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Prove its Gauss curvature K = 1

  1. Nov 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Assume that the surface has the first fundamental form as

    E = G = 4(1+u2+v2)-2

    F = 0


    2. Relevant equations

    K = [itex]\frac{-1}{2\sqrt{EG}}[(\frac{E_v}{\sqrt{EG}})_v + (\frac{G_u}{\sqrt{EG}})_u][/itex]



    3. The attempt at a solution

    Ev = -16v*(1+u2+v2)-3

    Gu = -16u*(1+u2+v2)-3


    When I take the partials and find K, I get something messy that doesn't lead me to the conclusion that K = 1.
     
  2. jcsd
  3. Nov 11, 2014 #2

    RUber

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    Start by substituting E=G and \sqrt(EG) = |E|.
    What did you end up with?
     
  4. Nov 11, 2014 #3
    Thanks for the reply. I found my error a little bit ago. For some reason, I kept using the denominator of E and G AS E and G. -_- It was late, so I'll credit it to sleep deprivation. Heh.
     
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