We consider a long co-axial hollow cylinder of radius r and thickness dr. The cross-sectional area of considered element is dA = 2​πrdr

∴   Electric resistance between ends of conductor of considered element is

$\text{dR}=\text{ρ}\frac{l}{\text{dA}}=\text{ρ}\frac{l}{2\text{π}\text{rdr}}=\frac{\text{α}l}{2\text{π}{\text{r}}^3\text{dr}}$       [as resistivity is $\rho =\frac{\alpha }{r^2}$ ]

The system may be assumed as a parallel combination of a number of such types of elementary resistance.

$\therefore \frac{1}{\text{R}}=\int \frac{1}{\text{dR}}=\frac{2\text{π}}{\text{α}l}{\int }_0^a{\text{r}}^3\text{dr}\text{or}\frac{1}{\text{R}}=\frac{2\text{π}}{\text{α}l}\left(\frac{a^4}{4}\right)$

$\therefore R=\frac{2\alpha l}{\pi a^4}$