Consider surfaces where the Gauss curvature $K$ is a 'simplest' function, namely, $K$ is constant on the surface. For example, in a plane $K=0$ and in a sphere of radius $r$, $K=1/r^2$. Other surface with constant Gauss curvature is the cylinder, where $K=0$ again. If one want to obtain more examples, one may consider this problem in the family of surfaces of revolution. In such a case, the equation $K=c$ is an ordinary differential equation, and by the general theory, for each initial conditions, there is a solution. Thus this provides us a huge family of examples.
If the profile curve is a graph on the $x$-line, that is, $\alpha(x)=(x,0,f(x))$, then
$$K(X(x,\theta))=\frac{f'f''}{x(1+f'^2)^2}.$$
We know that when $K=0$, then $f''=0$, obtaining planes, circular cylinders and cones. On the other hand, the equation $$\frac{f'f''}{x(1+f'^2)^2}=c$$ is not possible to integrate completely and only some particular cases can solved: for example, the sphere and the pseudosphere.
However, if we assume that the surface is compact, then panorama changes drastically, because we have
Theorem (Hilbert): Sphere is the only compact surface with constant Gauss curvature.
We observe that the topological assumption on compactness is essential in the result. It is known that in a compact surface there exists elliptic points, so the value of the constant $K$ must be positive. Then the key is based in the following result:
Lemma (Hilbert): Let $S$ be a surface and $p\in S$ an elliptic point. Consider $\kappa_1\geq \kappa_2$ the principal curvatures on $S$. If $p$ is a global maximum for $\kappa_1$ and it is a global minimum for $\kappa_2$, then $p$ is an umbilical point.
The proof of the Hilbert's theorem is then as follows. Since $K=\kappa_1\kappa_2>0$, the sign of the principal curvature is the same. After a change of orientation, we suppose $\kappa_i>0$. Let $p\in S$ a point where the (continuous) function $\kappa_2$ attains a global minimum. Since $\kappa_1=\frac{c}{\kappa_2}$, then $p$ is a global maximum for $\kappa_1$. By the lemma, $\kappa_1(p)=\kappa_2(p)$. Then we have for any $x\in S$, $$\kappa_2(p)\leq\kappa_2(x)\leq\kappa_1(x)\leq\kappa_1(p)$$ and thus, $\kappa_1=\kappa_2$ on $S$. Then $S$ is an open of a plane or a sphere. Since $S$ is closed, by connectedness, $S$ is a plane or it is a sphere. But sphere is the only one that is compact.
