The Infinite Vuvuzela Paradox

[jestingrabbit]

This is the shape about which, OP, you are talking? If so (if not, then correct me), I agree the surface is infinite, but how is the volume? It just keeps going up, getting infinitely tall, and it gets infinitely wide at the bottom.

If you take that shape and revolve it around the x axis, and then only look at the part with x>=1, you have the correct shape. A better page to look at is this one Gabriel's Horn.

[Mike_Bson]

This is the shape about which, OP, you are talking? If so (if not, then correct me), I agree the surface is infinite, but how is the volume? It just keeps going up, getting infinitely tall, and it gets infinitely wide at the bottom.

If you take that shape and revolve it around the x axis, and then only look at the part with x>=1, you have the correct shape. A better page to look at is this one Gabriel's Horn.

That's pretty funny. Thinking intuitively, the volume would be infinite (the end never actually closes up), but the formula shows otherwise. Weird.

Now, when it says it would be pi units in volume (one unit being one little cube on the graph), wouldn't that be a little small?

[jestingrabbit]

wouldn't that be a little small?

Its not just the two dimensional area of the graph, its the three dimensional volume of the solid of revolution. The area is in fact infinite.

[SlyReaper]

I just noticed the timestamp on the OP. I think this could possibly be the only mention of vuvuzelas on the Internet ever before this year.

[ConMan]

I just noticed the timestamp on the OP. I think this could possibly be the only mention of vuvuzelas on the Internet ever before this year.

I noticed that, too, and suspect that may even be why the thread was necro'd. I have no proof of this, unfortunately.

[makc]

However, with any depth at all, he's going to need infinite paint.

not at all, this is just difference of two volumes, each with slightly different radius, thus finite. of course, if he would keep painting at the same depth all along to +∞, _then_ it would be infinite.

[Maquox]

I hate how people break their heads with these questions, there is no paradox, you can't paint the "tip" of the vuvuzela as it not only doesn't exist, but is also smaller than an atom, thus smaller than the smallest molecule of paint, and even without that limitation, it would take infinite time for paint to reach the "tip"...

[jestingrabbit]

Its true that its not a real world situation, and nor is it particularly paradoxical that surface area and volume aren't related in any obvious way, but I don't think its wrong for people to try and think out what the mathematics is saying.

[tastelikecoke]

I just noticed the timestamp on the OP. I think this could possibly be the only mention of vuvuzelas on the Internet ever before this year.

I noticed that, too, and suspect that may even be why the thread was necro'd. I have no proof of this, unfortunately.

I proved your suspicion. I found it, and decided to revive it. Now I'm reviving it again.

[thoughtfully]

I just noticed the timestamp on the OP. I think this could possibly be the only mention of vuvuzelas on the Internet ever before this year.

Vuvuzelas are big now. They're on Slashdot!

[phlip]

Vuvuzelas are big now.

But are they infinitely big?

[thoughtfully]

Vuvuzelas are big now.

But are they infinitely big?

Not in surface area.

[HungryHobo]

I'm guessing that we are assuming paint particles are infinitesimal and don't clump together

as with all things infinity leave common sense at the door.

talking about litterally infinitesimal particles leads to madness,. example: how many such paint particles would it take to fill a 10 litre bucket?

An infinite amount.

no matter if you kept pouring more and more into your bucket it would remain almost totally empty.

It doesn't even really make much sense to talk about volumes of these paint particles at all, how many are in an single cm^3, how about in a meter cubed, what's stopping however many are in a meter cubed from being squished down to a cm cubed?

Forget anything about painting an infinite surface area with these particles, filling a finite recular bucket would take an infinite amount of these paint particles.

So no real paradox, if you talk about infinitesimal particles of paint then talking about it's volume makes no sense.
If you're not talking about infinitesimal paint particles then the surface area they can reach and thus paint is finite.

[sledDog]

The volume is finite but it's not definite. You can't tack on a set volume because it will always get a little more, further down the line, but you can say it will never be greater than V (too lazy to do the integration) because the increments become consistently smaller in proportion to the width of the increments and approach 0. It has infinite surface area because as x approaches infinity, the increments decrease in such a way that they do not approach 0 but rather 1. Take a vase with a similar shape and set on it's small end and slice it parallel to the ground with thickness super-dooper-small. The volume of each slice from top to bottom approaches 0. If the vase extended for infinity, each slice would be smaller than the last and eventually be 0. Now take a tailor's measuring tape and measure the length of material in the vase to go down distance super-dooper-small. Each segment is smaller than the last but it doesn't approach length=0, it approaches length=super-dooper-small. If it extended to infinity, each segment would be a little less larger than super-dooper small but it will never be less than the width of the interval (unless you know a shorter distance between two points than a straight line). I can see the paradox but think of it this way: way down there at infinity, the vuvuzela is closed. But no matter how far you go, there will be more of it to paint.

[phlip]

That's... wrong, in almost every respect.

(a) The volume is definite. It isn't "always getting bigger" - it's a specific shape, it's not changing. The only way its volume could change is if it changed shape. But it's just a static shape of infinite length. Not (as you seem to be misunderstanding) a changing shape of finite but increasing length. No, it's already infinite, it's not growing. You may be getting confused by the fact that, to calculate the volume and surface area of the infinite shape, we look at the finite-but-growing shapes, and take the limit as it grows to infinity. However, even though the finite-but-growing shapes merely approach the final volume V in the limit, the infinite shape's volme equals V (and, since the shape is static, doesn't change, so it can't be said to "approach" anything).
(b) When we look at the finite-but-growing shapes (which, again, we take the limit of in order to calculate the volume/surface area of the original shape), the volume/SA added by each additional "slice" approaches 0 for both the volume and the surface area. It definitely does not approach 1 for the SA (it's approximately a cylinder, which is 2*pi*r*h, h is a constant and r goes to 0, so the whole thing goes to 0). The statements "goes to 0" and "goes to a veryvery small number" which you try to draw a distinction between are equivalent. If the limit of a sequence is an infinitesimal, then the limit is zero. The only difference is that if you claim the limit is infinitesimal then it's evidence you're confused about something. Both the surface area and volume of each slice goes to 0, it's just that the SA goes down by 1/x, and the volume goes down by 1/x².
(c) The limit of the length of each slice is 1, but that's irrelevant, because we're not trying to find the length of the shape. We already know that's infinite.
(d) The "no matter how far down you go, there's always more to paint" line is also irrelevant - no matter how far down you go there's always more volume, too, yet the volume is still finite. So this line proves nothing.

[RonWessels]

@phlip, I have to call "wrong" on you as well.

... the infinite shape's volume equals V (and, since the shape is static, doesn't change, so it can't be said to "approach" anything).

Ignoring possible grammatical issues, you cannot talk about "the infinite shape's volume", since the infinite shape cannot exist. That is the problem with infinities - they are not a number that can be represented in the real world. They are a shorthand notation for limits as something gets larger and larger. Speaking about "the infinite shape" must therefore be a shorthand for a discussion of the limiting properties (ie. what the shape "approaches") as the length gets longer and longer.

[phlip]

you cannot talk about "the infinite shape's volume", since the infinite shape cannot exist

Sure it does, and sure I can. Let S = {(x,y,z) in R³ : x² + y² <= 1/z² and z >= 1}. This set is an infinite (by which I mean "unbounded; of infinite length") shape. The volume of S is (NB: not "approaches") pi.

No, you couldn't build this in real life, but we're not talking about real life, we're talking about mathematics. If your contention is that we can't talk about anything that's unbounded in size, then we can't talk about R³ at all. Or, hell, even R.

[xkcdfan]

@phlip, I have to call "wrong" on you as well.

... the infinite shape's volume equals V (and, since the shape is static, doesn't change, so it can't be said to "approach" anything).

Ignoring possible grammatical issues, you cannot talk about "the infinite shape's volume", since the infinite shape cannot exist. That is the problem with infinities - they are not a number that can be represented in the real world. They are a shorthand notation for limits as something gets larger and larger. Speaking about "the infinite shape" must therefore be a shorthand for a discussion of the limiting properties (ie. what the shape "approaches") as the length gets longer and longer.

Do you not understand what math is? What the fuck are you doing here?

Do you not understand what civility is? -jr

[redscourge]

Correct me if I'm wrong, but can't you do something like this to approximate the surface area of this shape:

let f(x) be a the function of the surface area of this shape, and e(x) be a function known to have slightly less surface area, and g(x) be a function known to have slightly more surface area, where e(x) and g(x) are both convergent. So, e(x) < f(x) < g(x). Now, can't you just reduce the difference between e(x) and g(x) so that they approach zero, and then use that to find the surface area of f(x) even though f(x) is not convergent?

[jestingrabbit]

Correct me if I'm wrong, but can't you do something like this to approximate the surface area of this shape:

let f(x) be a the function of the surface area of this shape, and e(x) be a function known to have slightly less surface area, and g(x) be a function known to have slightly more surface area, where e(x) and g(x) are both convergent. So, e(x) < f(x) < g(x). Now, can't you just reduce the difference between e(x) and g(x) so that they approach zero, and then use that to find the surface area of f(x) even though f(x) is not convergent?

The problem with that is that any figure that contains this one will also not have convergent surface area.