Mean curvature of a ruled surface in $\mathbb{R}^3$ I was trying to prove that non compact surfaces with constant mean and gaussian curvature in $\mathbb R^3$ are part of planes or cylinder. Here is how I worked out the problem so far,
First of all I proved that the gaussian curvature must be zero i.e $K=0$,
Then I used the fact that the surface will be develop-able and hence in this case a ruled surface. I considered a parametrization $x(u,v)=\beta(u)+v\delta(u)$ of the surface, I know that the gaussian curvature in this case is given by $K=\frac{-M^2}{EG-F^2}$ which will imply that $M=0$, Can someone show me how to explicitly compute $H$ in this case? and how can i use the fact that $H$ is constant to prove that $\gamma'(u)=0$, which will prove that it is a cylinder.
Any help will be greatly appreciated.
So here is what I think, since $H=\frac{EN-2FM+GL}{2(EG-F^2)}$ in general, and in this case, $N=0$, I will be left with only $H=\frac{LG}{2(EG-F^2)}$, for $H$ to be constant, $L$,$G$and $EG-F^2$ should be constant. How do i get further information form here?
 A: It may be relevant to examine the arguments here proceeding from scalar curvatures.
In terms of principal curvatures $ k_1,k_2 $, when H and K are both constant,
$$ k_1 + k_2 = 2 H , k_1 k_2 = K $$ solving,
$$  k_1  = 1/a,  k_2  = 1/b $$ where $a,b$ are constant.
To understand this in case of surfaces of revolution :
They are 
1) Toroids,Spheres  ($ k_1 = const $ ) and, 
2) Special cases of Weingarten surfaces where $ \dfrac {k_1}{k_2} = const. $
Examples are Spheres, Cycloids and  Elastica special case with meridional 
tangent are normal to axis of symmetry.
3) Minimal surface.
Applying Quotient Rule,
$$ k_2= const. = \dfrac{\cos \phi }{r}  =  \dfrac {-\sin \phi\; k_1} {\sin \phi} = -k_1 $$ 
or,
$$ k_1 + k_2 = 2 H =0 $$
which is a catenoid minimal surface, $ r/c = \cosh(z/c) $, an exceptional case for a ruled surface.
But note that these are not ruled surfaces in general nor do they belong only to $K=0$ class of developable surfaces (cones, developable helicoids).
EDIT1:
There is something disconcerting about taking $ K and H $ independent though, shall mention it.
For a surface of revolution $ \tan\phi = \frac{dr}{dz} $ 
Let $$ R_1 = 1/k_1, R_2= 1/k_2 $$
It can be verified that
$$ \dfrac{d R_2}{d\cos\phi} =- (R_1+R_2) $$
so they are not really independent.
