Derivation of resistance of coaxial cable
Starting from the basic formula for resistance:
R = \$\rho L/A\$ consider the concentric cylindrical shells, and cable length G
We want to calculate the resistance from one side to the other of a thin cylindrical shell of length G and thickness dr (current passing radially through the shell).
So L in this case is the infinitesimal radius change dr
We will use the thin slice in the radial direction because the radius changes as we go from inside to outside, and thus the resistance of a slice of the same thickness dr decreases as we go outward, and we want to integrate along that path to find the total resistance. We are integrating along the path the current follows from inside conductor to outside conductor. There is assumed negligible resistance along the center and outer conductors.
A is the length of the cable G multiplied by the circumference, which is \$2 \pi r G\$.
so \$dR = \rho dr/(2\pi r G)\$
Pulling the constants out of the integration we have:
The resistance R = \$\frac{\rho}{2\pi G}\int_a^b\frac{1}{r}dr\$
The definite integral we know to be ln(b)-ln(a) = ln(b/a), and we get the solution from the site.