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$$ [\text{size of boundary}] \times [\text{rate of motion of boundary}] = [\text{rate of change of size of bounded region}] $$ This differs from the fundamental theorem of calculus in that it does not mention antiderivatives and you can present it on the first day of a beginning calculus course, with concrete examples. It immediately appeals to the intuition of even relatively non-mathematically inclined students. I have a forthcoming paper on the use of this proposition in teaching.

My question is: Is there some standard name for this statement?

And does it appear in the literature somewhere other than in my paper?

Added in a later edit: Several answers say this is a special case of something else (just as I did above). But in each case the "something else" is far too abstruse to present on the first day of the freshman calculus course. So let's add a different question: What should this simple statement be called?

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The problem with an ellipse is different: if you want to keep your shape as an ellipse, different parts of the boundaries have to move at different speeds. –  JBL Oct 15 '10 at 22:45
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It seems like you ought to be able to make Green's Theorem (or, more generally, the Divergence Theorem) say something like this, since on one side of the equation is an integral over the boundary and on the other side is an integral over the region. –  Mike Spivey Oct 15 '10 at 23:43
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This works out as one-variable formulas for regular 3-D polyhedra if, instead of the edge length, the parameter used is the inradius, distance from the center of the polyhedron to the center of a face. So, for a cube, instead of $V = s^3, \; \; S = 6 s^2$ we are writing $ V = 8 r^3, \; \; S = 24 r^2.$ For the octahedron, $ V = 4 \sqrt 3 r^3, \; \; S = 12 \sqrt 3 r^2.$ Checking the other 3 and wondering about the truncated cube or truncated octahedron with constant inradius. For Archimedes' cylinder enclosing a sphere of radius $r,$ we get $ V = 2 \pi r^3, \; \; S = 6 \pi r^2.$ –  Will Jagy Oct 16 '10 at 2:39
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Hi, Michael. I keep a Latex file, if I have a long comment or elaborate answer I typeset it in between two \newpage commands. Once that previews correctly, I copy and paste into MO. I checked (easy) that the inradius version works for the regular $n$-sided polygon in the plane. I have a lovely book Platonic and Archimedean Solids by Daud Sutton. Willie's co-area formula trumps all other descriptions. But I am still curious about my inradius formulation for the rhombic dodecahedron, various polyhedra where the inradius is constant and each "radius" meets a regular polgon at its center. –  Will Jagy Oct 16 '10 at 19:55
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@Michael: Just copy the comment to the answer box to preview (but don’t submit the answer, of course). It is not 100% accurate (it seems that some of the MathJax/Markdown interference nuisance break in a different way in answers than in comments), but still it is more reliable and more handy than using actual LaTeX. –  Emil Jeřábek Aug 23 '11 at 14:17
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4 Answers 4

I believe what you wrote down is the differential form of the co-area formula. (Unfortunately, the version on Wikipedia is too "advanced" or "modern" so it is hard to see immediately that when you integrate both sides of your equation in time, you get precisely the co-area formula.)

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I think this is better than my original suggestion about using the Divergence Theorem (which I'm not so sure anymore will actually work). –  Mike Spivey Oct 16 '10 at 2:41
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Link to the Wikipedia article: en.wikipedia.org/wiki/Coarea_formula –  Hans Lundmark Oct 16 '10 at 13:59
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Well, this is quite enjoyable. In $\mathbf R^3,$ as long as we are taking polyhedra or other figures that are strictly similar, with some parameter $w$ to which any length measurement of the figure keeps a constant ratio, it follows that the volume is some $V_1 w^3$ and the surface area is some $ S_1 w^2.$ Furthermore, some choice of length scale or units causes $3 V_1 = S_1$ and we get a pure derivative law such as the earlier examples I gave in comments. For several very symmetric figures, the "inradius" or radius of an inscribed sphere gives such a fortunate relationship.

I thought it might depend critically on symmetry, but perhaps not as much as I thought. For the square pyramid with the square base on the $xy$-plane and one of the triangular faces within the plane $x + y + z = (1 + \sqrt 3 ) w,$ we get the center of the inscribed sphere at $(0,0,w), $ the point of tangency on that triangle $w(\frac{1}{\sqrt 3}, \; \frac{1}{\sqrt 3}, \; 1 + \frac{1}{\sqrt 3}), $ the length of any edge $ w (1 + \sqrt 3 ) \sqrt 2,$ the volume $ V = \frac{2}{3} (1 + \sqrt 3 )^3 w^3, $ finally the surface area $S = 2 (1 + \sqrt 3 )^3 w^2.$ Notice that the inscribed sphere does touch the square at its center but the triangles elsewhere.

I will fiddle with this some more. As I mentioned, there is either a truncated cube or a truncated octahedron which possesses an inscribed sphere that is tangent to all the faces, the amount of truncation must be just right. Interesting to check.

EDIT: it always works, symmetry is not important, and it is utterly trivial but cute. Take an irregular polyhedron that is convex, and is such that every face is tangent to the same sphere of radius $r.$ For each face, find the similar face that would be tangent to a sphere of radius 1, call that of area $A_j,$ so that the actual area of the original face $j$ is $A_j r^2.$ Because we use the same $r$ for all faces, the surface area is $$ S = \sum A_j r^2 = r^2 \sum A_j.$$ But the volume of a cone or pyramid is one third of the altitude times the area of the base, and the altitude is $r$ because of the tangency condition. So $$ V = \frac{1}{3} r^3 \sum A_j $$ and the derivative relationship holds.

Same for irregular polygons with all edges tangent to the same circle.

Who knew?

EDIT 2: To turn it around, the full set of polygons or polyhedra for which this recipe works comes from taking a finite set of distinct points on a circle or sphere, a sufficient condition for a closed bounded figure being that the set of points not be contained in any closed semicircle/hemisphere. For each point, take the tangent plane, and indeed the half space on the same side of that tangent as the sphere. The intersection of these half-spaces is the convex polygon/polyhedron. In the polyhedron case, it will not be obvious ahead of time how many sides each polygonal face will possess. Next, the example I called Archimedes' favorite cylinder can be thought of as a limit of regular $n$-sided prisms as $n$ increases.

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I second the idea that this, or something like this, is a version of Green's theorem. (For a family of simple closed curves c_t, anyway, a precise statement would replace "size of boundary times rate of motion of boundary" with the integral of the normal component of the flow velocity (ie the vector you get by differentiating c_t with respect to t, holding the curve variable fixed) over c_t. And the simplest--- maybe only--- way I can think of to prove that is to use Green's theorem to write the area enclosed by c_t as a contour integral.)

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It's also "a version" of the Fundamental Theorem of Calculus and of various other things, but it's so simple that freshmen can understand it on the first day of their calculus course. Is there a name for that simple freshman-friendly statement? –  Michael Hardy Oct 17 '10 at 17:39
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Various answers say this is this-or-that standard proposition that could not possibly be presented on the first day of a first-year calculus course, whereas the proposition as I stated it fits easily into that context.

Therefore I have coined a name: the boundary rule.

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