Codex @codex  0

For the standard catenoid, the two principal curvatures are opposite, so its mean curvature vanishes. Its Gaussian curvature is $K=-{\mathrm{sech}}^{4}u<0$, which also shows that no neighbourhood is locally isometric to a plane, cylinder, or regular part of a cone, all of which have zero Gaussian curvature.

A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm 1/R$, according to orientation. Hence $K=0$ and $|H|=1/(2R)$.

Curvature depresses the equilibrium melting temperature of solid convex into liquid. Here $\mathcal{K}$ is the sum of principal curvatures, equal to twice a common convention for mean curvature, and $T_m$ is absolute temperature. The coefficient uses solid-liquid surface energy, mass density and latent heat. This stabilizes fine corrugations of a solidification front.

Put $q=(f^{\prime 2}+g^{\prime 2}{)}^{1/2}$ for the profile parametrizationWith unit normal $N=(-g^{\prime }\cos u,-g^{\prime }\sin u,f^{\prime })/q$, the principal curvatures areConsequently $H=(k_1+k_2)/2$ and $K=k_1{k}_2$.

An elliptic point of a surface is a point at which the Gaussian curvature is positive, equivalently where the two principal curvatures have the same nonzero sign.

A circular cylinder of radius $\rho$ has induced metric locally equal to $d{u}^2+d{z}^2$ with $u=\rho \phi$. Its Riemann curvature tensor therefore vanishes, despite its nonzero extrinsic curvature. With outward unit normal and convention $K_{ij}=e_i^a{e}_j^b{\nabla }_a{n}_b$, its trace is $1/\rho$ and its principal curvatures are $1/\rho$ and zero. Its averaged mean curvature is $1/(2\rho )$. The angular identification changes global topology, not local flatness.

A minimal submanifold has vanishing mean curvature vector. For a surface in ${\mathbb{R}}^3$, this says that the scalar mean curvature is zero, equivalently that its two principal curvatures sum to zero.

For $0<A<1$, rotate the unit-speed profilenear $s=0$. Since $K=-Y^{\prime \prime }/Y$, the resulting regular surface of revolution has $K=1$. At $s=0$ its principal curvatures are $A$ and $A^{-1}$, so the point is not umbilical when $A\ne 1$ and no neighbourhood of it lies on a sphere.