According to wikipedia the mean curvature is define by an integral of the form
$$
H = \frac{1}{2\pi} \int_0^{2\pi} \kappa(\theta)d\theta
$$
but is also equal to
$$
H = \frac{1}{2}(\kappa_1+\kappa_2)
$$
where $\kappa_1,\kappa_2$ are the principals curvatures. I struggle to find a proof though, can any of you prove it or point out any reference please?
Best Answer
Let $e_1,e_2$ be the principal directions, so that if $L$ denotes the Weingarten operator, we have $\left\langle Le_i,e_i\right\rangle=\kappa_i$, and write $$e_{\theta}=\cos \theta e_1 + \sin \theta e_2 $$ Thus $$k(\theta)= \left \langle Le_{\theta},e_{\theta}\right\rangle=\kappa_1\cos^2\theta +2\kappa_{12}\cos \theta \sin \theta + \kappa_2\sin^2\theta$$ Where $\kappa_{12}=\left\langle Le_1,e_2\right\rangle$. Integrating $$\int_0^{2\pi}k(\theta)d\theta =\kappa_1\int_0^{2\pi}\cos^2\theta d\theta+2\kappa_{12}\int_0^{2\pi}\cos\theta\sin\theta d\theta+\kappa_2\int_0^{2\pi}\sin^2\theta d\theta =\pi(\kappa_1+\kappa_2) $$ (one can also prove that $\kappa_{12}=0$, but this is not needed here).