# Can I integrate to find resistance of an object of increasing radius?

1. Nov 30, 2013

### LiamG_G

1. The problem statement, all variables and given/known data
I have to find the ratio of RB/RA where R is electrical resistance.
Both objects A and B have circular cross section and length L, and are made of the same material.
A has radius r0. B has radius r0/2 at the bottom and increases linearly to 2r0.

2. Relevant equations

R=pl/A

3. The attempt at a solution
RA is simple enough to determine, but I'm stuck with RB
I have 2 potential theories as how to tackle this.
One is that the thin end and the thick ends will cancel, due to symmetry, leaving me with a ratio of 1.
The other involves R=pl/A becoming R=p∫0L1/A dx and somehow changing this to involve r0/2 and 2r0

2. Nov 30, 2013

### tiny-tim

Hi LiamG_G!

B is a lot of discs in series.

3. Dec 1, 2013

### rude man

Pick that one!
Your integral is OK, so what is A(x)?

4. Dec 1, 2013

### LiamG_G

So A=∏r2 and r ranges from r0/2 to 2r0

But with my integral, which I thought would be R=ρ∫0L (1/A)dx relates to the length of the cylinder, not it's radius.

I think I need to find an equation for r at a specific point in the bar in terms of x (which ranges from 0 to L)
Would this be somewhere along the right track?:
radius of disk, rdisk=(3r0/2)(x/L) + r0/2
This way, at x=0 (the bottom of the bar) the radius is equal to r0/2 and at x=L (the top of the bar) the radius is equal to 2r0
So then I would also have to differentiate that I can replace dx with dr, but then I'm sure I would still have x in the integral, which leaves me with two variables.

Am I at least on the right track? Where would I go next?

5. Dec 1, 2013

### tiny-tim

suppose you had n discs in series, of areas A1 … An (and all of the same thickness, ∆x)

what would the formula be for the total resistance R?​

(and i'm going out now )

6. Dec 1, 2013

### Staff: Mentor

Very nice reasoning. This is exactly what you should do.

7. Dec 1, 2013

### rude man

You are very much on the right track.

So now you have r(x), what is A(x)? A(x) contains r0 only, which is a constant. So you integrate with respect to x only. You never integrate w/r/t r.

8. Dec 1, 2013

### Staff: Mentor

Actually, you could do it either way. For example, dx=((2L)/(3r0))dr