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Since for a cylinder $R = \rho\frac{\text{length}}{\text{Area}}$

I assumed an equation of radius of this frustum: $\text{radius} = \frac{b-a}{\text{length}}x$

and then integrated to get the answer $$\int_0^L dR = \int_0^L \rho\frac{dx}{\pi\left(a+\frac{b-a}{L}x\right)^2}$$ $$R = \rho\frac{L}{\pi ab}$$

Are there other ways to find the Resistance without assuming an equation of radius (although it is not that difficult to assume such an equation using details given in the problem)

  • $\begingroup$ You are not assuming anything here. You are merely using a fact given in the question that radius varies linearly and wrote the equation for that. There are other more complicated ways of solving this using different infinitesimal elements and putting them in parallel combination. Your method is by far, the easiest. $\endgroup$ – sarat.kant May 11 '16 at 12:20

You aren't assuming anything. The problem states

The radius of the cross-section varies linearly from $a$ to $b$

Therefore, your "assumption" is just a fact given to you by the problem. If the radius didn't vary linearly than you would be dealing with an entirely different geometry.

I don't think there's another way to solve this problem, and if there is I can't imagine it being any simpler than the way you've already used.


Your solution is acuatlly wrong. Full explanation of the reason is given in this paper:


The reason is that you can't separate the resistor into slabs to do the integration as you did because the slabs surfaces are not equipotentials. There are better ways to approximate a result.

Though I havent seen an analytical result, the problem can be stated into a Laplace's equation with appropriate change of variables, since solving PDE is no simple task, computer solutions shows that the method you used gives 9% of error from the correct result.


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