Is there any shape (of 3 dimensions) with infinite volume that does not have an infinite surface area? At start, the negation of this question seems simple enough, almost axiomatic. It's near impossible to invision a shape with a finite surface area having bundled in it an infinite volume. But, then again, it's near impossible to invision a region of infinite area being rotated to form a finite volume. Has any work ever been done on this subject? It seems intuitive, but a proof ends up being very hard, even for revolved shapes.

Brad

Interesting... Maybe it could have something to do with fractals, but in 3 dimensions?

Olof

The thing with an infinite area being rotated to form a finite volume - isn't it similar to the idea of an integral having a finite value even when it's limits are taken from 0 to infinity ?

Olof

Brad,
i am trying to get to the answer by using these formulae,

V=pò_a^b y² dx
So far its not so good,so i am letting you know if you could get somewhere using this formulae.

love arun

That's the idea I had, but after about 10 minutes of manipulating equations to come up with a result that covers 1 example out of a billion, I decided another appraoch was needed. I think I have an idea for volumes of revolution, though:

dM=2py dx

then dM < dA (for A being surface area). You can convince yourself of this by drawing a picture. Also if V is volume,

dV=py² dx

Then for all y < 2, dV < dM < dA. So, for all regions in which y is under 2, the area will always be greater than the volume. Now here comes the tricky part we'll do a proof by contradiction. We know that for a volume of revolution, for voume to be infinite, the parameters must have an infinite difference. Without loss of generality, suppose we had a function f(x) always > 0, such that

ò₀^¥ f(x)² dx=z

for finite z, also

ò₀^¥ f(x)(1+f ' (x)²)^1/2 dx=¥

=ò₀^¥(f(x)²+f(x)² f ' (x)²)^1/2 dx

We've already shown that for f(x) < 2, A > V, and we also know that for the general case of when F(x) > 1, for any F(x), òF(x)² dx > òF(x) dx. Also note that when f(x) > 2, (f(x)²+f(x)² f ' (x)²)^1/2 > 1. Thus,

ò₀^¥ (f(x)²+f(x)² f ' (x)²)^1/2 dx < ò₀^¥(f(x)² +f(x)² f ' (x)² dx

Therefore,

ò₀^¥ f(x)²+f(x)² f ' (x)² dx=¥

As ò₀^¥ f(x)² dx is finite, we may subtract it to get

ò₀^¥ f(x)² f ' (x)² dx=¥

This requires (remembering the integral of f(x)² is finite)

f(x)² f ' (x)² > f(x)²

So

f ' (x)² > 1

Taking the square root and integrating,

f(x) > x

That means that

ò₀^¥ f(x)² dx > ò₀^¥ x² dx

Thus ò₀^¥ f(x)² dx=¥ contradicting the given that it is finite.
This really isn't very rigorous I don't think, and aside from that, it still leaves the far wider case of figures that aren't composed by circles unanswered. Any more ideas?

Brad

Wouldn't such a shape violate the 3-d isoperimetric inequality rather spectacularly? (This result states that among all shapes of the same volume, the sphere has the least surface area.)

Of course the proof of the said inequality probably doesn't cover certain perverse shapes. Does anyone actually know how it is proved?

David

Ah, of course! The statement "given a certain surface area, the figure with the most volume is the sphere" can be be proven through the calculus of variations, I'm sure. I've never seen a proof, though I'll try to find one. Of course, in proving this with the calculus of variations, we are assuming the the surface is continuous and differentiatable, but I bet someone has done a proof of this for non-differentiatable curves.

Brad

As far as I know it is true that for all shapes which have both a volume and a surface area (measure theoretically, that is) a sphere has the maximum volume for fixed surface area. Given that, it would be impossible to construct a shape with infinite volume and finite surface area. I don't know how to prove this theorem for the most general sort of shape. I could probably prove it for shapes bounded by a differentiable surface using the calculus of variations (there's a thread in Asked NRICH about this here . I think it would be quite easy to extend the proof to shapes which are non-differentiable at isolated points, or maybe even more general shapes than that, but I don't know how to do it in full generality. Anyone?

Actually, now that I think about it, the original proposition I made follows merely from the statement, given a certain surface area, there is a maximum volume, and the ratio between the surface area and volume is finite in this shape . Before doing a proof with variations, I think we'd have to assume this; i.e. before performing a maximization, we are assuming that there is a maximum. It's a fair assumption, but it's equivalent to assuming what's in my original question. Any ideas on how to prove the italicised part rigorously?

P.S. It's rather trivial now, but my proof can be made rigorous if one breaks the integral up into a section from 0 to a, in which f(x) is greater than 1, and from a to infinity in which it is less, or vice versa, or break it up into a hundred segments of the like if the curve oscillates, and so on.

Brad

BTW, a quick search on the internet will reveal that the proof that the sphere has maximum volume was proved in 1884, by H.A. Schwarz (what a quick search will not reveal is the actual proof). Such a recent date suggests that the proof might be either very difficult (or that mathematicians simply didn't care about the problem).

Hmm. Can I ask how you are defining the surface area? For instance having a space that is actually all space, then it has no edge, hence no surface area..... Just think about inifinities for a moment, and try and find and nice definition for surface area that excludes this case. Remember we are dealing with the case at infinity, not as it tends to infinity. I may be wrong, but I just thought I would put that in.

That's a good point, but we want to be able to include shapes like the one Brad mentioned above when you take the surface of revolution of y=1/x² and cap it at x=0. This has finite volume and infinite surface area.

I've put my thoughts on what constitutes a surface below, since they're quite technical and anyway not even conclusive. Assuming we have a definition of a surface and the area of it, let's continue.

I think the problem Susan mentions boils down to this: a surface S in R³ splits R³ into 2 partitions which we could call "inside" and "outside". As far as I can see, there's no consistent way of saying what is inside and what is outside. So we could say that the set V={x in R³ :|x| > = 1} has finite surface area because it has a boundary which is the same as the sphere of radius 1.

We could stop at this point and say that here is the example that Brad was looking for, but it seems like a bit of a cheat don't you think? I'm not sure how to resolve the problem though.

Perhaps we could try and find if there is a surface of finite surface that breaks R³ into two partitions both of which have infinite volume. If there is no such surface, then every surface of finite area breaks R³ into a finite and infinite partition, and so we could say that the shape defined by the surface is the one with finite volume. This would prove that Brad's example can't exist. The obvious example of a surface which breaks R³ into two partitions of infinite volume is any plane, but this has infinite surface area.

Technical aside follows:

A surface S is a subset of R³ which is locally homeomorphic to R² . For those who haven't sat a course on topology I'll explain what that means. It means for every point x on the surface S, there is a subset U_x containing x and a one-to-one continous function f:U_x -> D where D ={y in R² :|y| < 1} (the open disc). That's the standard definition of a topological surface. Unfortunately, I don't know how to define the surface area of such an abstract object. If we could assume that the surface was differentiable, or at least piecewise differentiable, then the surface would be a differentiable 2-manifold and we could define integration on it and hence give it a surface area. That's the limit of my knowledge of this area, so someone more knowledgeable will have to step in if we want a more general definition.

Would it work to define volume with respect to a surface with a given area if we were to put a plane through the object that would allow us to represent the object in a coordinate system, say the x,y,z coordinate system. It is a necessary condition that z=f_z (a,b),y=f_y (a,b), and x=f_x (a,b), and that there are at two z coordinates for every set of x and y, two x's for every y and z, and two y's for every x and z. Then we define the set of points composing volume to be the set of all points in which the z coordinate is less than the maximum z coordinate of the surface for a given x and y and greater than the minimum z for a given x and y. We then also define the set of points in which x and y may be selected from to be the set of points (at z=0) in which the x coordinate is greater than the minimum x on the surface and less than the maximum x on the body. That's very poorly described, hopefully it can give you an idea though. This would be nice as it allows for curves to noncontinuous, but somehow that defintion of volume seems at once contorted and simple-minded, so I wouldn't be surprised if there are some errors to be found with this definition.

I've just now had the thought that there could be more than two z coordinates for a given x and y. If so, it's pretty simple to see how this definition could be modified.

Brad

I think the proof of this may be possible, and not all too hard. First, here is the page that Dan speaks of above. It is proven there that a circle has the maximum volume given a certain length of a closed curve. It is a corallary of this that for extrusions, a cylinder has the maximum volume given a certain lateral surface area (*). Now, it is also a principle (Toricelli's or Tortellini's principle; something like that) that any solid abject can be considered a bunch of infinitesinal extrusions, and that the volume and surface area of the solid is equivalent to the summation of all the volumes and lateral surface areas of the extrusions, respectively. That means since we are worried about maximizing the volume with respect to a given surface area, and given (*), we may consider solids built of only disks.

It is also a consequence of Mr. T's aforementioned principle that a body composed of disks (even infinitesimal ones) can be sliced up and rearranged any way possible and still have the same volume. And that means we could rearrange any solid we need concern ourselves with to be a revolution of a given function. And, I've alreasy proven above that a revolution of a given function will not result in a finite surface area and an infinite volume. Also, since every other solid's ratio of volume to surface area would be less, no other solid will have an infinite volume and a finite surface area!

There are for sure problems with the rigor of this; any comments or suggestions?

Brad

Tortellini is a sort of pasta (a filling, commonly spinach and ricotta, wrapped up in a sheet of fresh pasta which is then folded over in a complicated way into a little parcel, mmmmm) so my guess is it's Toricelli's principle (kind of rings a bell, but I'm not sure). Or maybe I'm thinking of Tortelloni?

I don't see why this principle should be true, it doesn't seem intuitively obvious. It seems like it's probably a principle that produces valid answers for simple shapes (like a sphere), but maybe not so for more complicated shapes.

I wasn't convinced by the above, but I'll look again tomorrow because I'm a bit tired now.

Now that I think about it, the principle must apply only to volume; for surface area it is flawed. I don't think that was part of the principle, I must've made up the surface area bit somehow, and not realized the error. The surface area I think wuold be found by adding up slices of 'distorted cones' though (by distorted cones, of course, I mean practically any shape taken up to a point) . Can anyone think of a way to use this?

I don't see what the problem is with what Dan mentions above - take a sphere for example. Now consider the volume of this sphere to be the volume of that which is 'outside' of it (in the way we usually think of 'inside' and 'outside') - in which case you get an infinite volume, but the surface area of a sphere.

Regards,
Olof

Dan's example works. It's just that it doesn't really seem like the question has been answered. That's not generally the way we treat volume.

Dan, Does the shape you mentioned fall under the class of closed solids?

On a related note, suppose we are talking of the cone-like extrusions that I was talking about above.

Given that the base part-that part that is a curve of some sort is given by the parametric equations x=f_x (t);y=f_y (t), what is the volume of the whole cone? The surface area?

Thanks,

Brad

Brad, I agree that the "outside of a sphere" example is unsatisfactory. As I suggested above, to prove what you want (i.e. that finite surface area implies finite volume) I think we should try and prove that finite surface area implies that one of the inside or outside is finite. I don't know what you mean by a "closed solid", but the "outside of a sphere" is a closed set (if you include the boundary).

I think that the volume of the cone is (1/3)Ah where A is the area of the base and h is the perpendicular height of the cone. Not sure about the surface area.

It's not the way we usually treat volume, but that doesn't mean we should discount it. After all, we don't usually look at infinite volumes at all. I don't see how there's any other way such a shape can exist. If we say that the surface of the shape is what seperates the inside from the outside, then the inside must be bound (if the S.A. is finite), which means that it cannot have an infinite volume. I'm probably overlooking something here, but I can't quite see it...

Also, I was playing with the thought of a possible consequence of finding such a shape as you describe Brad - wouldn't it mean that the Universe (perhaps Space) could have some sort of surface containing it?

Regards,
Olof

Hmmm nice point OLOF.
i quite agree with you (though partially).
(i quite have some doubt but i think i will let them out later).
love arun