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

[LiamG_G]

Homework Statement

I have to find the ratio of R_B/R_A 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.

Homework Equations

R=pl/A

The Attempt at a Solution

R_A is simple enough to determine, but I'm stuck with R_B
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∫₀^L1/A dx and somehow changing this to involve r₀/2 and 2r⁰

 

[tiny-tim]

Hi LiamG_G!

B is a lot of discs in series.

 

[rude man]

Homework Statement

I have to find the ratio of R_B/R_A 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.

Homework Equations

R=pl/A

The Attempt at a Solution

R_A is simple enough to determine, but I'm stuck with R_B

The other involves R=pl/A becoming R=p∫₀^L1/A dx and somehow changing this to involve r₀/2 and 2r⁰

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

 

[LiamG_G]

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

But with my integral, which I thought would be R=ρ∫₀^L (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, r_disk=(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?

 

[tiny-tim]

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

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

(and I'm going out now )

 

[Chestermiller]

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

But with my integral, which I thought would be R=ρ∫₀^L (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, r_disk=(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?

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

 

[rude man]

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

But with my integral, which I thought would be R=ρ∫₀^L (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, r_disk=(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?

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.

 

[Chestermiller]

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.

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