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Posers and Puzzles

Posers and Puzzles

  1. 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
  2. 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
  3. 25 Nov '04 07:12 / 1 edit
    "What about such a function that yields a shape of finite volume but infinite surface area?"

    Gabriel's Trumpet.
  4. Standard member TheMaster37
    Kupikupopo!
    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. Donation Acolyte
    Now With Added BA
    25 Nov '04 11:40 / 1 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 member Nemesio
    Ursulakantor
    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. 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. Subscriber AThousandYoung
    It's about respect
    26 Nov '04 09:54
    I think yes and yes.
  9. 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. 27 Nov '04 17:37 / 1 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. 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.