surface in $R^3$ that has $ds^2 = du^2/v^2 + dv^2/v^2$

Question

For a 2D surface, if we have the first fundamental form of $$ ds^2 = du^2/v^2 + dv^2/v^2$$, can we integrate it out to get the parameter form of the surface embedded in $R^3$?

I tried something like $(x,y,z)=(\cosh(u)\cos(v),\cosh(u)\sin(v),\sinh(v))$ , which has a similar diagonal form of metric tensor, but I could not make one that has $g_{11} =g_{22} = 1/v^2$.

Answer 1

The pseudosphere is such a surface. However this can only be done locally. The global metric in the upperhalfplane cannot be embedded in 3-space by a difficult old result of Bernstein.