Finding the power loss in a wire of varying cross-sectional area

Glenn G said:

TL;DR Summary: Finding the power loss in a wire of varying area - my problem is I don't know how to set up the integral

Hopefully you can see in the diagram below that the area of the wire varies linearly with length. I know the equations for resistance and power loss and I can express the resistance of a thin slice but need to integrate over the whole wire to get the full resistance - to find power then it is trivial. I've thought about this for a good while now but can't get it. Please help, this isn't homework (I'm very old and doing this for fun!)

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DaveE said:

For a conical wire:

the radius ##r_1 = \sqrt{\frac{A_1}{\pi}}## and ##r_2 = \sqrt{\frac{A_2}{\pi}}## so ##r(x) = r_1 + x \frac{r_2-r_1}{L}## where ##x \equiv 0## at ##r_1## and ##x \equiv L## at ##r_2##

The resistance of a thin disk at ##x## is ##R(x) = \frac{\rho}{\pi r(x)^2}dx##

Then the total resistance is ##R = \int_0^L R(x) = \frac{\rho}{\pi} \int_0^L \frac{1}{r(x)^2} \, dx = \frac{\rho}{\pi} \int_0^L \frac{1}{(r_1 + x \frac{r_2-r_1}{L})^2} \, dx = \frac{\rho L}{\pi (r_1-r_2)} \left. \frac{1}{(r_1 + x\frac{r_2-r_1}{L})} \right|_0^L##

## R = \frac{\rho L}{\pi (r_1-r_2)} (\frac{1}{(r_1 + L\frac{r_2-r_1}{L})} -\frac{1}{r_1}) = \frac{\rho L}{\pi (r_1-r_2)} (\frac{1}{r_2 } -\frac{1}{r_1}) = \frac{\rho L}{\pi (r_1-r_2)} (\frac{r_1-r_2}{r_1 r_2 } ) = \frac{\rho L}{\pi r_1 r_2 } ##

## R = \frac{\rho L}{\pi \sqrt{\frac{A_1}{\pi}} \sqrt{\frac{A_2}{\pi}} } = \frac{\rho L}{\sqrt{A_1 A_2} } ##

haruspex said:

That's fine if ##L>>r_2-r_1##. Otherwise there is the complication that current near the surface of the wire has further to travel than that near the core.