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Excellent question: A=infinity, V=pi?

  1. Feb 20, 2006 #1
    My professor posed a brain teaser question today, and I cant get it out of my mind. I was hoping the forum can help me make sence of it.

    Area under 1/x = infinity:
    [tex]A = \int_{1}^{\infty} (1/x)dx[/tex]
    [tex]A = \lim_{t\rightarrow \infty} \int_{1}^{t} (1/x)dx[/tex]
    [tex]A = \lim_{t\rightarrow \infty} \ln t - \ln 1[/tex]
    [tex]A = \infty[/tex]

    but the volume of 1/x rotated around the x axis is equal to pi
    [tex]V = \pi \int_{1}^{\infty} (1/x)^2dx[/tex]
    [tex]V = \pi \lim_{t\rightarrow \infty} \int_{1}^{t} (1/x)^2dx[/tex]
    [tex]= \pi \lim_{t\rightarrow \infty} (-1/t + 1/1)[/tex]
    [tex]= \pi (1)[/tex]

    How can this be true?
    Last edited: Feb 20, 2006
  2. jcsd
  3. Feb 20, 2006 #2


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    Well, one way to break down your intuitive opposition to it is that for circles with small radii, pi times the square of the radius is smaller than the radius itself.
  4. Feb 20, 2006 #3


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    Why can't it be true :wink: ?

    It seems that popularly said, it is perfectly possible to create a mathematical 'object' with a finite volume but an infinite surface area (which is 2*pi times the integral you calculated first).
  5. Feb 20, 2006 #4


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    See here. Especially the paragraph about the "paradox" that you could fill this shape with a finite amount of paint, but cover an infinite surface area.
  6. Feb 20, 2006 #5


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    The solid of revolution formed by rotating y=1/x about the x-axis from x=1 to infinity is called Gabriel's Horn.
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