[Hilbert proved](http://en.wikipedia.org/wiki/Hilbert%27s_theorem_%28differential_geometry%29) that there's no complete regular ($C^k$ for sufficiently large $k$) isometric embedding of the hyperbolic plane into $\mathbb{R}^3$.  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 $\mathbb{R}^3$?

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 $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

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


[![disk in the pseudosphere in the upper half plane model][1]][1]
[![disk on the embedded pseudosphere][2]][2]


  [1]: https://i.sstatic.net/Bknxc.png
  [2]: https://i.sstatic.net/p1Wd4.png