A system is set up which is composed of a cylinder with hemispheres on either ends. The length of the cylinder is l and inner and outer radii of cylinder and hemisphere are $R_i$ and $R_f$. The thermal conductivity of both parts are same and equal to k. The temperature inside the setup is maintained at a constant temperature of $\theta$ while the surrounding temperature is $\theta_o$.

$\mathbf{Approach 1}$: Consider an infinitesimal shell at distance r from the axis of the cylinder and r distance from the center of the hemispheres. Let the thickness of the shell be $dr$.

The resistance of this shell is $$dR=\frac{1}{k}\frac{dr}{2\pi rl+4\pi r^2}$$. Since all the shells are in series connection, the net resistance is $$\int dR=\int^{R_f}_{R_i}\frac{1}{k}\frac{dr}{2\pi rl+4\pi r^2}$$ which gives $ R=\frac{1}{2\pi kl}(ln\frac{R_f}{R_f+l/2}-ln\frac{R_i}{R_i+l/2})$

Now consider an alternative approach.

$\mathbf{Approach 2}$:

(I am not going to show the calculations for the resistance of cylinder and hemisphere; I will directly use the results.) Since the the inner parts of the cylinders and the hemispheres are at the same temperature. Similarly, the outer parts are also at the same temperature. So we can say that the cylinders and hemispheres are arranged in a parallel connection. So the equivalent resistance is given by $$\frac{1}{R_eq}=\frac{2\pi kl}{ln\frac{R_f}{R_i}}+\frac{4\pi k R_f R_i}{R_f-R_i}$$ which on solving for $R_eq$ gives a completely different result from the first one.

On observing the first approach more carefully, it is clear that the equivalent resistance for parallel combination has been used before integrating the elemental resistance. While in the second approach, the elemental resistances have already been integrated and then the formula for parallel combination has been used.

Could you tell me which approach is wrong and why? I can't understand where I went wrong.