Math 2.0

Differential geometry Minimal and constant mean curvature surfaces Example

Catenoid

Example

For u∈ℝu \in \R and v∈ℝv \in \R, let x(u,v)=(cosh⁡vcos⁡u,  cosh⁡vsin⁡u,  v).x(u, v) = (\cosh v \cos u, \; \cosh v \sin u, \; v). Its image, the catenoid, is the surface of revolution obtained by rotating the catenary r=cosh⁡zr = \cosh z about the zz-axis. It is a minimal surface.

Check. The partial derivatives are xu=(−cosh⁡vsin⁡u,cosh⁡vcos⁡u,0)x_u = (-\cosh v \sin u, \cosh v \cos u, 0) and xv=(sinh⁡vcos⁡u,sinh⁡vsin⁡u,1)x_v = (\sinh v \cos u, \sinh v \sin u, 1). So E=cosh⁡2vE = \cosh^2 v, F=0F = 0 and G=sinh⁡2v+1=cosh⁡2vG = \sinh^2 v + 1 = \cosh^2 v, and the parametrization is conformal. For a conformal parametrization, xuu+xvv=2EHNx_{uu} + x_{vv} = 2 E H N, where HH is the mean curvature and NN the unit normal. Here xuu=(−cosh⁡vcos⁡u,−cosh⁡vsin⁡u,0)x_{uu} = (-\cosh v \cos u, -\cosh v \sin u, 0) and xvv=(cosh⁡vcos⁡u,cosh⁡vsin⁡u,0)x_{vv} = (\cosh v \cos u, \cosh v \sin u, 0), so xuu+xvv=0x_{uu} + x_{vv} = 0 and H=0H = 0.

Remarks

The principal curvatures are ±1/cosh⁡2v\pm 1/\cosh^2 v, and the Gaussian curvature is K=−1/cosh⁡4vK = -1/\cosh^4 v. Apart from pieces of planes, the catenoids are the only minimal surfaces of revolution. The catenoid is locally isometric to the helicoid, and one can be bent into the other through a family of minimal surfaces.

Further reading

  1. Web https://en.wikipedia.org/wiki/Catenoid
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Changed2 November 2026

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