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Integration using substitution

  1. May 6, 2013 #1
    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. jcsd
  3. May 6, 2013 #2

    mfb

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

    du= is missing a dx.
     
  4. May 6, 2013 #3

    HallsofIvy

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    How can you have a wire (one-dimensional) "in shape of a truncated cone" (three dimensional).
     
  5. May 6, 2013 #4

    mfb

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    A wire is not a line, it has a finite cross-section.
     
  6. May 6, 2013 #5

    HallsofIvy

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    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 [itex]A= \pi (a- b)^2(x^2/4L^2)+ \pi(a- b)(x/2L)b+ \pi b^2/4[/itex].

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

    Drawing a thin "lamina" perpendicular to the
     
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