# Prove the Gaussian curvature $K=0$

If two families of a geodesics on a surface intersect at a constant angle $\theta$, prove that the Gaussian curvature of the surface is zero, i.e. $K=0$.

Please explain how to show the question. Thank you for helping.

$K=k_1k_2$ for principal curvatures $k_1, k_2$.

EDIT: Here's what I mean: Take two geodesics $\gamma_1$ and $\gamma_2$ from family A and two geodesics $\sigma_1$ and $\sigma_2$ from family B. Define points $P$, $Q$, $R$, and $S$ by taking $P=\gamma_1\cap\sigma_1$, $Q=\gamma_1\cap\sigma_2$, $R=\gamma_2\cap\sigma_2$, and $S=\gamma_2\cap\sigma_1$. (Draw a picture of all this.) Apply Gauss-Bonnet to this "parallelogram" formed by the pieces of the four geodesics joining $P$, $Q$, $R$, and $S$.
• No, Gauss-Bonnet is applied to the region $\mathscr R$ inside $PQRS$. Keep track of the four exterior angles. This should show $\iint_{\mathscr R} K\,dA=0$. Now how do you deduce $K=0$ everywhere? Commented May 14, 2014 at 19:51
• If I tell you $\int_a^b f(x)dx = 0$ for a continuous function $f$ and all intervals $[a,b]\subset\Bbb R$, how do you prove that $f=0$ everywhere? Commented May 14, 2014 at 20:03