91Ó°ÊÓ

The two cylinders are connected to the terminals of a battery with potential difference $\mathrm{Δ}V$, causing current $I$to flow radially from the inner cylinder to the outer cylinder.

The metal cylinders are light and have an enormous cover area, and will ignore their resistance and only regard the inner material's resistance. To uncover the resistance, split the space between the cylinder into little shots of thickness$dr$. The differential resistance of one of these shells will be similar to
$dR=\mathrm{ÒÏ}\frac{dr}{A}$

$=\mathrm{ÒÏ}\frac{dr}{2r\mathrm{Ï€}l}$

The point that $2r\mathrm{Ï€}l$is the surface area of this cylindrical shell that recreates the cross-section's function via which the current flows. Now the total resistance is only the integral of $dR$where integrate $r$from $R_1$to $R_2$of these shells$r$:

$R=\mathrm{ÒÏ}\frac{1}{2\mathrm{Ï€}l}{∫}_{R_1}^{R_2}\frac{dr}{r}$

$=\frac{\mathrm{ÒÏ}}{2\mathrm{Ï€}l}\ln \frac{R_2}{R_1}$

An expression for the resistance of this device is $R=\frac{\mathrm{ÒÏ}}{2\mathrm{Ï€}l}\ln \frac{R_2}{R_1}$.