Resistance of cylinder with radially varying resistivity [closed]

Question

There is a cylindrical conductor of radius $R_0$ and length $l$. The resistivity $(\rho)$ varies as a function of $r$ the distance from the center of the conductor. $\rho(r)=\frac{a}{r^2}$ where $a$ is a positive constant. I have to find the resistance of the conductor per unit length.

What I have tried so far is that I have divided the cylindrical conductors into infinitely many thin hollow cylinders of radius $r'.

Hence the resistance of that hollow cylinder is $$R(r)=\frac{\rho(r)l}{\pi r^2}=\frac{al}{\pi r^4}$$

Now since the adjacent cylinders are of the same potential, we can treat it as though these are resistors in parallel. Hence the formula:

$$\frac{1}{r_{eq}}=\sum_{i=1}^{n}\frac{1}{r_i}$$

Here is the problem. I do not how to extend the idea to this format. Any hints on what to do further will be appreciated very much. My gut feeling says the answer is $$\frac{a}{\pi R_0^4}$$ but it is only a guess that may not be true

Answer 1

Instead of working with resistivity, switch to conductivity, which is the reciprocal. $$\sigma = \frac{1}{\rho} \qquad G = \frac{1}{R} = \sigma\frac{A}{l}$$ where $\sigma$ is the conductivity and $G$ is the conductance (analagous with resistivity $\rho$ and resistance $R$). The total conductance of resistors in parallel is the sum of their individual conductances. $$G_{eq} = \sum_i G_i$$ This should make it easier to integrate. Afterwards, $$R_{eq} = \frac{1}{G_{eq}}$$