Question about charge density in a solid sphere

[LostInToronto]

I've already posted this question in the advanced physics forum, but I really think it should go here. My apologies for the double posting.

Homework Statement

If we are given a uniformly charged solid sphere with total charge Q and radius R, then the volume charge density rho is given by

$$\rho = \frac{Q}{\frac{4}{3} \pi R^3}$$.

My question is: How do we express the surface charge density sigma of a spherical shell of infinitesimal width dr, located within this solid sphere?

Homework Equations

The Attempt at a Solution

I keep reading that

$$\sigma = \rho dr$$

but I really don't understand why this is the case. If someone could help clear this up, I'd really appreciate it.

 

[Dick]

Ok, the charge dq contained in a shell of radius r and thickness dr is rho(r) times volume dV=4*pi*r^2*dr. Now surface charge density is charge dq divided by surface area 4*pi*r^2. So I suppose that would give you rho(r)*dr. But I'm not sure I'd really call that a 'surface charge'. I think it's really just a slice of the volume charge.

 

[LostInToronto]

Thank you.