most recent 30 from http://mathoverflow.net 2013-05-26T07:11:02Z http://mathoverflow.net/feeds/question/89991 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89991/naive-definition-of-surface-area-doesnt-work Steven Gubkin 2012-03-01T21:01:17Z 2012-05-04T14:58:02Z

<p>A first stab at a definition of surface area might go like this:</p> <p>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).</p> <p>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?</p>

http://mathoverflow.net/questions/89991/naive-definition-of-surface-area-doesnt-work/89994#89994 Joseph O'Rourke 2012-03-01T21:26:29Z 2012-03-01T21:26:29Z

<p>Perhaps you are thinking of the <em>Schwartz Lantern</em>? 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, "<a href="http://mathoverflow.net/questions/75255/" rel="nofollow">Convergence of finite element method: counterexamples</a>." There is nice applet <a href="http://www.cut-the-knot.org/Curriculum/Calculus/SchwarzLantern.shtml" rel="nofollow">here</a> showing the lantern rotating. Here is an image from <a href="http://conan777.wordpress.com/2011/11/28/the-schwartz-lantern/" rel="nofollow">Conan Wu's blog</a>: <br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <img src="http://conan777.files.wordpress.com/2011/11/lantern-2.jpg" height="300" /></p>