**eklhad@ihnet.UUCP (K. A. Dahlke)** (11/19/84)

< sacrifice > Here is a question which some of you have probably heard before. What 3-dimensional shape has a finite volume, but an infinite surface area? You can fill it up with paint, but you can't paint it. I will give the answer in a week if it has not been spoiled. The second part is harder. Prove that no shape has a finite surface area and an infinite volume. I might give the answer to this part too, if I get the chance to think about it during turkey (chomp chomp). Enjoy continuously! -- Karl Dahlke ihnp4!ihnet!eklhad

**jfh@browngr.UUCP (John "Spike" Hughes)** (11/20/84)

SPOILER (sort of) **************************************************** The answer to the second part is that it's incorrect: the subset of space consisting of all points whose distance from the origin is greater than one is an example: its volume is infinite, but its surface area is 4 pi. By the way, the first problem seems to be a paradox unitl you consider the following algorithm for painting the plane: take 1 gallon of paint. Use the first half gallon to paint all the points whose distance from the origin is less than one. Now use the next quarter gallon to paint the points whose distance form the origin is between one and two. Now continue in this manner (of course the paint is much thinner as you move away from the origin). Nonetheless, you see that you can 'paint' and infinite area with a finite quantity of paint. This is just what happens when you fill the answer to part 1 with paint...

**pws@faron.UUCP (Phillip W. Servita)** (11/21/84)

In article <ihnet.176> eklhad@ihnet.UUCP (K. A. Dahlke) writes: >Here is a question which some of you have probably heard before. >What 3-dimensional shape has a finite volume, but an infinite >surface area? >You can fill it up with paint, but you can't paint it. >I will give the answer in a week if it has not been spoiled. >The second part is harder. >Prove that no shape has a finite surface area >and an infinite volume. >I might give the answer to this part too, >if I get the chance to think about it during turkey (chomp chomp). >Enjoy continuously! > >Karl Dahlke ihnp4!ihnet!eklhad as for the first part, consider the curve y = 1/x (x >= 1) rotate this around the x-axis. the resulting shape has volume PI, but its surface area integral: /~\ infinity | | | (1 + 1/x^4)^.5 diverges by inspection. (always 2*PI | ---------------- dx below the curve y = 1/x) | x | \_/ 1 BUT, neither you, nor i, nor the Pentagon could possibly keep one of these things in their bedroom. so here is something a bit smaller: construct a "cylinder" whose cross section is a snowflake curve. such a cylinder can be constructed so as to fit inside of a normal cylinder of arbitrarily small diameter and height, but STILL would have infinite surface area. happy munching, -phil

**hes@ecsvax.UUCP** (11/21/84)

I have heard this one before - you can't paint it, but can fill it with paint, pour out the paint and thus it is painted. My only problem is the amount of time it takes to fill it with a finite amount of paint. (Ref: FTL discussion in net.physics.) --henry schaffer north carolina state university

**eklhad@ihnet.UUCP (K. A. Dahlke)** (11/25/84)

Thanks for the many responses to "strange shapes". If you were not familiar with infinite-area finite-volume shapes, here is one example. Take the graph y = 1/x and rotate it around the x axis. Let x run from 1 to infinity, and close the shape at x = 1 with a unit disc. This hyperbola of revolution works because volume drops as the cube of R, and surface area drops as the square of R. The difference is just enough to produce a finite volume and an infinite surface area. Try the integrals yourself. One interesting response stated the shapes need not be unbounded. A snowflake (fractle) kind of shape might work. I guess I never said the shapes had to be smooth or continuous. I am a little skeptical, and very interested. Can anyone give an analytic definition for a bounded shape with these properties. As for the converse, I was a bit embarrassed by the appropriate response "all points whose distance from the origin >= 1" This shape indeed has infinite volume and finite surface area, but it wasn't quite what I had in mind. I better learn to phrase questions unambiguously. I wanted the infinite volume to remain inside the shape. Now define inside (aargggg). Anyways, I didn't think much about this part, too busy eating. I guess one could flatten non-convex portions, decreasing area and increasing volume. The shape must then be convex, and therefore bounded or infinite, and... (wave wave wave, it's intuitively obvious, won't bore you with details). -- Karl Dahlke ihnp4!ihnet!eklhad

**gjk@talcott.UUCP (Greg J Kuperberg)** (11/26/84)

> I have heard this one before - you can't paint it, but can > fill it with paint, pour out the paint and thus it is painted. > My only problem is the amount of time it takes to fill it with > a finite amount of paint. (Ref: FTL discussion in net.physics.) > --henry schaffer north carolina state university A little thought will give a better answer: Painting an object means putting a coat of paint *of uniform thickness* on the object. All objects with finite volume and infinite surface area have arbitrarily small cracks and crevices where a coat of paint "does not fit". As someone aptly pointed out, an example of such an object is a snowflake. Try putting a coat of paint on a snowflake.

**gwyn@brl-tgr.ARPA (Doug Gwyn <gwyn>)** (11/26/84)

> Painting an object means putting a coat of paint *of uniform thickness* on > the object. All objects with finite volume and infinite surface area have > arbitrarily small cracks and crevices where a coat of paint "does not fit". > > As someone aptly pointed out, an example of such an object is a snowflake. > Try putting a coat of paint on a snowflake. "On" an object in practice means "within cohesive distance" of the object. There are always tiny cracks in practice but they are covered by paint. This means that it is easy to paint a snowflake; just immerse it. And there you have the difference between a physicist and a mathematician. This is not as facetious as it sounds. If you admit conventional concepts of infinity, set theory, and the like, then you support such paradoxes as the Banach-Tarski dismantling of a sphere into a few congruent parts that can be reassembled into a smaller sphere, and so forth. There is more to reality than that, since in practice no real sphere can be so rearranged.

**jhputtick@watrose.UUCP (James Puttick)** (11/26/84)

The reply about "...all points whose distance is >= 1 unit from the origin" is not what is intended, as has been noted. However, note also that, if you are given a surface which encloses a finite volume, then there is an infinite volume also enclosed by that surface, which is the "outside". Any ideas on how to define the problem so that there is no ambiguity? My only idea is to propose a different problem: find a surface of finite area which has both "inside" and "outside" of infinite volume -- and these two volumes must be distinct. My hunch is that no such surface exists...but I don't want to go hunting for it in any case!

**gjk@talcott.UUCP (Greg J Kuperberg)** (11/28/84)

> The reply about "...all points whose distance is >= 1 unit from the origin" > is not what is intended, as has been noted. However, note also that, if > you are given a surface which encloses a finite volume, then there is > an infinite volume also enclosed by that surface, which is the "outside". > Any ideas on how to define the problem so that there is no ambiguity? > My only idea is to propose a different problem: find a surface of finite > area which has both "inside" and "outside" of infinite volume -- and these > two volumes must be distinct. > My hunch is that no such surface exists...but I don't want to go hunting > for it in any case! How about this: Find a two-dimensional surface with finite surface area which separates three-dimensional space into two regions, each of which have infinite volume. --- Greg Kuperberg harvard!talcott!gjk "Eureka!" -Archimedes

**jfh@browngr.UUCP (John "Spike" Hughes)** (11/28/84)

There is a nice picture of a bounded 'shape' with infinite volume inn the book "Calculus on Mandifolds", by Michael Spivak, pub. Benjamin; the shape is best described as a failure to extend the usual definitionn of arclength (sorry about that bouncy 'n' key): one defines arclength by subdividing with points, and then taking the polygonal curve that passes through the points in the same order as the curve did. Then one computes the length of the polygonal path, and takes a limit as the "gap" between points goes to zero. If you try to do the same thing for surface area, eg for a cylinder, by inscribing polygos within it, it turns out that the limit need not exist: the polygons can get arbitrarily small without the sum of their areas tending toward a constant. In fact one can make this sum-of-areas go to infinity. There are more details (I think) in Spivak's Comprehensive Introduction to Differential Geometry, Vol. 1 or 2, Publish or Perish Press.

**jfh@browngr.UUCP (John "Spike" Hughes)** (11/28/84)

I think that the proposal "Find a two-dimensional surface with finite surface area which separates 3-space into two regions, each of which has infinite volume" has some merit. Shall we assume that "2-dimensional surface" means a smooth or polygonal 2-dimensional manifold (i.e. a set that is locally homeomorhic to the plane)? Or is 'dimension' to be measured in some more abstract way? I propose that, initially at least, we restrict to sets which are 'parametric' surfaces, i.e. are the images of nice functions from 2-space to 3-space. (e.g. the sphere is the image of (x, y) -----> (cos x cos y, cos x sin y, sin x), where x goes from -pi/2 to pi/2, and y goes from 0 to 2 pi). I suspect that in this case, the answer is that there is no such surface.

**wpt@fisher.UUCP (Bill Thurston)** (11/29/84)

>> Painting an object means putting a coat of paint *of uniform thickness* on >> the object. All objects with finite volume and infinite surface area have >> arbitrarily small cracks and crevices where a coat of paint "does not fit". >"On" an object in practice means "within cohesive distance" of the object. >There are always tiny cracks in practice but they are covered by paint. >This means that it is easy to paint a snowflake; just immerse it. >And there you have the difference between a physicist and a mathematician. >This is not as facetious as it sounds. If you admit conventional concepts >of infinity, set theory, and the like, then you support such paradoxes as >the Banach-Tarski dismantling of a sphere into a few congruent parts that >can be reassembled into a smaller sphere, and so forth. There is more to >reality than that, since in practice no real sphere can be so rearranged. There is a huge difference between the kind of infinity in a snowflake curve, which serves as a good model for many things in nature, and the transfinite processes used to derive the Banach-Tarski paradox. Certainly, no actual physical model of a cylinder with a snowflake cross-section can exist, because of quantization phenomena at small scales. Even if it existed, there it could be painted by dipping it in a bucket of paint, but the paint would obscure the fine details. Nonetheless, there is a real interpretation to the idea that such a cylinder would have infinite surface area, and would require infinitely much paint -- in units of (volume of paint / thickness of the coat). The physical interpretation is that as the thickness of the coat goes down, and as the accuracy of the model correspondingly increases, this ratio goes to infinity: in fact, it obeys a very regular scaling law, increasing as a certain inverse power of the thickness (which depends on the fractional dimension of the snowflake curve, in a certain sense.) The Banach-Tarski paradox is red herring in this discussion. Bill Thurston, Mathematics Dept, Princeton University {princeton , allegra }!fisher!wpt

**dmrp@ubu.UUCP (David Park)** (11/30/84)

Am I being simple-minded, or is there not a surface of maximal volume with a given finite surface area -- a sphere ? So there is no surface with finite area containing infinite volume. Or is there some subtlety I missed ?

**jfh@browngr.UUCP (John "Spike" Hughes)** (12/02/84)

David Park asks: Isn't the surface of maximal volume with a given finite surface areas always a sphere? I expect so, but I'd like to see a proof. It's certainly true if the bounded region is a subset of a finite ball in 3-space (I mean the conjecture about finite area => finite volume), but I don't see any self-evident reason that it should be true for regions extending to infinity. For some cogent remarks on 'obvious' statements, see 'Proofs and Refutations', by Imre Lakatos. It's a fascinating book on deduction in mathematics... -jfh

**gjk@talcott.UUCP (Greg J Kuperberg)** (12/03/84)

> David Park asks: > Isn't the surface of maximal volume with a given finite surface areas always > a sphere? > I expect so, but I'd like to see a proof. It's certainly true if the bounded > region is a subset of a finite ball in 3-space (I mean the conjecture about > finite area => finite volume), but I don't see any self-evident reason that > it should be true for regions extending to infinity. This is a difficult theorem, even in the case surfaces that are contained in a finite-sized polyhedron. --- Greg Kuperberg harvard!talcott!gjk "Madam, there is only one important question facing us, and that is the question whether the white race will survive." -Leonid Breshnev, speaking to Margaret Thatcher.

**bs@faron.UUCP (Robert D. Silverman)** (12/13/84)

To prove that a sphere is the 3-manifold of fixed area that holds the largest volume is trivial. It is a simple iso-perimetric problem in the Calculus of Variations. See for example "Lectures on the Calculus of Variations", Bolza, Oskar, Dover Press, chapter 6. It amounts to finding a function G(x,y, x', y') such that the double integral over G is fixed and for which the volume integral is maximal.

**jlg@lanl.ARPA** (12/18/84)

> > To prove that a sphere is the 3-manifold of fixed area that > holds the largest volume is trivial. It is a simple iso-perimetric > problem in the Calculus of Variations. See for example "Lectures on the > Calculus of Variations", Bolza, Oskar, Dover Press, chapter 6. > > It amounts to finding a function G(x,y, x', y') such that > the double integral over G is fixed and for which the volume integral > is maximal. This assumes that the surface in question is continuously differentiable. Or at least that the differential has a finite number of discontinuities. If a solution to the stated problem exists, it maybe nowhere differentiable, in which case the above mentioned proof is irrelevant. (How do you compute the area of a surface that is nowhere differentiable? I suspect it depends upon the method used to construct the surface. I don't know.) Actually, I don't think there is a finite surface that 'encloses' infinite volume. And I think that a sphere probably does enclose the maximal volume for a given area. But there have been no proofs yet.

**colonel@gloria.UUCP (George Sicherman)** (12/20/84)

[_Black Hole_] > Actually, I don't think there is a finite surface that 'encloses' infinite > volume. And I think that a sphere probably does enclose the maximal > volume for a given area. But there have been no proofs yet. A finite surface that encloses an infinite volume? How about a Klein Bottle? If you insist on an orientable surface, see Kellogg on Potential Theory. His proof of Gauss's theorem will give you some ideas about what restrictions are necessary. -- Col. G. L. Sicherman ...seismo!rochester!rocksanne!rocksvax!sunybcs!gloria!colonel

**biep@klipper.UUCP (J. A. "Biep" Durieux)** (01/02/85)

>> Actually, I don't think there is a finite surface that 'encloses' infinite >> volume. And I think that a sphere probably does enclose the maximal >> volume for a given area. But there have been no proofs yet. In article <734@gloria.UUCP> colonel@gloria.UUCP (George Sicherman) writes: >A finite surface that encloses an infinite volume? How about a Klein Bottle? >If you insist on an orientable surface, see Kellogg on Potential Theory. His >proof of Gauss's theorem will give you some ideas about what restrictions >are necessary. > Col. G. L. Sicherman [] What was wrong with the idea of a sphere with negative orientation, so as to enclose the whole "outer world", only leaving the "inside" (actually it's outside now!) unenclosed? -- Biep. {seismo|decvax|philabs}!mcvax!vu44!botter!klipper!biep I utterly disagree with everything you are saying, but I am prepared to fight to the death for your right to say it. --Voltaire