Solution

(1) According to similarity

$\begin{aligned} \frac{d}{d + L} &= \frac{a}{b} \To d = \frac{aL}{b-a} \end{aligned}$

The lower bound of x is $d = \dfrac{aL}{b-a}$, and the upper bound is $d + L = \dfrac{bL}{b-a}$
(2) The area of cross section is

$\begin{aligned} \frac{x}{d + L} &= \frac{r}{b} \To r = \frac{(b-a)x}{L}\\ A &= \pi r^2 = \frac{\pi(b-a)^2 x^2}{L^2} \end{aligned}$

(3) The resistance can be calcuated by integration.

$\begin{aligned} dR &= \frac{\rho dx}{A} \quad(R = \frac{\rho L}{A})\\ R &= \int_{aL/(b-a)}^{bL/(b-a)} \frac{\rho}{\frac{\pi(b-a)^2 x^2}{L^2}} dx\\ &= \frac{\rho L^2}{\pi(b-a)^2} \int_{aL/(b-a)}^{bL/(b-a)} \frac{dx}{x^2}\\ &= \frac{\rho L}{\pi ab} \end{aligned}$