Let us call a *cap* the intersection of the boundary of 3-dimensional convex compact set $K$ in $\mathbb{R}^3$ with a half-space bounded by a plane $H$ such that the orthogonal projection to $H$ of this intersection is contained in $K\cap H$.


**QUESTION.** Given a metric on a closed 2-dimensional disk which has non-negative curvature in the sense of Alexandrov. Can the disk be isometrically imbedded into $\mathbb{R}^3$ as a cap?


A reference would be helpful.



Isometric imbedding of a 2-disk into Euclidean 3-space
Take doubling of the disc, we obtain a metric on the sphere.
By Perelman's theorem it had nonnegative curvature in the sense of Alexandrov.
Therefore, by Alexandrov's theorem, it is isometric to a convex surface in the Euclidean space.
This convex surface is unique up to congruence (Pogorelov's theorem). Therefore, the involution of our sphere extends to a reflection of the space.
In particular, the boundary of the disc lies in one plane.


**Postscript.** You may also proceed as suggested by Joseph O'Rourke --- approximate the disc by polyhedral space, apply Alexandrov's theorem for polyhedral surfaces (which has uniqueness for free) and pass to the limit.


