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Homework Help: Surface of constant curvature

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find a surface of revolution with Gauss curvature K of +1 at all points, which doesn't lie in a sphere.

    2. Relevant equations

    The surface is parametrized as [itex]\psi (t, \theta ) = ( x(t), y(t) cos \theta , y(t) sin \theta ) [/itex]

    I have the equation
    K = \frac{x' (x'' y' - x' y'')}{y(x'^2 + y'^2)^2}

    3. The attempt at a solution

    I am thinking it has to do with the curve [itex] \alpha (t) = (x(t),y(t)) [/itex] not having unit speed, but I am kind of stuck as to where to go from there.

  2. jcsd
  3. Dec 15, 2008 #2


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    Is there one? The only way out I can think of is to make it disconnected. A surface of constant guassian curvature is locally isometric to a sphere.
  4. Dec 16, 2008 #3
    I know that for a unit-speed [itex]\alpha (t)[/itex], the equation reduces to [itex]K = \frac{-y''}{y}[/itex], which does clearly represent a sphere.

    The way the question is worded on my homework seems to point to the fact that that reduction only applies to unit-speed curves, which is why I think that perhaps an [itex]\alpha (t)[/itex] that doesn't have unit-speed perhaps can give a surface of revolution with constant curvature +1 that isn't a sphere... but maybe there's another "gimmick" that I'm overlooking...

    Anyways, thanks for your help!
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