"I think I can see that a shell has the volume of dA= 2pi L dR Is that right?"
Be careful not to confuse your r (a radial distance) with your R (electrical resistance).
The surface area of the shell is 2pi*r* L (the volume would be 2pi*r*L*dr) - see below.
ρ is an intrinsic property of the material (unlike resistance, which depends of the size of the material).
The relationship between ρ and R is ρ = RA÷l where A is cross-sectional area and l is length in the direction of electrical current flow. In this problem l is not L, it is r. Or, rather, for the shell it is dr. The relevant cross sectional area of the shell (bearing in mind that current is flowing radially) is 2pi*r*L (this is stated incorrectly in the problem, though the correct value is used in the integral - perhaps this was the cause of xvxvxv's confusion), so the small increase in resistance across the shell is dR = ρl/A
or dR = ρdr/(2pi*r*L).
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