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One common definition of arclength is to just define it as a supremum of the set of lengths obtained by approximating your curve as a union of line segments (I was asked in the comments for a more precise definition; see https://en.wikipedia.org/wiki/Arc_length#Definition_for_a_smooth_curve). The natural analogue of this to the surface area of a surface in 3 space fails quite spectacularly thanks to constructions such as the Schwarz lantern, which shows we can approximate a cylinder by polyhedra whose surface areas approach infinity!

Is there an intuitive reason that polygonal approximation works so well for curves but fails so spectacularly for surfaces?

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    $\begingroup$ Can you carefully define the curve case? There is a classic example of approximating a circle from outside by recursively cutting corners out of a square, resulting in an apparent approximation to the perimeter of a circle of radius $1/2$ of $4$. I don't quite see how you are excluding this situation (or an analogous situation where you work from inside, perhaps) here. $\endgroup$
    – Ian
    Aug 25, 2019 at 2:10
  • $\begingroup$ Does this difficulty of triangulation arise only for Gauss curvature $K=0$ surfaces? I mean if $ K>0 $ or $ K<0 $ then, no matter how sub-divided it would not matter.. .is that correct? What requirement is to be implicitly obeyed in order that a given area integration be convergent? $\endgroup$
    – Narasimham
    Aug 25, 2019 at 3:28
  • $\begingroup$ static.nsta.org/pdfs/QuantumV1N4.pdf page 6. $\endgroup$ Aug 25, 2019 at 11:08
  • $\begingroup$ @Ian See en.wikipedia.org/wiki/Arc_length#Definition_for_a_smooth_curve. The counterexample of a circle you give is not problematic for this definition. $\endgroup$
    – user147556
    Aug 25, 2019 at 21:14

3 Answers 3

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I'd give an intuitive reason as follows: in the case of a smooth curve, if points $A$ and $B$ are on the curve, then line $AB$, as $B\to A$, tends to the tangent at $A$; while in the case of a smooth surface, if points $A$, $B$ and $C$ are on the surface, then plane $ABC$ needn't tend to the tangent plane at $A$ as $B,C\to A$: the limit depends on the way $B$ and $C$ approach $A$.

This is not surprising, after all, if you think how subtler limits in two or more dimensions are with respect to limits in one dimension.

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  • $\begingroup$ All the best.... $\endgroup$
    – Sebastiano
    Nov 11, 2020 at 23:50
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Even in the case of curves, convergence requires some care, as evidenced by the following meme:

enter image description here

In both the case of arclength and surface area, the key intuition is that it is not enough that your discrete geometry converges pointwise to the smooth one: both points and normals must converge under refinement for the length/area to converge. It's just that in the case of curves, sampling points on the curve and connecting them by segments is enough to guarantee convergence of normals as well, whereas as you point out, sampling points on the surface and connecting those point with triangles can have high error in the normals. (If you guarantee that your triangles stay "nicely shaped" (for example, intrinsic Delaunay) under refinement, then you can regain normal convergence and surface area convergence. Even in the case of the Schwarz lantern, you converge to the correct surface area if you refine at an equal rate in the axial and azimuthal directions.)

So I'd say that the curve case is just another of the many instances of getting "lucky" in low dimensions due to not having enough rope to hang yourself.

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    $\begingroup$ That's a bad example because the example of Schwarz tries to imitate the computing the length of a curve by inscribed polygons -- which doesn't happen in your example. $\endgroup$ Aug 28, 2019 at 7:56
  • $\begingroup$ Love that iterated square around circle and removing corners. It reminds me of something I used to give some of my classes. Draw a 45-degree line within a grid of horizontal and vertical lines. Use 100 times square-root(2) for the length, so line goes from (0, 0) to (100, 100). Now, approximate the line by going one right, one up in a zigzag. We have (0,0) to (0,1) to (1,1) to (2,1) to (2,2) ... ending at (100,100). Path is 200 long. Now repeat with movements 1/2 of a unit right, 1/2 up. Again, path is 200 long. Repeat. So, limit of path length (200) equals about 141.4 (100*root2), right? Ha! $\endgroup$ Aug 28, 2019 at 9:51
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    $\begingroup$ @MichaelHoppe My point is that normal-convergence is the key property, not being inscribed. In 1D it turns out that being inscribed luckily also implies normal-convergence. $\endgroup$
    – user7530
    Aug 28, 2019 at 15:31
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The increasing of the triangle-shaped surface of the polyhedra is due to the fact that those areas are very "creased", i.e., that those triangles are very "steep" compared to the tangent plane of the cylinder -- which cannot happen if we approximate a curve by polygons.

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  • $\begingroup$ Instead of "As two triangles" do you mean "Although two triangles"? And can an example of the "bad" definition of surface area be directly made from polyhedra illustrating Hilbert's third problem? I have not seen surface area examples described in this way. $\endgroup$
    – KCd
    Aug 27, 2019 at 23:22
  • $\begingroup$ Yes, "although" is what I meant. See my edit, please. $\endgroup$ Aug 28, 2019 at 7:43

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