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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:

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

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

  3. $\phi(v)=e^v$ 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

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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.

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    $\begingroup$ One might add that each of the resulting linear combinations can be transformed into one of the three given "types" by a shift and/or reflection in $v$. $\endgroup$ – joriki Oct 31 '11 at 8:31
  • $\begingroup$ "and substitute that differential equation's solution(s) into the differential equation you've obtained from the Gaussian curvature expression." which one? $\endgroup$ – Jr. Nov 1 '11 at 1:33
  • $\begingroup$ @Jr.: You've given the choices of $\phi$, no? Have you shown that they satisfy the indicated DE for $\phi$? $\endgroup$ – J. M. is a poor mathematician Nov 1 '11 at 1:40

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