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Electrical resistance of a paraboloid.

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    What would the electrical resistance of a paraboloid from y = 0 to L be?

    2. Relevant equations
    [tex]R = \rho \frac{L}{A}[/tex]

    3. The attempt at a solution
    Okay, so I'll put the parabola (that would rotate into the paraboloid) into the form [tex] y = \sqrt{x}[/tex]

    The function A(x) is just the area of the circle, at distance x.

    A = [tex]\pi y^{2} = \pi x[/tex]

    I'll break the paraboloid up first into a finite sum of discs, from 0 to L.

    [tex]R = \Sigma \rho \frac{\Delta x}{A(x)}[/tex]


    [tex]R = \int^{L}_{0} \rho \frac{dx}{\pi x}[/tex]


    [tex]R = \frac{\rho}{\pi} \int^{L}_{0}\frac{1}{x} dx[/tex]

    This integral resolves to:

    [tex]R = [\frac{\rho}{\pi}ln(x)]^{L}_{0}[/tex]

    Natural log of 0 is undefined, so this would resolve to a numerical answer, as at the limit of x ==> 0, the area approaches zero and this means infinite resistance. But is the maths correct? Or can someone suggest a better way to actually find an answer.
  2. jcsd
  3. Apr 21, 2010 #2
    bump. <_<
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