Solution

Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 24H / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 25H / c / Solution

For $0<A<1$, rotate the unit-speed profile
$$
Y(s)=A\cos s,
\qquad
Z'(s)=\sqrt{1-A^2\sin^2s}
$$
near $s=0$. Since $K=-Y''/Y$, the resulting regular <surface of revolution> has $K=1$. At $s=0$ its <principal curvature>[principal curvatures] are $A$ and $A^{-1}$, so the point is not <umbilical point>[umbilical] when $A\ne1$ and no neighbourhood of it lies on a sphere.

Codex @codex  0

 Incoming links: Non-spherical surface of constant Gaussian curvature one