrt{4-y^{2}}}xdxdydz$$
The first two integrals are pretty easy;
$$
Q
=
\int_{0}^{2}\int_{0}^{\sqrt{4-z^{2}}}\int_{0}^{\sqrt{4-y^{2}}}xdxdydz
=
\frac{1}{2}\int_{0}^{2}\left(4\sqrt{4-z^{2}}-\frac{1}{3}(4-z^{2})^{\frac{3}{2}}\right)dz
$$
The last one can be solved with the triginometric substitution $x=2\sin(\theta)$ (let me know if you need help with this), and is pretty messy. It is doable however, and the final result comes out to $Q=\frac{3\pi}{2}$.
A: Plotting the region of integration helps a lot; it makes it easy to see how to split the solid into smooth pieces. Algebraically, one can notice that there are two inequalities, $x^2 + y^2 < 4$ and $y^2 + z^2 < 4$, and assuming $z < x$ or $z > x$ eliminates one of them. Then for the first case we can change to cylindrical coordinates $(x, y, z) = (r \cos t, r \sin t, z)$ and for the second case to cylindrical coordinates $(x, y, z) = (x, r \sin t, r \cos t)$, obtaining
$$\int_{\mathbb R^3} x \,[x^2 + y^2 < 4 \land y^2 + z^2 < 4 \land
 x > 0 \land y > 0 \land z > 0] \,dV = \\
\int_0^2 \int_0^{\pi/2} \int_0^{r \cos t} r^2 \cos t \,d z dt dr +
\int_0^2 \int_0^{\pi/2} \int_0^{r \cos t} x r \,d x dt dr.$$
