The Infinite Vuvuzela Paradox

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Taejo
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The Infinite Vuvuzela Paradox

Postby Taejo » Wed Oct 25, 2006 7:01 pm UTC

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)
Last edited by Taejo on Wed Oct 25, 2006 7:26 pm UTC, edited 1 time in total.
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Postby Air Gear » Wed Oct 25, 2006 7:08 pm UTC

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.

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Re: The Infinite Vuvuzela Paradox

Postby wisnij » Wed Oct 25, 2006 7:22 pm UTC

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

Postby svk1325 » Thu Oct 26, 2006 12:26 am UTC

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.

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Re: The Infinite Vuvuzela Paradox

Postby Air Gear » Thu Oct 26, 2006 12:39 am UTC

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 p-series converges for p > 1, but the integral is going to diverge.

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Postby Gelsamel » Thu Oct 26, 2006 12:48 am UTC

Stuff wrote:http://en.wikipedia.org/wiki/Gabriel%27s_Horn

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Re: The Infinite Vuvuzela Paradox

Postby SpitValve » Thu Oct 26, 2006 2:03 am UTC

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 p-series 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/a-1/b
if [a,b]=[1,infinity], area = 1/1-1/infinity = 1 i.e. convergent.

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Postby moopanda » Thu Oct 26, 2006 9:47 am UTC

Just so everyone's on the same page, the area of a volume of revolution is pi*int([f(x)]^2 dx). So the area of revolution on 1/x is the integral of 1/x^2.

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Postby Ephphatha » Thu Oct 26, 2006 10:02 am UTC

Uh, just so we're clear, does this thing ever converge? Because if not it's gonna have a hole in the bottom and if you try fill it with paint it's just gonna go straight through.
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Postby Andy » Thu Oct 26, 2006 12:49 pm UTC

Well, the "volume" of a plane is 0, but the surface is infinite... it doesn't seem like a paradox to me.

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Postby Vaniver » Thu Oct 26, 2006 12:52 pm UTC

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.

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Postby Gelsamel » Thu Oct 26, 2006 1:03 pm UTC

The paint bit is an analogy to explain why finite surface area yet infinite volume is a paradox.

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Postby Hix » Thu Oct 26, 2006 1:53 pm UTC

There's a well-known 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.

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Postby Gelsamel » Thu Oct 26, 2006 2:00 pm UTC

Off the top of my head that should work, Hix.

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Postby yy2bggggs » Fri Oct 27, 2006 6:23 am UTC

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 cookie--we 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 amount--at 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 wait--I 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?

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.
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Postby svk1325 » Fri Oct 27, 2006 8:02 am UTC

Um... not to stray too far from the topic, I'll give my thoughts about the paint and stuff after a bit of re-thinking

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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Postby ikefalcon » Fri Oct 27, 2006 10:13 pm UTC

The volume contained within this object isn't finite. It's infinite. Calculus methods show this.

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Postby Gemini25RB » Sat Oct 28, 2006 12:07 am UTC

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 funnel-shaped 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.

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Clarification: V = pi

Postby Taejo » Sat Oct 28, 2006 5:48 am UTC

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).

The area under the curve does indeed diverge.[/i]
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Postby Air Gear » Sat Oct 28, 2006 10:15 am UTC

People seem to keep missing one thing...

This curve goes from 1 to infinity, not 0 to infinity. It works.

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Postby ikefalcon » Sat Oct 28, 2006 11:16 pm UTC

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

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Postby Gemini25RB » Sun Oct 29, 2006 12:41 am UTC

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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Postby GreedyAlgorithm » Sun Oct 29, 2006 1:15 am UTC

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/y-1) dy
V = 2pi int(0,1) (1-y) dy
V = 2pi (1 - 1/2)
V = pi
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Postby no-genius » Sun Oct 29, 2006 1:28 am UTC

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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Postby Gemini25RB » Sun Oct 29, 2006 1:32 am UTC

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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Postby silverhammermba » Mon Oct 30, 2006 6:16 am UTC

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.

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Postby Air Gear » Mon Oct 30, 2006 11:37 am UTC

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...

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Postby moopanda » Mon Oct 30, 2006 11:47 am UTC

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 :-)

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Postby Air Gear » Mon Oct 30, 2006 2:14 pm UTC

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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Postby Hix » Mon Oct 30, 2006 5:24 pm UTC

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.

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Postby LE4dGOLEM » Mon Oct 30, 2006 6:03 pm UTC

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 no-one could be in ur base killin ur plot points
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Re: The Infinite Vuvuzela Paradox

Postby tastelikecoke » Mon Jul 05, 2010 2:40 pm UTC

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*

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Re: The Infinite Vuvuzela Paradox

Postby scaliper » Tue Jul 06, 2010 5:38 am UTC

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.
Spoiler:
You could technically fill the volume with paint, but it would take an infinite amount of time, due to the infinite distance between the start-point and the end-point. With that line of reasoning, you could say that you could not, in fact, fill the volume with paint. On the other hand, you could also argue that all that is required to cover the outside with paint is an infinite amount of paint and an infinite amount of time. In effect, both come out with the answer that filling the horn with paint and covering it with paint are equally feasible. You can either do both or neither.

On the other hand, I'm not entirely sure that that's a valid answer to the question. I suppose that the question is more straightforward when phrased "how can it be possible that one can measure out enough paint to fill an object, but not enough to cover it?" This is further complicated that the surface of the shape has no depth, so filling the shape would at the same time cover the shape, thus leading to the conclusion that it would require less paint to cover it than to fill it.

However, that is mixing units. Let's say that the volume is pi cubic meters, and the area is thus infinity square meters. Assuming that the paint, when spread, spreads to a point where it is infinitely thin, we can put it into two dimensions. Let's consider a cube of this paint measuring one cubic centimeter(1cmx1cmx1cm) that we want to find the surface area of(i.e. the paint, not the cube. This area will henceforth, and more than likely incorrectly, be referred to as "inherent surface area"). The easiest way to illustrate this would be to go into a sort of Zeno's Paradox. In order to discover how much area this paint will cover, we can split the cube into halves, each measuring 1cmx1cmx0.5cm. By doing this, we have increased the surface area by 2 square centimeters, from 6 square centimeter to 8 square centimeters. We can then reduce the depth by 1/2^x cm, each time increasing the total surface area by 2^x square centimeters, for an infinite amount of time, at which point we will have an infinite number of sheets of paint measuring 1cmx1cmx0cm. Each of these sheets has an area of 1 square centimeter, and the aforementioned infinite number of them will yield an infinite surface area. This can be carried over to any starting dimensions for any 3-dimensional shape. In other words, any three-dimensional object has an infinite inherent surface area. This seems to stand to reason, and also the laws of geometry, what with there being an infinite number of points on any line. It follows, therefore, that the previous conclusion that it would take less paint to cover the shape than to fill it was correct, regardless of unit-mixing. It only seems illogical because of the use of the word "paint," since an actual layer of paint is three-dimensional. Actually, an interesting point is that the volume has decreased to zero, but only at the depth of zero. Until the depth of each sheet was reduced to zero, the volume remained at 1 cubic centimeter. That, however, is an interesting paradox for another thread. Anywaym we can now go back to the conclusion of the original paragraph, and(I think) find that, regardless a few initial incorrect premises, it is a valid answer, contrary to the first statement of my second paragraph.

If we say that any volume of paint except for 0 cubic centimeters has an infinite inherent surface area, then any amount of this paint could be used to cover an infinite surface. In fact, all that is needed is a volume of paint with the dimensions >0 x >0 x >0. Again, I'm sure the notation is wrong. I prefer to write each dimension in the invalid but easier to understand form of 0.Õ1*10(with the Õ representing a repeating 0. Silly computer doesn't have a command for that. Anyway, this is, in effect, an infinitely small but still existent number). It follows that the premise in the first paragraph that an infinite amount of paint would be needed is actually not true; You actually need only an infinitely small amount, and the rest is just runoff. Since any amount of paint can be used, we can say that the only truly required "unreasonable" resource is an infinite amount of time to cover an infinite area. This is actually the same requirement for filling the volume. So going back to the initial horn, you would need pi cubic meters of paint and an infinite amount of time to fill the area, and to cover the surface area, you would need only any volume of paint other than zero, as well as an infinite amount of time. So one can say that any finite amount of this theoretical paint greater than or equal to pi cubic centimeters could be used to either fill the horn or cover its surface. Both take a finite amount of paint(actually, does infinitesimal count as finite?), and both take an infinite amount of time. We can therefore say that the initial conclusion that filling the horn and covering it are equally feasible was correct. Both are possible with an infinite amount of time, and impossible with a finite amount of time.

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Re: The Infinite Vuvuzela Paradox

Postby redrogue » Wed Jul 07, 2010 3:05 pm UTC

@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."
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Re: The Infinite Vuvuzela Paradox

Postby jestingrabbit » Wed Jul 07, 2010 4:12 pm UTC

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

Postby redrogue » Wed Jul 07, 2010 6:10 pm UTC

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]
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Re: The Infinite Vuvuzela Paradox

Postby Vesuvius » Thu Jul 08, 2010 1:29 am UTC

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.

GeorgeH
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Re: The Infinite Vuvuzela Paradox

Postby GeorgeH » Thu Jul 08, 2010 4:08 pm UTC

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 mathy-speak*:

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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WarDaft
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Re: The Infinite Vuvuzela Paradox

Postby WarDaft » Sat Jul 10, 2010 10:09 pm UTC

GeorgeH wrote:This is exactly it. Restating what's already been said in (hopefully) slightly more illuminating mathy-speak*:

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.
Actually you could have any finite amount of infinitely divisible paint cover the whole surface with a non-0 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)
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Mike_Bson
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Re: The Infinite Vuvuzela Paradox

Postby Mike_Bson » Thu Jul 15, 2010 7:07 pm UTC

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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