Can someone help me understand how I can calculate the mean curvature of an infinitely long cylinder of radius $R$? I know the definition of mean curvature as

$H = \frac{1}{2}(\kappa_1 + \kappa_2)$

where $\kappa_i$ is the $i$th principal curvature. Since the cylinder is infinitely long, I think $\kappa_2 = 0$ (along the axis). Can someone confirm this?

Then the mean curvature of an infinitely long cylinder with radius $R$ would be simply

$H = \frac{1}{2R}$

Thank you for your help!

  • $\begingroup$ Principal curvature. Yes, you are correct. The tangent plane stays constant along the rulings, so $\kappa_2=0$. $\endgroup$ Commented Aug 17, 2020 at 23:58
  • $\begingroup$ $\kappa_2=0$, be it an infinite cylinder or not. $\endgroup$
    – lisyarus
    Commented Aug 18, 2020 at 4:20

1 Answer 1


Your result is correct.

Pick a point on the cylindrical surface. You must convince yourself that one principal direction is perpendicular to the axis of the cylinder (but starting from your point of the surface). Along this direction, the surface looks like a circle with radius $R$, so the principal curvature for this direction is $\kappa_1=\frac1R$. The other principal direction is parallel to the cylinder axis, and along this direction, the surfaces looks (locally near your point) like a straight line, so $\kappa_2=0$. So from the formula $H=\frac12 (\kappa_1+\kappa_2)$ you get the mean curvature you mention.

As TonyK says, this is the same for any point you pick. So if you regard $H$ as a function, mapping each point on the surface to a real number, then $H$ is constant, for the cylindrical surface.

As we see, mean curvature is a local property, so it does not matter if the cylinder is infinitely long or not; as long as there is a neighborhood around the point you consider, where the surface is a cylinder, then the mean curvature at that point is $\frac1{2R}$.


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