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Surface area of a solid of revolution

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider now a shape that is obtained by revolving
    the “infinitely long” function
    f (x) = 1/x , 1 ≤ x < ∞
    around the x-axis. Find both the surface area of the resulting object, and the
    enclosed volume of it, i.e. the volume of the solid obtained from revolving the
    area under the graph of f (x) around the x-axis.

    If you do this correctly, you will find an infinite surface area but a finite volume. This would mean that a certain (finite) amount of paint is sufficient to fill up our object, but that this
    amount would not be enough to actually paint it. Does this surprising result
    mean that there is something fundamentally wrong with our approach? Why
    or why not? Comment critically!

    2. Relevant equations
    I understand how to calculate the surface area and the volume of a solid of revolution, however it's the second part of the question I don't really undertand, the part about there being something fundamentally wrong with our approach. I was wondering if anyone could give me some hints to point me in the right direction.
     
  2. jcsd
  3. Mar 9, 2009 #2

    Mark44

    Staff: Mentor

    A good start would be to actually calculate the surface area and volume for this solid. Keep in mind that you are going to have improper integrals for both, since one of the limits of integration is [itex]\infty[/itex].
     
  4. Mar 10, 2009 #3
    Yeah, I did that:

    I got pi as the volume of the solid and undefined (infinity) for the surface area. However its the theoretical part that i'm unsure about, where its asking if there's something wrong with the method...
     
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