Relationship between volume and surface density I was not sure if it fits better here, or in the math forum, but this is the problem I am trying to solve:
Say you have a full cylinder of radius $R$, height $L$, and its uniformly charged with volume density $\rho$. And now I draw a hollow cylinder inside (with the same height, and radius $r < R\,$), is there any numerical relationship between its surface density $\sigma$(r) and $\rho$ ?
Thanks in advance.   
 A: Now let's see. 
The total charge, $C_t$, in the cylinder can be written as:
$$C_t=\pi  R^2 L\rho$$
The total charge on the hollow cylinder, $C_{thc}$, with radius $r$ is
$$C_{thc}=2\pi rL\sigma,$$ 
where $\sigma$ is independent of $r$, just like $\rho$.
The point though is that because the hollow cylinder has a two-dimensional surface (zero thickness), we can't relate $\rho$ to $\sigma$. If $\sigma \gt 0 $ then $C_t$ would be infinite. $\rho$ is only defined for three-dimensional cases, while $\sigma$ only for two-dimensional cases.
Maybe you could think that
$$C_t=\int _{r=0} ^R 2\pi r L \sigma dr,$$ 
but this gives for the dimension of $C_t$ $(Cm)$.
You cán get the volume though of the cylinder by considering the surface of the hollow cylinder and integrate from $r=0$ to r=R.
So maybe it is possible if we consider the inverse of $\sigma=\frac 1{\sigma}$, a constant, which has as its unit $(\frac{m^2}{C})$, and the inverse of $\rho=\frac 1 {\rho}$, also a constant in a uniform medium, with unit $(\frac{m^3}{C})$.
Now let's (for simplicity) consider a cube of which the sides L are $1(m)$, and proceed in the way we calculate the volume from the integral of surfaces:
$$\frac 1{\rho}=\int _{l=o}^{L}\frac 1 {\sigma}dl=\frac 1 {\sigma}\int _{l=0}^{L} 1 dl=\frac 1 {\sigma}|_{l=0}^L l=\frac 1 {\sigma}L ,$$ so
$$\rho=\frac{\sigma}{L}.$$
Off course, when $L=1$, this reduces to $\sigma=\rho$, but keep in mind the units are the same in this case.
You can try the same procedure for the cylinder and find out yourself the relation between $\rho$ and $\sigma$ (keep an eye on the units!).
A: I have seen this answer asked other times and always with unsatisfactory answers. Here's an attempt to clarify, probably also unsatisfactory. The following are some examples related to your question, choose your favourite and hopefully it will give you some insight!
The easy way out?
Physically, there always are volumes, however thin. So an easy way out is to take a thin cylinder at distance $r$ from its axis, with thickness $\Delta r$. It has a volume $v(r)=2 \pi L r\Delta r$ so it has a charge
$$Q(r)=\rho v(r) = \rho 2 \pi L r\Delta r$$.
Because we want the surface density, we take the surface of such cylinder $S \approx 2\pi r L$ (the $2$ is there because we have two sides) and we just compute
$$\sigma = Q/S = { \rho 2 \pi L r\Delta r \over 2\pi r L}$$
i.e.
$$\sigma= 2 \rho \Delta r $$
This is not the right answer. As it depends on how thin you make your slice, you get to the contradictory results that the thinner you make your slice, the smaller the surface charge despite the fact that the surface is roughly the same: this is because we are equating surfaces and volumes in a weird way.
If we include the surface due to the thickness in our computation (we add the contribution coming from the top and the bottom), we get
$S=S \approx 2(\pi r L +[\pi(r+\Delta r)^2-\pi r^2]) = 2\pi(r L + 2 r \Delta r +\Delta r^2)$
that means
$$\sigma = Q/S = { \rho 2 \pi L r\Delta r \over 2\pi(r L + 2 r \Delta r +\Delta r^2)}$$
which again is $0$ as $\Delta r$ becomes small.
This is because we are comparing an infinitely small surface to a finite volume, so it will vanish in the $\Delta r\to 0$ limit.
So no luck, but it is a first approximation.
A case we can solve analitically
The cylinder is hard because its local volume depends on the radius (see the expression for the volume before). Let's try with a cube of side $L$.
Imagine: if you split your cube in $N_v$ smaller cubes, you get $n$ cubes on each edge such that $n*n*n=N_V$ i.e. $n=\sqrt[3]{N_v}$ and thus ona side of the cube you get $n^2=\sqrt[3]{N_v^2}$ elements.
This means that your surface density of "small cubes" is $\sigma =\sqrt[3]{N_v^2}/L^2$.
Now if you have a charged cube, you have a total number of charges $Q_v=\rho L^3$ which means (by the above reasoning) that you have $Q_s=\sqrt[3]{Q_v^2}$ charges on the surface and thus a surface density of
$$\sigma = \sqrt[3]{Q_v^2}/L^2 = \sqrt[3]{Q_v^2/L^6}$$
we now use $Q_v=\rho L^3$ and get
$$\sigma = \sqrt[3]{(\rho L^3)^2/L^6} $$
which eventually becomes
$$\sigma = \sqrt[3]{\rho^2}$$
Now this solution is right: it scales in the right way (if you put 8 times more particles in your cube, so that $\rho_1=8\rho$, on the surface you only get $\sigma_1=4\sigma$ because not all of the new particles end up on the surface - that's $\sqrt[3]{8^2}=4$). The above solution from Deschele Schilder is mainly wrong in the case of the cube for this reason: it does not take into account the $different$ between volume and surface. One [the volume] scales as $L^3$ (if you double the size, the volume becomes 8 times bigger) and the other one [the surface] as $L^{3/2}$ (if you double the size, the surface become 4 times bigger, and the same must happen to the surface density which must get 4 times lower while the volume one has to become 8 times lower!).
And the cylinder...?
Can we do the same for the cylinder? Not easily, because it does not have such a simple volume, but we can imagine it will scale in the right way. So the charge distributes unevenly on the surface and in the volume and it's harder to get a precise scaling law: a thin elongated cylinder will have a different surface to volume ratio of a short large one, so $\sigma$ will have a weird scaling with $r$ and $L$ which I can not think of at the moment.
