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Resistance: Non-constant Surface Area

  1. Sep 7, 2003 #1
    I think I'm on the verge of a breakthrough on this problem, but it's just not coming.

    Please tell me where my approach goes wrong or whether I'm correct but don't know how to integrate the resulting equation properly.

    The problem reads: A circular disk of radius r and thickness d is made of material with resistivity p. Show that the resistance between the points a and b (a is the front of the curved side of the cylinder and b is the backside) is independent of the radius and is given by R = [pi]p/2d.

    I start out with R = pL/A where L is length and A is the surface area perpendicular to the direction of the flow of the charges.

    dR = pdx/A

    A = d * the width of the rectangular surface area for the little sliver of the cylinder, which is given by the equation of the circle y = 2[squ] (r^2-(x-r)^2).

    Is this correct?

    I then integrate R = p/2d[inte] dx/[squ](r^2-(x-r)^2) from 0 to 2r?

    This gives me R= [pi]pr^2/8d.

    Obviously wrong!!!

    Any suggestions would be greatly appreciated.

  2. jcsd
  3. Sep 8, 2003 #2

    Tom Mattson

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    I am a little fuzzy on where a and b are, so I don't know if your setup is correct, but...

    I don't know how you did that integral, but when I do it I get exactly what you say I am supposed to get.


    from 0 to 2r.

    Let u=x-r
    so du=dx



    from -r to r.

    I get:


    When you evaluate that, you will get R=ρπ/2d, as advertised.
  4. Sep 8, 2003 #3
    It's been a year since I've taken my last calculus class so I need to brush up. I used a table of integrals and I might have misread something.

    I'm sorry that I didn't include a diagram. Imagine the cylinder sitting in front of you on its flat side. a to b is the front side to the back side or left side to right side. In this case the width of the rectangular surface area increases as you move from a to b according to the equation of a circle of radius r centered at (r,0).

    So it looks like I set the integral up correctly, but evaluated it incorrectly.

    Thanks so much for your help. It came just in the nick of time.:smile:
  5. Sep 8, 2003 #4

    Tom Mattson

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    That must be it. As for me, I just happen to be teaching Calculus II this semester, and I taught my class that exact integral last week, so it is still pretty fresh in my mind. If not for that, I probably would have gone to an integral table myself!
  6. Sep 11, 2003 #5


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    I suspect that your problem is that you "lost track" of the fact that the square root is in the denominator. The way I would do this, by the way, is integrate from -r to r rather than 0 to 2r with the origin at the center rather than in at p. That precisely the same as making the substitution "u= x-r" in your integral and gives

    int(u= -r to r) (r^2- u^2)^(-1/2) du

    A standard way to integrate something like that is to make the substitution u= r sin[theta]. Then du= r cos[theta] d[theta] and
    r^2- u^2= r^2(1- sin^2[theta])= r^2 cos^2[theta] so that the square root gives r cos[theta]. Because those are in the DENOMINATOR, they will precisely cancel the "r cos[theta]" in the numerator. Your integral becomes int([theta]= -[pi/2] to [pi]/2) d[theta]= [pi].

    That gives the result you want.
  7. Sep 11, 2003 #6
    Thanks for tip.

    I'm sure I'll plenty of use for it during this semester's Electricity and Magnetism class.
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