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A first stab at a definition of surface area might go like this:

Let S be a surface. Select finitely many points from S and make a bunch of triangles having these points as vertexes. Add up the areas of all of the triangles. This is a finite approximation to the surface area. Now to get the actual surface area, we increase the number of points so that they "densely cover" the surface (Maybe a good working definition would be that any open set of S should eventually contain some vertexes of triangles).

I seem to remember reading about a counterexample to this naive definition, but I can't find a reference. I believe there is even a natural looking polygonal approximation to the cylinder whose surface area diverges to infinity. Can anyone help me out?

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Are you really just randomly making triangles from a finite collection of points on the surface? One dimension lower, if you choose a finite collection of points on a curve and randomly make intervals out of them (ignoring the linear ordering along the curve), adding up the length of the interval, you almost always get a number very far from the length of the curve. Don't you want to triangulate the surface $S$? – Ryan Budney Mar 1 '12 at 21:14
I have a related question for anybody to answer: Is there any deep reason why this attempt at extending "rectifiability" fails when the Hausdorff dimension is greater than 1? – Christopher A. Wong Mar 1 '12 at 21:54
@Ryan: When I think about a "triangulation" of a surface, I usually think about the triangles living on the surface, but here the triangles are passing through the ambient space. So I didn't have a good word for it. You are right that my language is imprecise, but I hoped it would conjure the same mental image for my readers as it did for myself. – Steven Gubkin Mar 1 '12 at 22:17
@Christopher: Although requiring a sequence of polygonal approximations to converge pointwise to the surface isn't by itself enough to get the definition of surface area to work correctly, imposing a mild additional requirement, such as convergence of normals, does suffice. So IMO, dimension 1 is special only because the limited amount of wiggle room means that pointwise convergence automatically implies some stronger convergence properties, so that you don't have to specify those stronger conditions explicitly. – Timothy Chow Mar 1 '12 at 22:29
Your strategy, to succeed, should rely on an additional hypothesis: each triangle should ultimately be almost parallel of the tangent plan of any close point on the surface. – Benoît Kloeckner Mar 2 '12 at 10:34
up vote 25 down vote accepted

Perhaps you are thinking of the Schwartz Lantern? It converges to the cylinder in the Hausdorff metric but its area can be arranged to head toward $\infty$. It was mentioned in the earlier MO question, "Convergence of finite element method: counterexamples." There is nice applet here showing the lantern rotating. Here is an image from Conan Wu's blog:

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page 129, Calculus on Manifolds, by Michael Spivak. – Will Jagy Mar 1 '12 at 21:51
@Joseph Excellent! Thanks a lot. @Will Yes that is where I must have seen it. – Steven Gubkin Mar 1 '12 at 22:22

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