1. Account suspended
    Joined
    11 Nov '04
    Moves
    77
    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. Account suspended
    Joined
    11 Nov '04
    Moves
    77
    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. Joined
    29 Feb '04
    Moves
    22
    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
    Joined
    25 Oct '02
    Moves
    20443
    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
    Joined
    04 Jul '02
    Moves
    3790
    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
    Pittsburgh, PA
    Joined
    05 Mar '02
    Moves
    32455
    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
    29 Feb '04
    Moves
    22
    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. SubscriberAThousandYoung
    All My Soldiers...
    tinyurl.com/y9ls7wbl
    Joined
    23 Aug '04
    Moves
    24791
    26 Nov '04 09:54
    I think yes and yes.
  9. Joined
    29 Feb '04
    Moves
    22
    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
    Joined
    11 Nov '04
    Moves
    77
    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
    19 Jun '04
    Moves
    2930
    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.
Back to Top