:frown: Normal curvature integral proof

[Mathman23]

Homework Statement

I need to show that the mean curvature $$H$$ at $$p \in S$$ given by

$$H = \frac{1}{\pi} \cdot \int_{0}^{\pi} k_n{\theta} d \theta$$

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

Homework Equations

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

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

The Attempt at a Solution

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

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

 

[Mathman23]

undeletable