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Infinte Area but finite Volume

  1. Nov 11, 2004 #1
    Whence evaluating the area under the curve

    [tex]y=\frac{1}{x} \\\ \mbox{for} \\\ 1 \leq x < \infty [/tex]

    it evaluates to [tex]\infty [/tex]

    But when evaluating the volume using

    [tex] Volume = \pi \int y^2 dx \\\ \mbox{on} \\\ a \leq x < b [/tex]

    hence


    [tex] Volume = \pi \int \frac{1}{x^2} \\dx [/tex]

    hence

    [tex] Volume = \pi [-\ \frac{1}{x}] \\\ \mbox{on} \\\ 1 \leq x < \infty [/tex]

    hence

    [tex] Volume = \pi [0 - - 1] = \pi [/tex]

    A finite value!

    Im having trouble comprehending such concepts and ideas.
    Can someone please explain?
     
    Last edited: Nov 11, 2004
  2. jcsd
  3. Nov 11, 2004 #2

    for a more direct "blow your mind" property, the surface area of that solid of revolution is infinite (and the volume, like you said, finite).

    things dealing with infinity get pretty strange.

    there are fractals that exhibit similar properties, like infinite surface area but zero volume and such.

    i deal with it by just casting it off as math. :biggrin:
     
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