What is the Surface of Revolution with Constant Curvature +1?

[mooshasta]

Homework Statement

I'm trying to find a surface of revolution with Gauss curvature K of +1 at all points, which doesn't lie in a sphere.

Homework Equations

The surface is parametrized as \psi (t, \theta ) = ( x(t), y(t) cos \theta , y(t) sin \theta )

I have the equation
<br /> K = \frac{x&#039; (x&#039;&#039; y&#039; - x&#039; y&#039;&#039;)}{y(x&#039;^2 + y&#039;^2)^2}<br />

The Attempt at a Solution

I am thinking it has to do with the curve \alpha (t) = (x(t),y(t)) not having unit speed, but I am kind of stuck as to where to go from there.


Thanks!

 

[Dick]

Is there one? The only way out I can think of is to make it disconnected. A surface of constant guassian curvature is locally isometric to a sphere.

 

[mooshasta]

I know that for a unit-speed \alpha (t), the equation reduces to K = \frac{-y&#039;&#039;}{y}, which does clearly represent a sphere.

The way the question is worded on my homework seems to point to the fact that that reduction only applies to unit-speed curves, which is why I think that perhaps an \alpha (t) that doesn't have unit-speed perhaps can give a surface of revolution with constant curvature +1 that isn't a sphere... but maybe there's another "gimmick" that I'm overlooking...

Anyways, thanks for your help!