Assuming I have a ruled surface parametrized as $x(u,v)=\beta(u)+v\delta(u)$, with zero Gauss curvature, which in this case is given by $K=-\frac{m^2}{EG-F^2}$=$\frac{- (\beta'\delta \times \delta')^2 }{|\beta' \times \delta +v\delta\times \delta|}=0$, we know that $H$ is a constant so how can I use this information along with the curvature to prove that is a right cylinder? Any help will be greatly appreciated.
P.S I know that flat ruled surface is a part of plane, generalized cone or generalized cylinder.
Ok, here is some addition. I think I got the idea, but still i am missing something. I found out that for ant ruled surface I can find a parametrization such that $\beta'\cdot\delta=0$ , Can I choose $\delta$ such that $\delta\cdot\delta=0$, this would imply $\delta'\cdot\delta=0$, so the metric elements are $E=(\beta'+v\delta')^2, F=0, G=\delta^2=1$ and since $n=0$ because $x_{vv}=0$, the mean curvature is reduced to $H=\frac{Gl}{2EG}=\frac{l}{2E}$, I am unable to move from here, what i know is that $H=\frac{1}{2r}$ for right circular cylinder. I have calculated the expression for $l=\frac{(\beta''+v\delta'')((\beta'+v\delta')\times\delta)}{|(\beta'+v\delta')\times\delta|}$, How can i use the fact that $H$ is constant to conclude the result??
Please help from here
