# Taking length/area element when trying to find E.resistance

I am trying to find out which 'element to take' (and why), When trying to find the resistance of some material with non-uniform resistivity $\rho$.

I'll give an example:

Say we have a co-axial cable with length $L$, inner radius $a$ and outer radius $b$ with resistivity $\rho(r)$

• If we look at the cable's resistance when the current is parallel to it's symmetry axis, we take the element $dA=2\pi r dr$ and the length as a (constant) function $\ell (r) = L$.
• If we look at the cable's resistance when the current is radial, we take the element $d\ell = dr$ and $A$ as a function : $A(r)=2\pi L r$

if the resistivity would be $\rho (z)$ - where $z$ is the axis of symmetry - would we switch that? I mean, in the second case would we take $\ell (r) =r\Rightarrow \ell_{total} =b-a$ and $dA= 2\pi(b-a) dz$

My reasoning is that we take $dA$ or $d\ell$ in respect to where the resistance grows: parallel to the current - we take $d\ell;$ otherwise - $dA$

I would be glad to hear if I'm right or wrong, and be corrected if needed.

Edit: This isn't really a homework question, it's tagged as one because it is homework related. I want to know how I should get the resistance in any way, and to conceptually understand why. a textbook refrence will be great too.

Thanks in advance, Daniel.

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I would go on with $A=\pi \left(b^2 - a^2 \right)$ and $dl = dz$, so I can (at least in principle) take the integral! – Ali Jul 7 '13 at 18:44
You're right, I made a mistake - it should be $dA=2\pi(b−a)dz$ , but I don't think you can say $d\ell =dz$ because the current doesn't flow in the $z$ direction, it flows in the $r$ direction. we can still take the integral: $\frac{1}{R}=\int\frac{dA}{\rho (z) \ell}=\int^{L}_{0}\frac{2\pi(b-a)}{\rho(z)\cdot (b-a)}dz$ this is similar to what we do in the first case I mentioned. – Bitx Jul 7 '13 at 22:25
My bad, I thought the current is similar to the first example(current in the $z$ direction). If the current is radial and $\rho$ is $z$-dependent, then $dA=2 \pi r dz$ and $dl = dr$. This gives a simple dual integral. – Ali Jul 8 '13 at 1:34
So can I conclude my reasoning is right? or does it just happens to work? – Bitx Jul 8 '13 at 10:21
In the second case, you cannot say $dA=2\pi(b-a)dz$, because even if the resistivity is constant radially, the resistance changes because the area through which the current is passing increases with every increment of $dr$. The equation @Ali suggests for $dA$ is correct and it will give you a dual integral, first through $r$ fro every $dz$ and then through $z$ – udiboy1209 Jul 8 '13 at 16:07