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:frown: Normal curvature integral proof

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data

    I need to show that the mean curvature [tex]H[/tex] at [tex]p \in S[/tex] given by

    [tex]H = \frac{1}{\pi} \cdot \int_{0}^{\pi} k_n{\theta} d \theta[/tex]

    where [tex]k_n{\theta}[/tex] is the mean curvature at p along a direction makin an angle theta with a fixed direction.


    2. Relevant equations

    I know that the formel definition of the mean curvature is [tex]H = \frac{k_1 + k_2}{2}[/tex]
    where k1 and K2 are the maximum and minimum normal curvature.

    I know that the normal curvature is defined as [tex]k_n = k \cdot cos(\theta)[/tex]. where [tex]\theta[/tex] is defined as the angle between the eigenvectors e1 and e2 of [tex]dN_{p}[/tex]

    3. The attempt at a solution

    do I then claim that [tex]<dN_{p}(\theta), \theta)> = k_n \cdot \theta[/tex] ??

    Could somebody please help me along here? with a hint or something? or this there is some theory that I have missed here??

    Best regards
    Fred
     
    Last edited: May 7, 2007
  2. jcsd
  3. May 9, 2007 #2
    undeletable
     
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