Surfaces with constant Gaussian curvature Must the surfaces embedded in $\mathbb{R}^3$ with constant Gaussian curvature be (a part of) surfaces of revolution? It seems that on the text book, the examples are only those surfaces of revolution if one talks about constant Gaussian curvature. 
Any counter-examples or proof? Thanks.
 A: Every "generalized cylinder" or "generalized cone" has Gaussian curvature identically zero. Most of these are not surfaces of rotation.
More interestingly, for each surface of rotation having constant Gaussian curvature, there is a one-parameter family of "helical" surfaces of constant Gaussian curvature, the parameter being the "pitch" of the helix. (The family arising this way from the pseudosphere is Dini's surface.)




There are also surfaces of constant negative Gaussian curvature having no ambient symmetries. This is physically clear if you imagine a hyperbolic patch made from a sheet of rubber or paper: The patch is "floppy", and most of its physical configurations cannot be "slid along themselves".
Edit: Parametric formulas for helical surfaces of constant curvature (joint work with J. M. Antonio, 2008, unpublished) are given in terms of differential equations, just as for surfaces of rotation.
Fix real numbers $k$, $C > 0$, and $B > 0$, and let
$$
G(u) = \begin{cases}
\phantom{\pm} B^{2} - u^{2} & K = 1, \\
\phantom{\pm} B^{2} + C^{2}u & K = 0, \\
\pm B^{2} + u^{2} & K = -1.
\end{cases}
$$
If we define
\begin{align*}
  h_{2}(u) &= \sqrt{G(u) - k^{2}}, \\
  h_{1}'(u) &= \sqrt{\frac{G(u)}{G(u) - k^{2}} \left[\frac{1}{G(u)} - \frac{G'(u)^{2}}{4(G(u) - k^{2})}\right]}, \\
  \psi'(u) &= -\frac{k}{G(u)}\, h_{1}'(u),
\end{align*}
then
$$
\mathbf{x}(u, v) = \left[
  \begin{array}{@{}c@{}}
    h_{1}(u) + k(v + \psi(u)) \\
    h_{2}(u) \cos(v + \psi(u)) \\
    h_{2}(u) \sin(v + \psi(u)) \\
  \end{array}\right]
$$
parametrizes a helical surface of constant Gaussian curvature in each interval where
$$
0 < G(u) - k^{2}\quad\text{and}\quad
\frac{G(G')^{2}}{4(G - k^{2})} \leq 1.
$$
The surface is immersed, but not necessarily embedded, and is not geodesically complete unless $k = 0$ and $G(u) = 1 - u^{2}$ (the sphere), or $C = 0$ in the flat case (a cylinder). Taking $B = 0$ in the negative-curvature case gives helical surfaces converging to the pseudosphere as $k \to 0$.
