What are the only possible surfaces with zero mean curvature?

[raopeng]

Homework Statement

Prove that the only surfaces with zero mean curvature are either planes or hyperbolic curves with the equation: $y = \frac{\cosh (ax+b)}{a}$ rotating alone the x axis.

Homework Equations

The Attempt at a Solution

I made an attempt by devoting the equation of the surface as r = r(x) then take this back to the definition of mean curvature which ended up with a very complicated differential equation. Then I worked out the expression of mean curvature using Vieta's formula only to find myself facing a even more complex differential equation again. However there must be an easy way to prove the statement.
Thanks!

 

[raopeng]

Eh solved it myself. Let be $x = u (r)$ substitute this in the Mean Curvature expression $(1 + \frac{\partial u}{\partial y}^{2})u_{zz} - 2 u_{x}u_{y}u_{xy} + (1 + \frac{\partial u}{\partial z}^{2})u_{yy}$
Integrate the expression obtained so one can find out the expression of the surface is to be {itex}\frac {e^{az} + e^{-az}}{2a}{\itex}

 

[raopeng]

Eh solved it myself. Let be $x = u (r)$ substitute this in the Mean Curvature expression $(1 + (\frac{\partial u}{\partial y})^{2})u_{zz} - 2 u_{x}u_{y}u_{xy} + (1 + \frac{\partial u}{\partial z}^{2})u_{yy}$
Integrate the expression obtained so one can find out the expression of the surface is to be $\frac {e^{az} + e^{-az}}{2a}$