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Problem about mean curvature

  1. Sep 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that the only surfaces with zero mean curvature are either planes or hyperbolic curves with the equation: [itex]y = \frac{\cosh (ax+b)}{a}[/itex] rotating alone the x axis.


    2. Relevant equations



    3. The attempt at a solution
    I made an attempt by devoting the equation of the surface as r = r(x) then take this back to the definition of mean curvature which ended up with a very complicated differential equation. Then I worked out the expression of mean curvature using Vieta's formula only to find myself facing a even more complex differential equation again. However there must be an easy way to prove the statement.
    Thanks!
     
  2. jcsd
  3. Sep 25, 2012 #2
    Eh solved it myself. Let be [itex]x = u (r)[/itex] substitute this in the Mean Curvature expression [itex](1 + \frac{\partial u}{\partial y}^{2})u_{zz} - 2 u_{x}u_{y}u_{xy} + (1 + \frac{\partial u}{\partial z}^{2})u_{yy}[/itex]
    Integrate the expression obtained so one can find out the expression of the surface is to be {itex}\frac {e^{az} + e^{-az}}{2a}{\itex}
     
  4. Sep 25, 2012 #3
    Eh solved it myself. Let be [itex]x = u (r)[/itex] substitute this in the Mean Curvature expression [itex](1 + (\frac{\partial u}{\partial y})^{2})u_{zz} - 2 u_{x}u_{y}u_{xy} + (1 + \frac{\partial u}{\partial z}^{2})u_{yy}[/itex]
    Integrate the expression obtained so one can find out the expression of the surface is to be [itex]\frac {e^{az} + e^{-az}}{2a}[/itex]
     
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