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

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