Differential Geometry and Its Application - Exercise 3.1.6

Problem:

Using Euler's formula (Corollary 2.4.11) to show

(1) The mean curvature $ H $ at a point is the average normal curvature

$ H = \frac{1}{2\pi}\int\limits_{0}^{2\pi}k(\theta)d\theta $.

(2) $ H = \frac{1}{2}(k(v_1) + k(v_2)) $ for any two unit vectors $ v_1 $ and $ v_2 $ which are perpendicular.

Solution:

The Euler formula is $ k(\theta) = \cos^2(\theta)\lambda_1 + \sin^2(\theta)\lambda_2 $.

For part (1), we simply integrate it.

Note that $ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} $, so $ \int\limits_{0}^{2\pi}\cos^2 \theta = \pi $.
Same for $ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} $, so $ \int\limits_{0}^{2\pi}\sin^2 \theta = \pi $.

Putting them together, we get

$ \begin{eqnarray*} & & \frac{1}{2\pi}\int\limits_{0}^{2\pi}k(\theta)d\theta \\ &=& \frac{1}{2\pi}\int\limits_{0}^{2\pi}(\cos^2(\theta)\lambda_1 + \sin^2(\theta)\lambda_2)d\theta \\ &=& \frac{1}{2\pi}(\lambda_1 \int\limits_{0}^{2\pi}\cos^2(\theta)d\theta + \lambda_2 \int\limits_{0}^{2\pi}\sin^2(\theta)d\theta) \\ &=& \frac{1}{2\pi}(\lambda_1 \pi + \lambda_2 \pi) \\ &=& \frac{\lambda_1+ \lambda_2}{2} \\ &=& H \end{eqnarray*} $

Part (2) is also simple, we have

$ \begin{eqnarray*} & & \frac{1}{2}(k(v_1) + k(v_2)) \\ &=& \frac{1}{2}(k(\theta) + k(\theta + \frac{\pi}{2})) \\ &=& \frac{1}{2}(\lambda_1 \cos^2\theta + \lambda_2 \sin^2\theta + \lambda_1 \cos^2(\theta + \frac{\pi}{2})+ \lambda_2 \sin^2(\theta + \frac{\pi}{2})) \\ &=& \frac{1}{2}(\lambda_1 \cos^2\theta + \lambda_2 \sin^2\theta + \lambda_1 \sin^2(\theta)+ \lambda_2 \cos^2(\theta)) \\ &=& \frac{1}{2}(\lambda_1 + \lambda_2) \\ &=& H \end{eqnarray*} $