Cylindrical vs azimuthal symmetry What is the difference between cylindrical and azimuthal symmetry, are they the same?
I have not been able to find a clear explanation of these terms wrt. a cylinder. I need to know which length I should integrate over to solve the problem below.

A cylinder of radius $R$ and length $L$ has a cylindrically symmetric volume charge density of $\rho(x)$. Calculate the total charge on the cylinder.


 A: From my understanding of azimuth, it describes the angle $\phi$ in cylindrical or spherical coordinates. Therefore I'd say that the cylindrical and azimuthal symmetry are the same thing: The object is invariant under arbitrary rotations around the $x$-axis.
Your charge density $\rho(x)$ is a volume density. Therefore it has units of $\mathrm{Charge / Length^3}$. You will need to integrate over the whole volume to get the total charge. What I find a bit strange is that the problem says “total charge on the cylinder” as that suggests that the charge is only on the surface and therefore one needs to integrate over the surface only.
You have to integrate over the whole volume of the cylinder. The function $\rho(x)$ seems to only depend on the $x$ value, but not on radius $r$ nor the azimuth angle $\phi$. Then use cylindrical coordinates to set up your integral. You will need to know the Jacobian determinant for this, which just happens to be $r$.
The solution follows, but since this looks like a homework problem, I'd ask you to try with the information I have given you so far. When you get stuck, you can take a look at the “spoiler” blocks below and/or leave a comment.

 The integral to solve is this:
 $$ \displaystyle Q = \int_0^L \mathrm dx \int_0^{2\pi} \mathrm d \phi \int_0^R \mathrm dr \, r \, \rho(x) \,. $$

Can you simplify this?

 The integrals over $r$ and $\phi$ are “trivial” because the function of interest, $\phi(x)$ does not depend on those variables. Therefore we can compute those integrals. We get $\pi R^2$ from those integrals which is just the cross sectional area of the cylinder.

How far can you simplify it now?

 The furthest this can be solved will be
 $$ \displaystyle Q = \pi R^2 \int_0^L \mathrm dx \, \rho(x) \,. $$

