Why the Gauss curvature is zero?

Suppose a parameterized surface in $\mathbb{R}^3$ is given by $S: r=r(u,v)$, $(u, v)\subset D\subset \mathbb{R}^2$. I wanna show that if $e_1=r_u$, $e_2=r_v$, then the Gauss curvature of $S$ must be zero. I've tried to do some $K=\frac{LN-M^2}{EG-F^2}$ work but failed to find the exact formula of these amounts. How can I show that the Gauss curvature is zero?

Solution 1:

If $r_u= e_1$, then $r= e_1u + f(v)$, so that if furtermore $r_v=e_u$, $r= u e_1+ve_2+ r_0$, $r_0$ a fixed vector; and your surface is an affine plane directed by $e_1,e_2$.

Another proof is that the Normal map $N(u,v)$ is constant, so that the Weingarten endomorphism is $O$, so total curvature and the Gauss curvature are 0.