Why is the half-torus rigid?

The half-torus surface that results from slicing a torus like a bagel, depicted below (left), is isometrically rigid. I know this from a remark of Alexandrov in Mathematics: Its Content, Methods and Meaning (Chapter 7. Curves and surfaces, p.101): For example, it has been shown that a surface in the form of a circular trough (...), does not admit continuous deformations (this explains, among other things, the familiar fact that a pail with a curved rim is considerably stronger than one with a plain rim) ... He gives no hint of a proof, nor a reference. Could someone supply either? I would like to understand this enough to generalize to, e.g., the bottom quarter of a torus (above, right), or to cross-sections or sweep curves other than circles (ellipses, smooth convex curves, ...). Are there general conditions known for a surface in $\mathbb{R}^3$ with boundary to be isometrically rigid, i.e., to admit no continuous deformation that "preserves the length of all curves on the surface" (to quote Alexandrov's definition)?