The Infinite Vuvuzela Paradox
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The Infinite Vuvuzela Paradox
Consider the surface described by rotating the curve y=1/x in [1, infinity) about the line y=0, which resembles an infinite vuvuzela. The volume contained within the surface is finite (= pi), but the surface area is infinite (if we had TeX forums, I'd show it).
Where's the paradox? You can fill the vuvuzela with paint, but you can't paint it.
Can you explain this? Or at least join me in my ponderings?
(Or perhaps I was fooled, and did the calculus wrong, too)
Where's the paradox? You can fill the vuvuzela with paint, but you can't paint it.
Can you explain this? Or at least join me in my ponderings?
(Or perhaps I was fooled, and did the calculus wrong, too)
Last edited by Taejo on Wed Oct 25, 2006 7:26 pm UTC, edited 1 time in total.
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My take on this one is that it's a units problem. So the volume is finite and the surface area isn't. Ok. Thing is, it's finite in, say, meters cubed, while it's infinite in, say, meters squared. So consider painting it. How thick is the paint? Very thin. In fact, if you're going to paint the whole thing, it's going to have to be an infinitely thin layer of paint.
Or, in measure theory terms, you could say that, if your measure gives a finite result for the volume, the "volume of the paint" is going to be of measure zero. So it's not quite that it'd take an infinite amount of paint. It'd take a volume of paint with measure zero.
Or, in measure theory terms, you could say that, if your measure gives a finite result for the volume, the "volume of the paint" is going to be of measure zero. So it's not quite that it'd take an infinite amount of paint. It'd take a volume of paint with measure zero.
Re: The Infinite Vuvuzela Paradox
Taejo wrote:Consider the surface described by rotating the curve y=1/x in [1, infinity) about the line y=0, which resembles an infinite vuvuzela. The volume contained within the surface is finite (= pi), but the surface area is infinite (if we had TeX blogs, I'd show it).
Where's the paradox? You can fill the vuvuzela with paint, but you can't paint it.
As you go out towards infinity from the origin, the volume becomes infinitesimally small, so you need increasingly smaller amounts of paint to fill the portion of the interior outside each given radius. But we're presuming the layer of paint used to coat the whole thing is uniform all over, so that quantity goes up more or less constantly no matter how far from the origin you are.
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Re: The Infinite Vuvuzela Paradox
Taejo wrote:The volume contained within the surface is finite (= pi), but the surface area is infinite
Uh... correct me if I'm wrong, but doesn't area under 1/x diverge? It's 1/x^2 that converges right?
I guess this is somewhat like infinite series, in the way of adding the terms. It would take you an eternity to add an infinity of terms, but you can use a different method to find the sum without having to add the terms.
Re: The Infinite Vuvuzela Paradox
svk1325 wrote:Taejo wrote:The volume contained within the surface is finite (= pi), but the surface area is infinite
Uh... correct me if I'm wrong, but doesn't area under 1/x diverge? It's 1/x^2 that converges right?
Having just looked it up, you're supposed to rotate the part of the curve from 1 to infinity around the axis. If you went all the way to 0 or further, it'd diverge. And forgot to say, I think all of them diverge from 0 to infinity, but I'm forgetting a lot of my calculus. The pseries converges for p > 1, but the integral is going to diverge.
Re: The Infinite Vuvuzela Paradox
Air Gear wrote:svk1325 wrote:Taejo wrote:The volume contained within the surface is finite (= pi), but the surface area is infinite
Uh... correct me if I'm wrong, but doesn't area under 1/x diverge? It's 1/x^2 that converges right?
Having just looked it up, you're supposed to rotate the part of the curve from 1 to infinity around the axis. If you went all the way to 0 or further, it'd diverge. And forgot to say, I think all of them diverge from 0 to infinity, but I'm forgetting a lot of my calculus. The pseries converges for p > 1, but the integral is going to diverge.
Integral(1/x)=ln(x)+c
Area under 1/x from a to b = ln(b)ln(a) = ln(b/a)
if [a,b]=[1,infinity], area = ln(infinity) = infinity, i.e. does not diverge
Integral(1/x^2)=1/x+c
Area under 1/x^2 from a to b = 1/b+1/a = 1/a1/b
if [a,b]=[1,infinity], area = 1/11/infinity = 1 i.e. convergent.
w00t!
Well, if it's real paint, it's not going to fit through an infinitessimally small hole. Also, if it's real paint, you'll have a glob of finite radius clustered around a horn of infinitessimally small radius, so if you take the limit to infinity you'll require infinite paint.
But, real paint assumes a real horn if the horn doesn't have thickness, it's not made of atoms, and then I would argue it doesn't actually exist.
But, real paint assumes a real horn if the horn doesn't have thickness, it's not made of atoms, and then I would argue it doesn't actually exist.
There's a wellknown shape that demonstrates this "paradox" that can be found by stepping back 1 dimension  it's a little easier to think about since it lies within a bounded region (i.e. doesn't go "off to infinity"). I'm referring to the Koch snowflake, which is formed by starting with an equilateral triange, and repeatedly replacing the middle thirds of the sides with smaller equilateral triangles.
At each step, the amount of area added to the shape is (4/9) the area added during the previous step, so the total area added is the sum of a geometric series, and thus remains finite, even after performing the operation an infinite number of times. However, each step in the operation increases the total perimiter by a factor of (4/3), so the perimeter increases without bound, and the final shape has an infinite perimeter.
I'm pretty sure that if we take the Koch snowflake idea and step it back up a dimension, we'll get another example of a shape with finite volume and infinite surface area. Specifically, start with a regular triangular pyramid, and sucessively replace the "middle triangle" of each face with a smaller pyramid. You'd end up with something that fits in a finite amount of space, but I wouldn't touch it  the infinite surface area of this shape makes it very pointy and sharp.
At each step, the amount of area added to the shape is (4/9) the area added during the previous step, so the total area added is the sum of a geometric series, and thus remains finite, even after performing the operation an infinite number of times. However, each step in the operation increases the total perimiter by a factor of (4/3), so the perimeter increases without bound, and the final shape has an infinite perimeter.
I'm pretty sure that if we take the Koch snowflake idea and step it back up a dimension, we'll get another example of a shape with finite volume and infinite surface area. Specifically, start with a regular triangular pyramid, and sucessively replace the "middle triangle" of each face with a smaller pyramid. You'd end up with something that fits in a finite amount of space, but I wouldn't touch it  the infinite surface area of this shape makes it very pointy and sharp.
Hix wrote:I'm pretty sure that if we take the Koch snowflake idea and step it back up a dimension, we'll get another example of a shape with finite volume and infinite surface area.
But the Koch snowflake is stranger than this.
Imagine a Koch cookiewe probably need to make these anyway for Halloween. Just take some mathematically continuous cookie dough, and roll it out to a perfect 1 unit thick layer. Get your Koch snowflake shaped cookie cutter, and cut a cookie out of it.
Now, add some orange frosting on the top. Imagine you have a large stockpile of frosting, but only a finite amount. No problem, huh? But what if you add frosting to the sides? How much will it take?
Well, obviously, an infinite amountat first thought, anyway. After all, we have a curve of infinite length, and we drag it through space, so we wind up with infinite surface area to cover with frosting. This means we don't have enough frosting.
But waitI say we do have enough! Put your koch cookie in a larger, circular cookie cutter. The volume of frosting we can put inside, to the side of the cookie, is certainly constant, and since this circular cookie cutter is comfortably wider than the Koch cookie, and we all know the koch cookie is bounded, we know we have way more than enough frosting there. Just cut away the excess.
You're left with a finite amount of frosting frosting an infinite area, and we're not even leaving any bald spots on the sides! So, what's the problem?
Actually, the more points you have, the less likely it will cut.You'd end up with something that fits in a finite amount of space, but I wouldn't touch it  the infinite surface area of this shape makes it very pointy and sharp.
Um... not to stray too far from the topic, I'll give my thoughts about the paint and stuff after a bit of rethinking
My thoughts wrote:I'm guessing that we are assuming paint particles are infinitesimal and don't clump together, to keep the problem somewhat simpler. You also have to remember that the surface has no thickness. (Wait, doesn't that mean no physical or structural support for the horn to keep it's shape? Math terms... go figure.) It may have a finite volume, but it should take forever for the paint to fully fill the volume. The paint would take an eternity to hit infinity, and technically it wouldn't actually reach it anyways. You might be able to fill it up with the amount of paint you have, but you can't actually fill it up. That's with 1/x^2 anyways. 1/x has infinite surface area and volume.
So we've cleared up the fact about filling it. What about painting it? You would need infinite paint, and of course infinite time to do it. Or, for the lazy people, you can point the horn downwards (infinite amounts if massless particles?) and just let the paint "run" downward the horn. Still infinite time, but you can do what you want instead of painting for eternity. I'd pick the lazy way. Heh.
Meh, whatever. We can't fully comprehend infinity. That's what leads us to topics like this  our curiousity.
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ikefalcon wrote:The volume contained within this object isn't finite. It's infinite. Calculus methods show this.
Doing some scratch work (mind you, I haven't done calculus in a while), I'm pretty sure that the volume converges on PI.
But back to the original question... wrote:What if we constructed this funnelshaped object out of a tight mesh fabric. We can then presumably fill the volume with paint. But there are wholes all along the sides, so some paint will leak out. Of that paint, most will drip off, but some will wrap around the funnel, eventually covering the surface in paint.
Clarification: V = pi
I would just like to clarify that it does have finite volume (V = pi), just like I said. Remember (or discover now) that volume of a solid of revolution is pi * integral of f(x)^2 dx. (This is because each slice is a circle with area proportional to the square of the radius).
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The area under the curve does indeed diverge.[/i]
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ikefalcon wrote:Using the shell method:
V = 2pi int(1,infinity) p(x) * h(x) dx
V = 2pi int(1, infinity) x * 1/x dx
V = 2pi int(1,infinity) dx
V = lim d>infinity [x](1,d)
V = infinity
...errr...
V = int{a,b}[ Pi * (radius)^2 ]dx
V = int{1,infin}[ Pi * 1/x^2]dx
V = Pi * int{1,infin}[1/x^2]dx
V = Pi
But I don't think we can use the shell method for this very easily. We are integrating along the x axis, so we need to integrate with respect to y. I can't figure out how to represent x=0 in terms of y.
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ikefalcon wrote:Using the shell method:
V = 2pi int(1,infinity) p(x) * h(x) dx
V = 2pi int(1, infinity) x * 1/x dx
V = 2pi int(1,infinity) dx
V = lim d>infinity [x](1,d)
V = infinity
Using the shell method correctly:
V = 2pi int(0,1) p(y) * h(y) dy
V = 2pi int(0,1) y * (1/y1) dy
V = 2pi int(0,1) (1y) dy
V = 2pi (1  1/2)
V = pi
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Gemini25RB wrote: I can't figure out how to represent x=0 in terms of y.
x=0 is just all y. which probably fails the vertical line test!
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I was just thinking, if we wanted to step it down a dimension, we could consider the same paradox for the Mandelbrot set graph. We can obviously color the entire shape (it all fits in a 4x4 square, I believe), but cannot possibly outline it all.
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It seems like sort of a moot point to me.
We want surface area of a rotated shape, well that's the integral of arc length. So we're finding the arc length of a function that exists for all x not equal to zero on the interval [1,infinity). Hell, the function y=x would give you the exact same problem of having infinite surface area.
So really, your question just becomes "How the heck does an infinite function have a finite area!?" Which is sort of just a fundamental concept of calculus that can be verified through series.
Basically it's just one of those things that makes calculus so awesome.
We want surface area of a rotated shape, well that's the integral of arc length. So we're finding the arc length of a function that exists for all x not equal to zero on the interval [1,infinity). Hell, the function y=x would give you the exact same problem of having infinite surface area.
So really, your question just becomes "How the heck does an infinite function have a finite area!?" Which is sort of just a fundamental concept of calculus that can be verified through series.
Basically it's just one of those things that makes calculus so awesome.
silverhammermba wrote:It seems like sort of a moot point to me.
We want surface area of a rotated shape, well that's the integral of arc length. So we're finding the arc length of a function that exists for all x not equal to zero on the interval [1,infinity). Hell, the function y=x would give you the exact same problem of having infinite surface area.
So really, your question just becomes "How the heck does an infinite function have a finite area!?" Which is sort of just a fundamental concept of calculus that can be verified through series.
Basically it's just one of those things that makes calculus so awesome.
Except series isn't intuitive and that's one of the major problems. Well, that plus the fact that you have to realize that we're also talking an object defined by a set with measure zero in R^3...
Gemini25RB wrote:I was just thinking, if we wanted to step it down a dimension, we could consider the same paradox for the Mandelbrot set graph. We can obviously color the entire shape (it all fits in a 4x4 square, I believe), but cannot possibly outline it all.
Or if we don't like mathematical constructs, perhaps we can just settle for the coastline paradox. But we do like mathematical constructs
moopanda wrote:Or if we don't like mathematical constructs, perhaps we can just settle for the coastline paradox. But we do like mathematical constructs
We only like CERTAIN mathematical constructs. For example, fuck the Vitali set. In fact, fuck everything that only has a nonconstructive form which requires the uncountable version of the Axiom of Choice to describe...and everything which requires the uncountable version of the Axiom of Choice to prove, on top of it.
WILL YOU JOIN ME IN THE WAR AGAINST THE UNCOUNTABLE VERSION OF THE AXIOM OF CHOICE?
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Hix wrote:Air Gear wrote:WILL YOU JOIN ME IN THE WAR AGAINST THE UNCOUNTABLE VERSION OF THE AXIOM OF CHOICE?
Never! Bad things happen when my vector spaces don't have bases.
At least noone could be in ur base killin ur plot points
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Re: The Infinite Vuvuzela Paradox
I think there's not much rules governing on necromancy other than high discouragement.
If you think about it, blowing such an infinite span of vuvuzela would be plausible because it is an air column, while because of infinite surface area would have infinite friction, it converts all the energy into noise.
*shivers*
If you think about it, blowing such an infinite span of vuvuzela would be plausible because it is an air column, while because of infinite surface area would have infinite friction, it converts all the energy into noise.
*shivers*
Re: The Infinite Vuvuzela Paradox
Arighty, for those still interested, the shape is known as Gabriel's Horn.
The following is sort of thinking out loud. I find it easiest to answer such questions when I write out my own thought process, and experience tells me that others find my thoughts easier to understand that way. I'm sure that certain parts, if not all of them, are bound to be invalid, but it'll be interesting to see how I do.
The following is sort of thinking out loud. I find it easiest to answer such questions when I write out my own thought process, and experience tells me that others find my thoughts easier to understand that way. I'm sure that certain parts, if not all of them, are bound to be invalid, but it'll be interesting to see how I do.
Spoiler:
Re: The Infinite Vuvuzela Paradox
@scaliper: You seem to have discovered that volume and surface area don't mix. It's like asking "How many square feet of water do you have in that bucket?"
When Gabriel's horn was explained to me, I was told "You can fill it with a finite amount of paint, but you'll never have enough paint to coat the INSIDE of the horn." To which I questioned, "So, when trying to paint it, why don't you just tip it on end and pour pi units of paint in there? If find yourself unable to paint any more, and you've run out somewhere between X=1 and X=infinite, you obviously have some volume LESS than pi inside."
When Gabriel's horn was explained to me, I was told "You can fill it with a finite amount of paint, but you'll never have enough paint to coat the INSIDE of the horn." To which I questioned, "So, when trying to paint it, why don't you just tip it on end and pour pi units of paint in there? If find yourself unable to paint any more, and you've run out somewhere between X=1 and X=infinite, you obviously have some volume LESS than pi inside."
Is 'no' your answer to this question?
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Re: The Infinite Vuvuzela Paradox
redrogue wrote:When Gabriel's horn was explained to me, I was told "You can fill it with a finite amount of paint, but you'll never have enough paint to coat the INSIDE of the horn."
It seems to me that that presentation doesn't work, as you have already realised. Consider that to paint something you want to cover it with paint at a more or less uniform depth. So now think about the pointier part of the horn. At some point the width is less than the paints thickness. At that point you aren't doing a surface area calculation to work out how much paint you need, you're doing a volume calculation to work out how much paint you need to stop up the horn with paint.
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Re: The Infinite Vuvuzela Paradox
jestingrabbit wrote:It seems to me that that presentation doesn't work, as you have already realised. Consider that to paint something you want to cover it with paint at a more or less uniform depth. So now think about the pointier part of the horn. At some point the width is less than the paints thickness. At that point you aren't doing a surface area calculation to work out how much paint you need, you're doing a volume calculation to work out how much paint you need to stop up the horn with paint.
Obviously, a vuvuzela full of paint won't make that lovely buzzing noise. You'd need to use less than pi units of paint, of course. That way, you'd leave a nice hole in the middle for those magical notes to float through. (Warning, Lazy Engineering Answer Ahead):
It wouldn't be *perfectly* uniform, but close: Fill the gap between the edge of the horn and any curve y = B /x [1, infinite], as rotated around the X axis, for any value of B between 0 and 1. As B approaches 1, you get a thinner, more uniform spread of paint on the inside, eventually approaching zero paint covering an infinite surface area. A similar curve for B > 1 (but approaching 1) will let you paint the outside.
[All of this hinges on some hypothetical Math Paint (tm) that can be spread thinner than known molecules]
Is 'no' your answer to this question?
Re: The Infinite Vuvuzela Paradox
Think about two fellows  one filling the vuvuzela with paint, the other painting it. If you could come back at the end of infinity, and magically teleport all the paint filling the thing into your hand (at which point the filling dude will snap because you've just removed all his infinitely hard work) you'll have pi cubic units of paint. Easy. However, I wouldn't reccommend teleporting all the paint on the outside of the vuvuzela into your hand. Not only will the painting guy come after you with an axe, but unless he painted it infinitely thin you have infinite paint in your hands.
If he can paint infinitely thin, he didn't need bother about how much paint he took; he could take just as much as you could balance on the tip of a pin and paint the whole thing if it was that thin. However, with any depth at all, he's going to need infinite paint. And if you teleport it off the thing you're going to be sucked into a black hole in an instant.
If he can paint infinitely thin, he didn't need bother about how much paint he took; he could take just as much as you could balance on the tip of a pin and paint the whole thing if it was that thin. However, with any depth at all, he's going to need infinite paint. And if you teleport it off the thing you're going to be sucked into a black hole in an instant.
Re: The Infinite Vuvuzela Paradox
jestingrabbit wrote: So now think about the pointier part of the horn. At some point the width is less than the paints thickness.
This is exactly it. Restating what's already been said in (hopefully) slightly more illuminating mathyspeak*:
Volume = Area * Depth
In the case of the horn, we have:
Area = ∞
Depth = 0
Therefore Volume = ∞ * 0. Like any "∞*0" problem, the answer could be anything between 0 and ∞; in this case it happens to be π.
In sentence form, this means that if you unrolled the horn and chopped up it up into a bunch of 2D squares laid flat and joined together, you would need an infinite amount of paint to cover the surface. If you then took that surface and rolled it back up into the horn shape, you'd have an infinite amount of paint come squirting out  unless you initially covered the 2D squares with a layer of paint having 0 depth.
*Some basic mathematics will be raped for clarity. If "reader"="mathematically pedantic", close your eyes and hum.
Re: The Infinite Vuvuzela Paradox
Actually you could have any finite amount of infinitely divisible paint cover the whole surface with a non0 depth. The depth just has to decrease asymptotically as you approach the extremities. With real world paint, it is obviously impossible with a finite amount of paint, even if you merely approximated arbitrarily thin paint by averaging the depth across increasingly large areas defined with increasingly sparse paint particles as the vertices (there are a finite number of particles, so without a point at infinity, there is a finite distance that the furthest paint particle can be from the origin, and thus any area defined by any subset of the paint particles can be shown to be inside some subset of the total surface area defined by the arrangement of the paint particles. (If we allow averaging over arbitrary portions of the area for each point on it, we can do it trivially with a single paint molecule)GeorgeH wrote:This is exactly it. Restating what's already been said in (hopefully) slightly more illuminating mathyspeak*:
Volume = Area * Depth
In the case of the horn, we have:
Area = ∞
Depth = 0
Therefore Volume = ∞ * 0. Like any "∞*0" problem, the answer could be anything between 0 and ∞; in this case it happens to be π.
In sentence form, this means that if you unrolled the horn and chopped up it up into a bunch of 2D squares laid flat and joined together, you would need an infinite amount of paint to cover the surface. If you then took that surface and rolled it back up into the horn shape, you'd have an infinite amount of paint come squirting out  unless you initially covered the 2D squares with a layer of paint having 0 depth.
*Some basic mathematics will be raped for clarity. If "reader"="mathematically pedantic", close your eyes and hum.
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Re: The Infinite Vuvuzela Paradox
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.
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