1. Account suspended
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    25 Nov '04 00:58
    Can you construct a function f(x) such that when rotated around the x-axis, it yields a three-dimensional curve whose surface area is finite yet whose volume is infinite?

    What about such a function that yields a shape of finite volume but infinite surface area?

    Kribz
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    25 Nov '04 01:08
    Time saving hint: don't try to construct them from scratch. Either you know the answers or you don't. They are intended to be answered Yes or No, not by the construction of f(x).

    Kribz
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    25 Nov '04 07:121 edit
    "What about such a function that yields a shape of finite volume but infinite surface area?"

    Gabriel's Trumpet.
  4. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
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    25 Nov '04 09:18
    Originally posted by Kribz
    Can you construct a function f(x) such that when rotated around the x-axis, it yields a three-dimensional curve whose surface area is finite yet whose volume is infinite?

    What about such a function that yields a shape of finite volume but infinite surface area?

    Kribz
    Yes & Yes
  5. DonationAcolyte
    Now With Added BA
    Loughborough
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    25 Nov '04 11:401 edit
    Originally posted by TheMaster37
    Yes & Yes
    I get No & Yes. Finite volume, infinite area is fine, but not the other way around. For revolution purposes we may assume that f(x) is non-negative. Then:

    Area = S(2pi*f(x)*sqrt(1+f'(x)^2))dx (where S means integrate over the domain of f)
    >= 2pi*S(|f(x)|)dx

    So for finite area, f must be 1-integrable.

    Volume = pi*S(f(x)^2)dx

    So for infinite volume, f must not be 2-integrable.

    However, all 1-integrable functions are 2-integrable by Hoelder's inequality in the case where the exponents are both 1/2:

    S|fg| =< sqrt(S|f|)sqrt(S|g|)

    Hence there is no function for which a finite-area surface and an and infinite-volume solid of revolution exist.
  6. Standard memberNemesio
    Ursulakantor
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    25 Nov '04 17:25
    Originally posted by THUDandBLUNDER
    "What about such a function that yields a shape of finite volume but infinite surface area?"

    Gabriel's Trumpet.
    http://curvebank.calstatela.edu/torricelli/torricelli.htm

    This math is so neat and so inaccessible to me that it blows my mind.

    Cool stuff!
  7. Joined
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    26 Nov '04 01:21
    Originally posted by THUDandBLUNDER
    "What about such a function that yields a shape of finite volume but infinite surface area?"

    Gabriel's Trumpet.
    I meant Gabriel's Horn.
  8. Standard memberAThousandYoung
    or different places
    tinyurl.com/2tp8tyx8
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    26 Nov '04 09:54
    I think yes and yes.
  9. Joined
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    26 Nov '04 20:54
    Originally posted by AThousandYoung
    I think yes and yes.
    As for objects with infinite volume but finite surface area, I don't think this is possible. One well-known property of a sphere is that it maximizes volume for a given surface area. If infinite volume in finite surface area were possible, then you could improve on a sphere.
  10. Account suspended
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    27 Nov '04 17:371 edit
    Originally posted by THUDandBLUNDER
    One well-known property of a sphere is that it maximizes volume for a given surface area. If infinite volume in finite surface area were possible, then you could improve on a sphere.
    Good observation. I think this closes this problem, as we have a proof by construction for second question, and a proof by contradiction against the first question.
  11. Joined
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    03 Dec '04 10:01
    Originally posted by Kribz
    Can you construct a function f(x) such that when rotated around the x-axis, it yields a three-dimensional curve whose surface area is finite yet whose volume is infinite?

    What about such a function that yields a shape of finite volume but infinite surface area?

    Kribz
    Finite volume with infinite surface area is possible.
    The simplest example is the Kronecker delta function, rotated about the
    x-axis.
    The other way round is not possible.
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