Hilbert proved that there's no complete regular (C^k for sufficiently large k) isometric embedding of the hyperbolic plane into R^3. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or C^2, say) isometrically embedded in R^3?

edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region *PS*={z | Im z ≥ 1, -π < Re z ≤ π} on the upper half plane model of H^2. Let z=x+iy, so that ordered pairs (x,y)∈ H^2 when y>0.

Next, Euclidean circles drawn on in the upper-half plane model with center (x,y\cosh r) and radius y\sinh r correspond to hyperbolic circles with center (x,y) and radius r. I can fit a Euclidean circle of radius π centered at (0,1+π) into the region PS. This corresponds to a hyperbolic disk of radius arctanh(π/(1+π)) ~ 0.993.

Surely one can do better?

edit2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures: