TITLE: Revolution surfaces of constant Gaussian curvature
QUESTION [5 upvotes]: I'd like help with the following question:
Prove that all revolution surfaces $(\phi(v) \cos u ,\phi(v) \sin u,\psi(v)) $ of constant Gaussian curvature $k = -1$ is one of the following types:

$\phi(v)=C\cosh v$ and $\psi(v)=\int_0^v \sqrt{1-C^2\sinh^2v} dv$
$\phi(v)=C\sinh v$ and $\psi(v)=\int_0^v \sqrt{1-C^2\cosh^2v} dv$
$\phi(v)=e^v$ and $\psi(v)=\int_0^v \sqrt{1-e^{2v} dv}$

Suppose $(\phi')^2+(\psi')^2=1$ and you know that $\phi''+k\phi= 0$
thanks

REPLY [6 votes]: Using the usual Gaussian curvature formula on the form of the surface of revolution you presented yields the expression
$$\frac{\psi^\prime (v)(\psi^{\prime\prime}(v) \phi^\prime (v)-\psi^\prime (v)\phi^{\prime\prime}(v))}{\phi(v)\left(\psi^\prime (v)^2+\phi^\prime (v)^2\right)^2}$$
With one of your assumptions, this simplifies to
$$\frac{\psi^\prime (v)}{\phi(v)}(\psi^{\prime\prime}(v) \phi^\prime (v)-\psi^\prime (v)\phi^{\prime\prime}(v))=-1$$
You already said you know that $\phi$ satisfies $\phi^{\prime\prime}+k\phi=0$; solve that differential equation and substitute that differential equation's solution(s) into the differential equation you've obtained from the Gaussian curvature expression.