Determine the rate of heat flow across a cross section of the rod...

Consider an elemental disc of thickness $dx$ at a distance $x$ from the end with radius $r_1$.

The radius of this disc is given by $r = r_1 + \left(\dfrac{r_2 - r_1}{L}\right)x$.

The thermal resistance of this elemental disc is $dR = \dfrac{dx}{K A} = \dfrac{dx}{K \pi r^2}$.

The total thermal resistance $R$ of the rod is the integral of $dR$ from $x = 0$ to $x = L$:

$R = \int_{0}^{L} \dfrac{dx}{K \pi \left[ r_1 + \left(\dfrac{r_2 - r_1}{L}\right)x \right]^2}$

Let $y = r_1 + \left(\dfrac{r_2 - r_1}{L}\right)x$.

Differentiating with respect to $x$, we get $dy = \left(\dfrac{r_2 - r_1}{L}\right)dx \Rightarrow dx = \dfrac{L}{r_2 - r_1} dy$.

The limits of integration change from $x=0 \to L$ to $y=r_1 \to r_2$.

$R = \int_{r_1}^{r_2} \dfrac{L}{K \pi (r_2 - r_1) y^2} dy$

$R = \dfrac{L}{K \pi (r_2 - r_1)} \left[ -\dfrac{1}{y} \right]_{r_1}^{r_2}$

R = \dfrac{L}{K \pi (r_2 - r_1)} \left(\dfrac{1}{r_1} - \dfrac{1}{r_2} \right)

$R =$\dfrac{L}{K \pi (r_2 - r_1)} \left( \dfrac{r_2 - r_1}{r_1 r_2} \right) = \dfrac{L}{K \pi r_1 r_2}

The rate of heat flow is given by $H =$\dfrac{\Delta \theta}{R}.

$H =$\dfrac{\theta_2 - \theta_1}{\dfrac{L}{K \pi r_1 r_2}} = \dfrac{K \pi r_1 r_2 (\theta_2 - \theta_1)}{L}