# Integration using substitution

1. May 6, 2013

### cris9288

1. The problem statement, all variables and given/known data

I have a wire in the shape of a truncated cone. One side has radius a and the other has radius b. The wire has resistivity ρ and length L. I am supposed to find the resistance of the wire using R = ρL/A

2. Relevant equations

R= ρL/A

3. The attempt at a solution

So far I have set up my integeral. I know the answer, but I am having trouble with the actual integration by substitution.

ρ∫ dx / (π * (a + (b - a)(x / L))^2)

u = a + (b - a)(x / L)

du = ( b - a ) / L

This just gets me into a huge mess. The answer is ρL/ab. Thanks for any help.

2. May 6, 2013

### Staff: Mentor

How does this huge mess look like? It should be quite simple.

du= is missing a dx.

3. May 6, 2013

### HallsofIvy

Staff Emeritus
How can you have a wire (one-dimensional) "in shape of a truncated cone" (three dimensional).

4. May 6, 2013

### Staff: Mentor

A wire is not a line, it has a finite cross-section.

5. May 6, 2013

### HallsofIvy

Staff Emeritus
Oh, I see. Thanks. So we are just talking about the truncated cone.

Since the sides are lines, the diameter at each point is a linear function of ditance along the axis. Taking x= 0 where the diameter is "b" and x= L where it is a, we have d= (a- b)(x/L)+ b so that the radius is d/2= (a- b)(x/2L)+ b/2 and the cross section area is $A= \pi (a- b)^2(x^2/4L^2)+ \pi(a- b)(x/2L)b+ \pi b^2/4$.

Since $R= \rho L/A$, with a small section of length "dx", $R=\rho dx/A= \rho dx/(\pi (a- b)^2(x^2/4L^2)+ \pi(a- b)(x/2L)b+ \pi b^2/4)$ and the total resistance is the integral of that, with respect to x, from 0 to L.

Drawing a thin "lamina" perpendicular to the