Yo, making an approximation of this horn and filling it up then measuring the volume would be a great new addition to Matt's weird ways of calculating π
Yeah, use approximation techniques to see how long your horn would have to be to be accurate out to...say like 10 decimal places, and then make a prototype. Man, I want a 3D printed one of these now.
For any who don't know this reference, this is a reference to a 3Blue1Brown video where he calculated pi based on how many times blocks with different mass ratios (all multiples of 10) hit each other before the bigger block and smaller block diverged. It's an amazing video
Yeah, the issue is that at some moment the width of the horn is less than an atom, so you can't really "paint the inside". But this is a physisists' problem that math people can't be bothered by.
I think the real key to the paradox is that these are different units. Any finite amount of volume can theoretically spread into an infinite area if you can spread it infinitely thin, so you CAN paint the horn (e.g. the inside, by filling it) with a finite amount of paint.
There is a limit to how thinly paint can be applied. 1 molecule thick is the minimum. So theoretically, Any finite amount of volume can not be spread over an infinite surface area. These type of problems are where pure math separates from applied maths. The paradox makes the same assumption you made here. The math isn't wrong, its the assumption that you can limit ds as dx goes to zero in the physical universe, which you can't.
If you're saying paint can only be applied at one molecule thick and not infinitely thin, than the diameter of the horn also has to have the limitation of a one molecule thick wall, and it can only get as small as one molecule at its point, making it finite. They are both infinite or both finite
The serious problem consists of a minimal thickness of ink. So, from outside, you cannot paint it with a finite amount of ink (pi). Now, considering the minimal thickness, filling it from inside will not guarantee that the interior surface area will be all painted. There comes a point where the regional volume will not suffice to paint the surrounding surface area. It's simple. I don't think this has to do with the boundary between applied and pure math. Also, the volume is clearly finite and surface area infinite.
It's probably not the solid that you imagine ... it is not defined at the center. ←----------O-----------→ You can't rotate something around the very something that doesn't exist and expect it to represent the volume defined by the exterior surface area, because it is undefined at the axis. That undefined center has infinite volume. That is what he leaves out to produce the 'paradox' ... by using a slight of hands you are tricked into believing it. I don't know if it is by accident ... maybe he is tricking himself. But I suspect he is smart enough to know better.
@@ejrupp9555 Disc integration is a completely legitimate method of computing the volume of a solid of revolution. The axis in the center doesn't matter; it doesn't contribute any amount of volume. If you want, you can use disc integration to calculate the volume of a finite solid, such as a cone, and you will get the correct result. You can even use disc integration to calculate the volume of a finite section of Gabriel's horn, which could theoretically be verified with a physical model, and as you take bigger and bigger sections, the volumes you get will asymptotically approach π.
Interestingly, the word “frustum” is taken directly from the Latin which means “crumb, or morsel” so comparing it to a pizza crust was more accurate than he ever imagined!
"You can't argue with the maths." Well. I CAN argue with the maths. I'll lose. They kicked me out of the math club a long time ago and told me never to return.
AN idiot will argue the math based on their own ignorance... perfect case in point... the fool that thinks one over infinity is equal to zero... mic drop.
@@captainoates7236 His point is that is not "out of nowhere". Pi is defined by the geometry of a circle. When there is a circle, pi usually appears some way or another. In some cases we can all agree that it is hard to see where there is a circle in the problem, but in this case it was really obvious. It is a figure you get by rotation along an axis. Pi was bound to show up
A nice way to think about this (and resolve the paradox), is that we are basically thinking about 2 different types of paint as if they are the same. When you fill the horn you are thinking of 3-dimensional paint particles, but when you are painting the horn's surface you are only thinking of 2-dimensional paint plates. If we would try to fill the horn with 2D paint, we would never finish. In fact, we wouldn't even be able to fill a tiny bit of it, because our 2D paint plates have height zero. This matches our intuition that something that fills out a volume must be more than something that only covers the surface. Contrary, if we would try to paint the surface of the horn (let's say the outside) with the 3D paint, we are making it unintentionally thicker everywhere as well (presumably by a constant epsilon > 0). If a covered surface is epsilon > 0 thicker than before, then the thickness of the horn doesn't converge to zero like explained at the beginning of the video, but converges to a constant > 0. Now, of course the paint we would need to cover the surface is still infinite, but the paint we would need to fill the entirety of the horn including the epsilon-thick surface is also infinite.
But the volume inside wouldn't change no matter how thick of a layer of paint is on the outside, wouldn't it? But then again, it wouldn't be surprising anymore that you would need an infinite amount of 3D paint to paint the outside when the outside is always bigger than a constant > 0.
Yeah, Imagine if you would pour a finite amount of (“mathematical”) liquid on an infinite plane. If it spread across only a part of the plane it would have some thickness. You could then take the top half of the liquid and spread it across another part of the infinite plane. The liquid will now have twice amount of area but half the thickness. This can be repeated infinite times such that the thickness becomes infinitely small but the area become infinitely big. Finite paint can paint any infinite area in that sense, and in that sense there is no paradox
I like the abstract theoretical dimensional analysis of paint, but volumes and surface areas aren't types of paint, theres no reason we can't have finite-volume-filling infinite-areas, finite-area-filling infinite-lengths, finite-length-filling points, etc., space filling curves and convergent series are just a part of calculus and without them there would be no calculus
Exactly. It's like how a coastline has infinite length but the area of the country is not infinite. This "paradox" may get you into Oxford but you would lose marks in grade school for not including the units!
You get Pi cubic-units of paint in a paint can that is 2 units wide and 1 unit high. Not standard paint can dimensions at Home Depot, but easy enough to imagine.
Absolutely loved it where you skipped the pain and substituted for “bigger than 1” just to get to the proof. Something I didn’t really encounter in maths at school. Mind blown in lots of ways on this video
This was used in my Calc 2 class in things like the comparison test, where we can compare a very complicated series to something that is easy to compute to prove divergence or convergence. For example we know that 1/((n^2)+1) < 1/(n^2), and since we know an infinite series of 1/(n^2) converges, we then know the first converges as well. the same can also be applied to divergence
I remember going over this in college. It wasn’t called Gabriel’s Horn, we didn’t talk about paint, and it wasn’t called a paradox. We simply derived and noted the relationship between the two solutions as being counterintuitive. A surface of revolution with a finite volume and an infinite surface area. The infinite surface area is no problem at all. The horn is infinitely long, of course the surface area is infinite. It’s the volume that’s tricky. But this may help- remember that this is a bottomless pit. The volume approaches a limit, but never gets there. You can always add a fraction of what you just added, dividing it forever and never getting to the volume. So the sense of infinity is still present.
@@Mcboogler Precisely- it is not that mysterious. There will always be a little bit more of Pi, just as there will always be a little bit more volume to fill. In that sense, the infinity is present
I am a Mechanical Engineering drop-out of 17 years, and I literally followed every single step of this. Thank you, Mr. Walton, Mr. Roy and whoever my calculus professor was in college.
It should work. Since we're doing surface area, we'd have the area of the cylinder, which is 2пr dx. With the limiting/integral process, this should become exactly the surface area. When you do the actual integral, you get 2п ln|x|, with the bounds from 1 to infinity. Plugging the bounds in yields infinity.
Especially since the area of the conical frustum is the same as the area of a rectangle, A * B. (I realize now I was wrong, but he does end up approximating the area with a rectangle anyway, so the slant does not matter, since even the rectangle approximation diverges..)
This is the approach I take when I show this to A-Level students. But I also liked the approach here showing off a neat trick for proofs, whilst also avoiding having to go into integrals of arc lengths.
@@gabrielhermesson9926 yes both approaches yield infinity but they are not equal (e.g. if the problem were finite). For the volume it doesn't matter. I will try to explain this intuitively. Let dV be the volume of the truncated cone and dV'=(1/x)^2пdx the volume of the (less accurate) cylinder. There is some factor v(x) that depends on x, such that dV=v(x)*dV'. For dx->0 it is clear that v(x)->1 and dV=dV'. However, the situation is different for the lateral surface area. Let dA be the lateral surface area of the truncated cone and dA'=2п(1/x)dx the lateral surface area of the (less accurate) cylinder. There is some factor a(x) that depends on x, such that dA=a(x)*dA'. For dx->0 the factor is still a(x)->sqrt(1+(d(1/x)/dx)^2) and dA≠dA'. As you can see, the factor that correlates the differentials is what matters and if you do the integration (=summation), the factor will still affect the whole sum, even if dx->0. (e.g. by factoring out the factor if it were constant, you get the idea)
@@markussteiner1105 That makes sense, though it feels a little hand wavy with the factors. Also, it feels intuitive to my math sense, but not my common sense. I was ABOUT to argue that, due to that factor a(x) that the integral with cylinders would provide a smaller surface area--and by solving that you can then state that the actual surface area would be larger. But then I realized that is exactly what they did in the video around 14:50 (though they didn't state it as such with the cylinders).
does it count as fitting if it's less? I feel like if it doesn't fill the space it's a partial fit. You can put a person that's too sizes too small into a shirt, but there's extra space so it doesn't fit unless you get more person (I guess we usually think about that one the other way around).
@@CaptainOblivion looks like the problem here is that "fit" can mean either "is the right size for" and also "can be placed inside of". I was using the latter (e.g. "two units of paint would fit in the horn").
how 2 paint the horn: fill the inside with paint dump the paint out, this will result in a coating of paint inside the horn. turn the horn inside out refill the horn with paint and dump it out again
@@Bosstastical Just like the concept of the horn is theoretical, i could easily say go create an infinity long object to begin with, and then take an infinity long amount of time to get the paint inside. So finding the end and doing my process would take no time at all because you already have to use up an infinite amount of time finding the end and getting the paint in there in the first place. My point is just that the paint can in fact cover the surface area you just need an infinity thin layer of paint by squishing it.
I thought of that too, but then it occurred to me that the molecule size would affect the painting of the outside of the horn as well. An infinitely small horn would be smaller than a molucule. But then, what is the horn made of? Since you are the chemist you can say, what is bigger, a molecule of tin/copper or one of paint?
So if I make Lucifer's horn from a revolution of 2/X, put 3π of paint in it, put Gabriel's horn inside with π paint in it, I seemed to have accidentally painted Gabriel's horn inside and out. 😂
I think I can resolve the paradox. When we normally talk about "painting" a surface, we choose a thickness of paint, say 1 millionth of a meter. Then the amount of painted needed to paint the surface would be the surface area times the thickness. If the surface area is infinite, that's an infinite amount of paint, no matter how thin the layer of paint is. Does that mean you can't paint the surface? No, not really. You can instead use a variable thickness of paint. Suppose you paint the region from x=1 to x=2 to a thickness of 1 micrometer. Then the region from x=2 to x=3 to a thickness of 0.5 micrometers, and the region from x=3 to x=4 to a thickness of 0.25 micrometers, etc. The region from x=n to x=n+1 is given a thickness of 1/n micrometers. Then this only requires a finite amount of paint. That's what's happening when you fill Gabriel's horn with paint. You're painting the inside of the horn, but the thickness of paint gets less and less as you go.
Nope. That's contradicted by the video. Nowhere in the video do they talk about "normal" paint with a thickness. Only about *surface area*. The surface area of the outside of the horn is infinite! (The same is true of the inside, too, because the horn has no thickness, therefore the two surface areas are the same.) So even an infinite amount of zero-thickness paint is needed to coat the horn, never mind your "thinner and thinner" paint.
@@ModestJoke The comment of Daryl only deals with the apparent paradox that arises when one imagines actually filling the horn and sees that by doing so paint will cover its whole inner surface. What is brought by Daryl's comment is that the mathematical area does not equal the amount of paint needed to cover it: the thickness of your coat of paint is a critical value to determine the amount of paint you will use. Therefore if you want to apply Gabriel's horn problem to a somewhat more realistic setting, you need to consider the thickness of your coat of pain, and if your thickness decreases as suggested by Daryl, the amount of paint needed to cover the horn will remain finite. Therefore the paradox is solved.
I feel like part of the reason that this is confusing is that it implies that it is reasonable to compare sizes of different dimensions. In some sense, any 3-dimensional size larger than zero is larger than any 2-dimensional size, even infinite. A single drop of paint that has no limit to how thinly it can be spread can cover an entire plain.
That's what I was going to say. What is the surface area of the interior of a volume? If you reduce it to molecules then it's no longer infinite, and neither is the surface of the horn. This is essentially Hilbert's Curve in another form.
Exactly this. The whole thing is only paradoxical if we assume that the paint layer has a fixed thickness. Of course, when painting the inner side of the horn this is not possible. The thickness of the paint layer must go to 0 as the diameter of the horn does.
Yes, this is the intuitive resolution to the paradox. Real paint is made of small particles of finite size. If you had a paint made of infinitesimally small particles, then it would only take an arbitrarily small amount of it to paint Gabriel's horn. Since Pi is larger than many arbitrarily small numbers, you could use that volume of paint to cover the surface of Gabriel's horn and still have enough left over to fill it up like a cup.
This kind of rational explanation is refreshing to read. I'm getting _really_ tired of so called paradoxes that always seem to involve algebra with infinities. Once they even mention infinity, it should be obvious that all logic based on real-word physics flies out of the window.
It occurred to me that painting - in mathematical terms - involves a layer of paint that is infinitely thin. This would cancel out the infinitely large surface area if we were using even a very small amount of volume of paint. In other words, even the tiniest volume of paint can paint even the largest area - if the thickness of the paint is zero.
Indeed. In fact, this comparison of volume to surface area is nonsensical unless there is some defined thickness to the layer of paint. A mathematical "painting" of the surface with no thickness would literally require no paint at all in terms of volume (thickness times area).
@@shadowcween7890 No, the surface area is infinite whether your imagine it painted or not. This "thought experiment" implies that painting that surface would require an infinite amount of paint. The "visualization" of the infinite surface is the horn of infinite length. None of these purely mathematical constructs can be related to a volume of paint (i.e., an amount of paint) except a measurement of volume. Even this is pure hypothetical in this situation since at some point the molecules of paint will no longer fit in the ever decreasing space between the walls of the horn. The horn is not real and could not be constructed physically and could never be filled with something that will not fit inside. The whole thing with the paint is to force your mental concepts of purely mathematical constructs to be confounded by notions regarding actual physical objects in the real world.
@@googleyt2622 Yeah but if the wall of the horn is infinitely thin then the internal surface area of the horn would be equal to the exterior surface of the horn. If the horn is filled with paint then clearly, there is more than enough paint to coat the entire internal surface _and_ the external surface too. In fact, you would have enough paint to cover the external surface in a layer of paint that is almost half as deep as the radius of the horn at the points being painted. Right? I mean; okay, we can slice the horn to give us a way to calculate volume and surface area but we are missing the fact that each slice represents a disc of paint. And only the paint at the circumference of that disc is in contact with the internal surface of the horn. And of course, the amount of paint in contact with the internal surface of the horn is very small compared to the amount of paint contained in the rest of the disc. If you were to transfer all the paint _not_ in contact with the internal surface to the external surface, you would end up with a torus with almost twice the radius of the original disc, wouldn't you? And both the internal _and_ external surfaces would most definitely be coated with paint, right? The real question is: how many infinite horns could be coated with paint, inside and out, using paint from just one horn?
I was thinking about Brady’s question about being able to fill up the volume with a finite amount of paint but needing an infinite number of paint for the surface. I think the idea is this: in ANY finite amount of VOLUME of paint, we can cover an infinite surface area. If you put your finite volume of paint in a can, there are an infinite number of cross sections we can take of the can of paint, so we can cover an infinite surface area.
You can put the thing differently like this. If the horn is filled with paint, each cross section has different thickness of paint. As the horn narrows down to nothing as we reach infinite, the cross section of paint will narrown down to zero. The further down you go, the less paint you need. No matter how large surface area you cover with it, the less paint you need. The area goes to infinite, but the amount of surface area a volume of paint covers goes to infinte too. As you divide those two infinities with each other, they will cancel each other. The total volume of paint needed to paint the whole inner surface is π.
I remember when they covered this trick in a module on proof, my mind was blown. Such a simple idea. Similar to the monotone convergence theorem, just with divergence instead.
@CHARLEE SANTOS 2^1024 is a finite number. Meaning it is a measurable static value. Infinity is the concept of being infinite ie not finite, where there is no countable number to describe the situation, and it has no “end” or “value” so to speak
for those who can not wrap their head around this - imagine a pancake. The thinner you make it the bigger pane you need. As the thickness will approach 0, the area covered by the pancake will approach infinity. However, the volume is still the same.
While calculating the volume he ignored the curved edge of the infinitesimal slice, but while calculating the surface area he took into account the curve of the slice edge. Wondering why is that
I’m learning from Numberphile! I was like “one series will converge, but the other will diverge”. All thanks to Numberphile and other similarly awesome channels.
CORRECTION: At 8:55, Area of "Net of Conical Frustum" = A*B, but "A"arc should be taken at mid section of Frustum and not at the uppermost portion as shown in the video.
Yeah, I also didn't get why he chose not to derive it. Just have the area of the ring with radii a and b: pi(b^2-a^2) and multiply it by the proportion that the angle makes to 2pi, so pi(b^2-a^2)*theta/(2pi). Cancel the pi, expand the difference of squares and note that a*theta is the length of the inner curve, b*theta - outer curve.
Thank you. His statement didn't seem to match any kind of stretching I could imagine to transform the shaded shape into a rectangle of equivalent area. Area = 0.5 * (sector angle in radians) * (outer_radius^2 - inner_radius^2) Area = 0.5 * (sector angle in radians) * (outer_radius + inner_radius) * (outer_radius - inner_radius) Area = slant_height * (sector angle in radians) * (outer_radius + inner_radius) / 2 Area = slant_height * (outer_arc_length + inner_arc_length) / 2 Area = slant_height * average_arc_length As (slant height) approaches zero, the difference between (the outer arc length) and (the inner arc length) approaches zero. So the ( *_average_* of the outer and inner arc lengths) is approximately equal to either (the inner arc length) or (the outer arc length). He chose the inner arc length as the approximation he carried forward because that is where he was calculating 'y'. I went on a web search because I didn't want to do the "four pages of algebra" he described. Nothing I found seemed to suggest that the area of a similar shape of finite size would be the (inner arc length) * (slant height).
I tried to repeat the calculation and got AB(1+B/(2R)). Where R is the distance to the center. Since B is dx, and goes to 0, this doesn't impact the calculation, but the statement that its just AB seems wrong.
@@yoelcalev2763 Definitely. If you test A = R, AB as shown would give the area of any circle to be 0. My hunch is he sloppily labelled the wrong side of the shape and B should be the outside curve, but I haven't done the math.
@@yoelcalev2763 a) You need to expand again so you'll get [Adx +(dx)^2]/(2R) the (dx)^2 is sloppily said nothing compared to dx (which is already infinitesimally small). Provided that your formula was derived properly. b) If you go by arc lengt sector formula you'll find also a cure to the paradox: let C be the arc length of the outer circle so area is the difference between [C*(R+B)]/2 and [A*R]/2. Integration over both, individually culminates in a difference infinity minus infinity. Something quite often measured "finite" by physicist but highly frowned upon by mathematicians. c)Try to calculate {(A+C)/2}*B ;-)
@@yoelcalev2763 I got the same thing. Perhaps the shape is incorrect and it's not the area of a big sector minus the area of a small sector, but a rectangle in which the ends are shifted up into a smile, so those lengths are still vertical instead of slanted like in the video
@@harleyspeedthrust4013 To me it's like watching a French video when you don't speak a word of French. Just more headache-inducing because you get a liiitle bit of it and your brain automatically tries to understand the rest (and fails spectacularly, every single time).
I think you could actually paint the horn. The paradox comes from assuming that a finite volume of paint can't cover an infinite area but the horn proves that you can't assume that. Simply fill the horn with paint, freeze it, and take it back out of the horn. The paint is now a solid copy of the horn and thus has finite volume but infinite surface area, therefor a finite volume of paint can have and cover an infinite area.
@@BrazilianImperialistWouldn’t its volume always be infinity minus a lesser degree of infinity though? The horn’s external boundaries would still have to be infinitesimally slightly larger than its interior, no?
Explanation is that you CAN paint the horn. By pouring the paint into the horn you've painted it, though the coat of paint gets thinner and thinner and thinner as you go down the horn. This is one example of how you can paint an infinite surface with a finite amount of paint.
Why can't we just fill the horn with paint and let it dry? I think to answer this we need to ask what does it mean to paint a surface? One answer is: Painting a surface is to cover the suface with paint such that the paint forms a 1mm thick coating on top. (1mm is not important and can be any non-zero amount). By this meaning, to paint a surface with surface area S we need S*0.1mm amount of paint. Thus we need infinity*0.1mm=infinity amount of paint to paint the surface of the horn. Now let us try painting by pouring pi amount of paint into the horn. Because the horn becomes thiner and thiner as we go, after a point the horn will become thinner than 1mm and the paint inside it cannot form a 1mm coating required for 'painting the surface'. To sum up: If we try to paint the horn by pouring paint into it, as we go further near the mouth piece the coating of paint becomes smaller and smaller and the paint becomes fainter and less visible as we go and thus pouring the paint will not 'paint' the horn in the narrower regions.
Think of it this way: What is the surface area inside the paint in the bucket? It's a volume so it's like a larger order of infinite because it has an additional dimension to it. Take a peek at Hilbert's Curve (infinite in length but folded into a finite area), or see that the probability of selecting any single point (1D) from a continuous number line (2D) is exactly 0% even though you obviously did pick one, to get a sense of how moving between dimensions makes them not really comparable. You could use any ridiculously small 3D volume and cover the horn completely. The paradox here is setting up the expectation that you could never paint a 2D surface. The paint only runs out if you are considering it as molecules of paint bonding to an infinite brass surface, but if it's just a 3D, infinitely divisible volume, then it could be spread to nearly ( but still >0) thickness and cover any infinite surface. In fact, 0% of the volume makes contact with the interior of the horn, it's all still leftover to fill the horn to the rim.
Well, I won't be aplying, but I do feel bad that my professors were nothing like you when explaining. Yes, I can function in life without the higher math knowledge, but after finding your and Numberphile videos, my hunger increases with each one.
What I found puzzling is why the slant was taken into account for the area but not for the volume. Will the result be the same if the slant is used even for the volume?
So, someone like Matt Parker could, for instance, construct a long enough Gabriels horn, in order to fill it all the way up such that he gets an approximation for Pi? 🤔
No. This is pure maths and assumes that the horn has no width to its surface. If you were to construct something like it it wouldnt be infinite and the surface width would make it never come near a pi aproxymation.
@@egggge4752 yeah but for the real world wouldn't it be close enough? like anything after 3.14 for most things is just an academic exercise. Like if you make Gabriel's horn 100 units long your height is 0.0100, adding another 100 units only changes that height to 0.0050 units, another 100 height is 0.0033 units. The returns are diminishing. You've effectively already converged on the volume within a precision level of 3-4 sig figs.
It would depend on how close you get the interior of your horn to the ideal 1/x cross section and, if you get it decently narrow at its end, upon the viscosity and adhesion of your paint.
For those curious -- If a horn was constructed with a mouth radius of 1cm, it would need to be roughly 5 meters long (or longer) to achieve 3.14~ ml precision.
Yeah, any finite volume of paint, if you can spread it infinitely thin, can cover an infinite area. The paradox is the intuitive thinking that an infinite area requires infinite paint.
But if the coat of paint has a thickness of 0, then you haven't put any paint on. Zero paint means unpainted. In order for a surface to be painted, it has to have some positive thickness of paint on it. Thus, Gabriel's Horn takes an infinite amount of paint to paint. But since the interior becomes smaller and smaller, it eventually gets thinner than any thickness of the coat of paint you care to define. At some point, the "thickness" of the paint filling it up, will be less than the thickness of the coat of paint you want to apply to the outside. And it will get thinner and thinner, in a way that converges to a finite number. So the volume is finite, but the surface is infinite. You can reach a similar paradox with fractals. The Koch Snowflake has a finite area, but an infinite perimeter. Which means you can paint over the whole thing, but you can't trace its outline. The interesting thing about Gabriel's Horn is that you can create this paradox without having to make it a fractal. It's a very simple shape.
@@PhilBagels I think this is purely about how you interpret words that have no everyday meaning in the situation you are trying to describe here, and that's the full source of the paradox. Let's say that if I fill a volume completely bounded by surfaces with ideal liquid - there are no places where the liquid is not touching the surface(s), otherwise there would be void to fill. If I interpret that liquid as paint, I have definitely painted all inner surfaces that bound that volume for a very reasonable definition of painiting of an inside of anything, I don't have to reflect an arbitrary number for thickness of paint, if it's full, it is painted from inside (and for usual objects it's more than just painted, I could pour something out). Then for Gabriel's horn I have just painted it's infinite area (from inside) with finite amount of paint (and also could pour some out). Well, that is exactly what mathematically happens here. Its just that for infinitely thin and long horn neither filling the inside nor painting from outside is something you can do with physical paint, and that's where the "intuition" breaks and paradox emerges. All the horns pictured in the video are infinitely shorter and (on average) infinitely thicker than Gabriel's horn (the mathematical one, not the physical one), so they also don't help with building any reasonable intuition about Gabriel's horn's properties. Mathematically, it's just properties of the geometry, that are (invented/discovered to be) very different from everyday objects. And same goes for the fractals you mention.
Surface area and volume is two different units, it's like comparing apples to oranges. To know if we have enough paint to cover the area we need to state the amount of paint as an area. Thus if the area of the horn is infinite m2 then we need to get the, for arguments sake, pi m3 volume to a surface area. How do we do this? Let's spread the paint out onto a surface, to get the volume to become a surface we need to devide by the thickness of the paint on the surface that we are covering. Since this is a purely mathematical construct and we are not dealing with phisics we can say that the thickness will tend towards 0. Thus lim x->0 (pi/x) and since the thickness will always be positive we can say that the area that the paint can cover tends towards infinity.
Yeah, the paradox runs on the assumption that the amount of surface a perfect mathematical paint can cover is dependant on it's volume, when it's not. I'm surprised the video didn't mention this...
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I like how the paradox is making people assume you cannot paint an infinite surface. Yes, you can, just have to make the pain infinitely thin
Exactly, they make a horrible mistake claiming that you can never paint the outside, or even inside because the surface area is infinite. If course you can, in fact it requires no paint at all (an infinitesimal fraction of PI)
@@ADHD9009 The calculation didn't account for surface tension, it just gave the volume of the horn, which is finite, meaning it can be completely filled up using paint leaving no part of the horn untouched. But at the same time the surface area is infinite which means it shouldn't be possible to touch one entire side of the horn.. Because that would require an infinite amount of paint. But it is possible because the volume is finite
@@rysea9855 It does actually account for surface tension. Just in an abstract way. At some point x the diameter will be to narrow for paint or even protons to fit into. But it will still continue on towards infinity.
This is an awesome production - very clean and extremely well explained by Mr. Crawford. The derivation of dS I would never come up with on my own. It's elegant to see the various tools of mathematics applied to a problem like this. The animations on this channel are strictly top drawer.
I guess this is similar to cutting a square in half infinitely, with the formula 1/(2^n). The total area is 1, but you can always add total perimeter length when you cut the next square in half.
A non-paradoxical paradox: the reason there seems to be a paradox here is because of an unstated (and absurd) assumption, that the thickness of the paint is uniform throughout the length of the trumpet! If the paint thickness tapers off as 1/x, adding another 1/x term to the integral, the volume of surface paint will be finite. This is exactly why the interior surface is coated with a finite volume of paint: that paint must taper off at least as fast as 1/x to fit inside the trumpet.. I didn’t read all 3,904 comments to see if anyone else mentioned this and apologize if they did. And I find it disappointing that the author of the post didn’t realize this. It is not at all unique to have a finite volume(surface) enclosed with an infinite surface(boundary), e.g., the Mandelbrot set is a finite surface area but its boundary is (hideously) infinitely long.
Really appreciated the walkthrough of the calculus, I always enjoy explanations of calculus that really focus on logical thought and physical characteristics and don't just go "this is the answer"
Not if, for example, he had the tattoo wrap around his wrist. That would make the smallest part touch the largest part (because it's wrapping around), so he could just stop there and say that it keeps going.
@@BlackKillerGamer If you mean viewed along the x axis, yeah, finite ink then. It might be hard to know what one's looking at, might just look like a filled circle depending on the rendering choices, but then the story.
Is the first Numberphile video to calculate in full an integral using antiderivatives? After the video from a couple weeks ago with ‘e’ where they did a derivative. Real calculus is so rare on Numberphile!
Assuming the horn is of negligible thickness, this claim asserts that the finite volume of paint that fills the horn will not coat the horn. However, if the horn is of infinitesimal thickness, the surface area of the inside and the outside are the same. Since the inside is holding the finite volume of paint, that volume of paint is touching the entire surface area. CLEARLY that volume of paint can also completely cover the surface of the horn.
It's the same paradox. You just painted an infinitely large surface with a finite amount of paint. There are many versions of this, Zeno's paradox, Koch's paradox.
The thing is, paint needs a thickness to be seen. It doesn't matter how arbitrarily thin you coat of paint is, there will be an infinite amount of horn where the diameter is too small to fit that much paint inside, so it'll look unpainted.
@@DanielHarveyDyer if you want to get into actual physical limitations of paint then at some point the diameter of the horn is smaller than a paint molecule leaving an unfillable void.
Yeah, imagine if you would pour a finite amount of (“mathematical”) liquid on an infinite plane. If it spread across only a part of the plane it would have some thickness. You could then take the top half of the liquid and spread it across another part of the infinite plane. The liquid will now have twice amount of area but half the thickness. This can be repeated infinite times such that the thickness becomes infinitely small but the area become infinitely big. I claim there is no paradox in that sense. You can pain ANY infinite area with finite paint
Jo a You are dealing with two infinities so you have to show that they are the same size of infinity. There are infinite kinds of infinity and they are not equal to each other.
7:23 can't we still assume the small limiting part is still cylinder, use 2πrdx and then just integrate it from 1 to ∞? S.A=integral of(2πdx/x) from 1 to ∞ would give S.A=2π(ln∞) which would still tend toward infinity, but an easier calculation
Except here it's not really out of nowhere, since the entire thing is circular. The general difference between calculating a plain old area under a curve and this kind of rotational volume calculation is literally just a pi in front of the integral. Same paradox arises with jsut the plain intagration, no 3D needed, and there's no pi. But this version showcases the bizzareness more, I guess.
For anybody bothered by the lack of intuition in the result, maybe this will help. What we have here is a finite volume contained within an infinite surface area. Consider the fractal coastline problem, wherein the coast length of a piece of land increases towards infinity as you account for smaller and smaller bumps, but the area of the land contained within the coastline stays approximately the same. This is an analogous situation where we have a finite area contained within a boarder of infinite length. Just increase everything by one dimension to get back to Gabriel's horn. If you were comfortable with the fractal coastline problem, now maybe you're comfortable with Gabriel's horn. If you aren't comfortable with the fractal coastline, then I've just doubled your anxieties; you're welcome.
How about the reverse? Can we have the infinite measure of (n+1) dimensional figure enclosed in finite measure of (n) dimensional figure? Seems obvious, but how do one goes about proving it? Or just definition of higher measure relies on finite measure of lower one?
The trouble comes when thinking of the paint as a surface. It isn't, but it's tempting to think of it that way as if it's equivalent to the surface area. But the paint has some volume, even if we make it infinitely thin, otherwise we couldn't fill the horn. Think of the paint being between your original horn and a slightly larger horn, and the paint is filling the gap between the two horns. We know the paint can't have infinite volume, or it wouldn't fit in the horn. Note that making the outer horn a little bigger does not make its volume infinite. We can make the outer horn as big or as small as we want as long as it's less than infinity, so the paint can be as thick or thin as we want it. But as long as the paint is between the gap between the two horns, it must be a finite amount, since it takes up finite volume. If we took off the horn when the paint is dry, it would be clear we still don't have an infinite amount of paint. And yes, the paint would still have infinite surface area, but still finite volume, just like the horn.
This is not a paradox; any finite amount of volume of paint can be painted on an infinite surface, as long the thickness of the paint is infinitely thin....
Yep. This. It's a mixing of two limits (numbers approaching ends). One goes to infinity, and one goes to a smaller number (possibly still infinity, but a smaller one ;) ).
the "paint" is a metaphor. the paradox is that if this object existed it would have a finite volume but an infinite surface area. don't think of it as paint if that confuses you, but it's there to make the problem easier to think about
The area for the side of the conical frustum is incorrect!! The "A" should be the length of the arc measured in the middle of the frustum, kind of like its midriff.
I was thinking the same thing when I saw it but didn't bother to math it out. It would make sense that if you're "averaging" the slope you would also "average" the distance from the center point, so to speak, intuitively. That makes perfect sense.
I was looking for this comment. I did the derivation because he said it was 4 pages of algebra. I did it on a sticky note but got an extra term of angle*A²/2, which accounts for that extra length if you moved B to the middle. I guess it still works out in the video because as the slices get infinitesimally small, the inner length approaches the middle length.
Thank you so much for pointing this out. I did the algebra 3 times using A times B and get getting an extra term. I thought I was going nuts till you pointed this out. Works out perfectly using the "midriff" times the length.
I haven't worked this out but I suspect that if B is very small relative to A then the area approximates to AB. And that approximation gets better and better as B gets smaller which it does as B is delta x which is taken to the limit, ie towards zero.
One really cool detail about this: Early on, the voice behind the camera points out that the volume of the horn appears infinite because you can always keep adding a little bit more at the end, infinitely, which is technically true. So the fact that the answer resolves to Pi, a number that famously has an infinite number of digits, really fits this perfectly. You can always keep adding smaller and smaller amounts at the end, and yet it is a finite constant. The volume is finite, but indefinite.
Why filling the horn with paint doesn't mean you can paint it: Although it isn't mentioned here, it has to be assumed that painting the horn means giving it a coat of paint with non-zero thickness (otherwise you don't need any paint to do it). Let's call the thickness t. Since the area of paint held by a section is pi r^2 dx and the paint required to paint that section is 2pi r t dx, there r
I think it makes sense somehow. In fact, with real world paint, made out of atoms of finite size, it would not be possible to fill the horn either. At some point, as x gets bigger and the diameter shrinks, it would get stuck (and probably before that because of surface tension anyway). That's the same reason why it would be impossible to paint the outside area: a finite volume of paint cannot be infinitely spread as particles cannot be spread. However, if one considers a mathematically ideal paint, made out of infinitely small particles, then the horn could be both filled and painted with it !
In real world paint, the smallest radius would be the radius of a paint molecule. That's big enough to make the horn extremely short compared to how much paint you need to fill it. For example, if the horn opening has a radius of 1m, and for argument's sake I'll say a paint molecule is about 0.1 micrometer in radius, then the horn is an impressive 1000km long. However, the surface area is just 2*pi*13.816 square meters = approximately 87m^2 (roughly the walls of an apartment) and you have 3142L (or 830 gal.) of paint to do the job. You can easily give it a few coating (thousands of atoms thick) and have 3000L of paint left.
But I'd like to ask how one does mathematically "paint" the surface of something? When we suppose that "painting" is a surjection from the set of points of a 3-d object (e.g. a bucket of paint) to the set of points of a 2-d surface, it is surely possible to "paint" the infinite AREA with finite VOLUME of paint. This answers the question from 16:02 too.
I agree, that a 3d coating of paint on the horn would have finite volume. However, again that paint would have infinite area. I find it perplexing, but basically my intuition of infinity is not applicable. Comparing area to volume doesn’t really make sense anyway. Volume provably finite, area provably infinite.
@Dalton Long There is no essential difference between 3D paint (which fills the horn) and 2D paint (which covers the surface); the whole horn and its surface are both subset (also infinite set) of R^3, and there exists a surjection from the horn to its surface. It means it possible to cover all the surface (infinite area) with the points from the horn (finite volume, but infinite points). That was my point.
@@rc5989 I like how you describe the paint as a 3d coating. It might be a better representation of real-life paint to consider the coating of paint has non-zero thickness. In this case, you are right; after we painting the surface of the horn, we will end of making another weird 3d object with finite volume and infinite surface area. Anyway, I was talking about what if we apply a paint with zero thickness as an absolute 2d object.
It is possible to paint the infinite area with any volume of mathematical (zero thickness) paint. However, if you do this at a finite rate (m^2 painted per hour), it still takes you infinite time. Then again, if you pour paint into an object at a finite speed, it still takes infinite time for the paint to reach the (un)end of the object, so you can't fill it either... ;-)
@@junyoungshin4719 a useful equivalence between the horn and its surface being a bijection sounss wrong to me. pure set bijections can take random paints and map them to random points just based on the cardinality of both these being 2^alef0. a useful idea of equivalence would have to at least be a homotopical equivalence, maybe even a homeomorphism, so that you can spread your paint continuously. neither of those equivalences happen tho.
I initially learned about Gabriel's horn from a book I'm reading called "The Math Book" and came to this video to learn more about it. I watched this video twice now and just noticed the book I'm reading is on the top shelf in the video! Great explanation by the way!
Explanation of the paradox for anyone who is curious: The paradox is the result of using two seperate models for how it is treating the paint in the two different scenarios. In the "Fill" scenario we treat the paint as a pure mathematical volume; which can be compressed and shrunken infinitely. This is not how paint actually exists in the real world. Which is to say, in the real world paint is made up of a finite number of paint particles with their own, immutable volume. However, in the "Paint the surface" scenario he switces from treating the paint as the mathematical idea of volume to treating it like actual paint, which can only be spread over a finite area because it is made up of a finite number of paint particles, and once they run out, you are out of paint. So now we can choose between two different paint models: If we treat paint as it actually is (as a finite collection of particles with finite, immutable volume) then you can neither paint the surface area nor can you completely fill the horn. You cannot paint the surface because you will eventually run out of particles, and you cannot fill the horn completely because eventually the horn will be too thin for the paint particles to fit. However, if we stick to the "Fill" example's model and treat paint like a pure mathematical volume. Then you can easily both fill the inside and also paint the surface area. In fact, when treating paint like this, *Any* amount of "paint" would be sufficient to paint the surface, both inside and out trivially. This is because you can (as demonstrated in his explanation) morph the "paint" into a shape wherein it maintains it's volume, but has infinite surface area and is wrapped around the horn. Interestingly enough, while treating paint consistently in this manner, you could not both fill the horn completely And paint the outside if you only had π units of paint. However if you had any amount of paint more, say π+0.0000000000000001 units of paint, you could both fill the horn completely and paint the outside completely as well.
You can treat them as fractions for the most part and you'll be gucci. If you ever solve differential equations you basically abuse the differentials and beat them into submission with math hacks
The solution of the paradox is simply that there are pi volumetric units, but infinite surface units. However, since there are infinite surfaces in a volume, you can "paint" infinite surfaces with any finite volume of paint.
@@pedroaleb i'll argue that since my calc professor completely skipped over the 'why' of a lot of things (for example, he never talked about ds or the area of a conical fruntum, simply gave the formulas as needed with no explanation) and as a result I barely retained the info. This guy managed to actually explain it in a 17 min video which I applaud
@@reilandeubank i agree. he does it very well. but there is certainly a difference between school, wich you have to attend even if you dont want to, and youtube videos and other content that you actively go after and chose to watch
Regardless of how many angels can dance on the head of the pin, the music they're dancing to is definitely played by an angel blowing on the infinitesimally small end of an infinitely long trumpet
If it assumed that the Horn has no thickness then the inside surface area and the outside area are the same therefore the horn is already painted inside and out. Also if you have 1 unit of thickness for the paint and slice the Horn to this thickness then the maths for a cylinder of this height reduces to :- Paint volume inside r^2 , Paint required outside 2r.
When we counted the volume we sliced the horn into thin cylinders, and when we counted the surface we sliced it into less approximated shapes (that have bigger volume). That doesn't feel right. If you ask me, we counted volume for different solid and surface for a different one. I know it was done most surely right, but I don't get it why we didn't slice the horn for same shapes counting both volume and surface. Would love an answer or a link for further reading. Thanks.
The idea is that in the limit as dx goes to 0, the approximation of the volume gets arbitrarily close to the actual volume of the slices, since the volume not counted looks like π(1/x^2-(1/x+dx)^2)dx ~ dx^2 -> 0. On the other hand, if we were to calculate the surface area of the slice approximated as a cylinder, it would not approach the correct surface area, since the error is proportional to sqrt(1+(dy/dx)^2). Incidentally, since we throw away that term in the end, approximating the surface area with cylinders is sufficient for this proof, since it is an underestimate of the true surface area that still diverges.
That's actually not important, because he gets rid of the square root anyways. The point is that the volume is proportional to 1/(x^2) and the surface is proportional to 1/x. Now if you integrate you get 1/x for the volume and ln(x) for the surface. If x gets infinite 1/x goes to zero but ln(x) goes to infinity. Therefore the volume is finite and the surface is infinite.
(copied from my reply to another comment): As I mentioned in the video, I did this calculation (if only to convince myself) which was several pages of algebra, so we decided to exclude it from the video. If you want to try it for yourself then use the slanted edge for the volume also, and then when taking the limit as dx tends to zero (ie. doing the integral) you'll find that we ignore the terms in dx^2 at leading order - and these are exactly the terms that come in when using a slanted edge for the volume as opposed to a straight edge. For the surface area, the 'slanted edge' terms are order dx and so we do need them.
@@TomRocksMaths I think the question is, though, why bother calculating the exact integral when you could instead calculate a lower bound using the innermost cylinder at each approximation? In other words you know the length of ds at each slant has to be greater than if you simply cut horizontally straight across at the inner base with width dx. So the surface area has to be greater than 2pi I(1/x) dx - the same inequality you end up with in the video only without having to actually figure out the approximate length of ds.
"wouldn't filling the horn paint the inside" We're defining painting as covering the surface with a layer of paint of finite thickness. When you fill the horn, the 'thickness' of the paint covering the inside tends to an infinitesimal
If you declare a finite thickness of paint, then the problem of painting the inside solves itself. Because once the horn gets too small, that layer of paint won’t fit inside.
Honestly that doesnt make any sense of it. Filling it would technically be a maximum thickness layer of paint on the entire inside. Filling any container with zero wall thickness and no wall touching areas would definitely mean you could paint the surface. Inside area is the same and you can cover the entire area with paint when filled.
@@noahtraylor525 I am unsure which side you are arguing. There isnt even a "thickness" paint in this thought experiment, just area. The op makes it sound as though filling cannot also be coating, which it definitely can be. If you fill a container, it would also be coated.
I’m just trying to say that it doesn’t seem that paradoxical to me. You can fill it, but you can’t cover it. Well this is because it is an infinite horn. So there is always another another length to be covered. But you can fill it because the volume converges to pi. Basically the surface area to volume ratio is infinite. It’s not that you don’t have enough paint, because any volume of paint can cover an infinite area. It’s just that you will never finish covering it. So I’m just trying to say that it doesn’t seem that paradoxical to me
Exactly. I thought the area would be an integral of 2*PI(1/x)dx right a way, you are basically adding up circumferences of the slices with dx so small that it does not matter. Apparently, it does.
(copied from my reply to another comment): As I mentioned in the video, I did this calculation (if only to convince myself) which was several pages of algebra, so we decided to exclude it from the video. If you want to try it for yourself then use the slanted edge for the volume also, and then when taking the limit as dx tends to zero (ie. doing the integral) you'll find that we ignore the terms in dx^2 at leading order - and these are exactly the terms that come in when using a slanted edge for the volume as opposed to a straight edge. For the surface area, the 'slanted edge' terms are order dx and so we do need them.
The longer the horn gets, the higher its volume is, and the higher its surface area is. The paradox only arises if you conceive of a limit as being the actual measure of something. If you state it as "as the horn gets longer, the relative increase in its surface area diminishes at a slower rate than the relative increase in its volume, because surface area increases with r and volume increases with r squared," then it doesn't seem paradoxical at all.
That's why it's called a Gabriel's horn. I mean how many angels can fit on a pin's head right? Angels may very well be differential beings. A differential being wouldn't have any issues with travelling to infinity to blow a horn.
If people get too serious, they don't notice they take things far too literal. Gabriel is an angel. And angels can do miracles. In this case, he could fly an infinite length and blow you a horn partita by Johann Sebastian. And he even flies back to you, to say "And that's what you can learn in heaven, dude! And now I got to fly there again, I forgot my mouthpiece." And now you'll start protesting the music would never reach the bell of that horn. You're beginning to understand.
@@DeadlyDanDaMan How about quantum effects? Entanglement and two connected things happening at the same time? And the first quantum computers are running. Still not perfect, but it IS an application of theory, and it will explode processor performance in the near future. A quantum computer is entirely different from the digital ones. And you can call it supernatural, for it is hard to explain WHERE the calculus is happening. In parallel universes? Still, there are some WORKING small quantum computers around the globe.
I feel like this, much like the perimeter of a fractal, can be painted. You would need to cover an infinite surface area in this case, but that's entirely possible. You would have to dump the entire horn in paint and the outside would be painted. If you made another infinite horn say, twice as large at every point so your horn fits inside it, you could fill it with paint while your horn is there (wouldn't take much paint) and your horn would be painted. This reminded me of like an infinitely long, higher dimensional fractal. You can easily have a fractal with an infinite perimeter and very finite area, and you could easily surround the whole thing with paint to paint the edges.
If you have paint that is so infinitely thin that it is able to run down into the tip you are also able to use a finite amount of it to spread an infinitesimally thin layer across the whole surface.
I’ve thought of a second paradox that arises when you try to fill the horn with paint. Assume we are in a frictionless, surface tension-less, atmosphere-less etc., etc. world but gravity is the same as earth’s. You start pouring the paint and the first bit is continually accelerating down the horn. However, because the horn is infinitely long, it will never reach the bottom in a finite amount of time. That will mean there is always space to be filled further down the horn, so you actually cannot empty your pi units of paint from its can. But, since the first paint poured will always be accelerating, it can’t “get in the way” of the paint you continue to pour. So you should be able to empty the can. But, of course that would mean there is still unfilled space inside the horn. Which would mean, actually you can’t empty the can and so on.
That's the problem with using real world analogies for mathematical concepts that approach limits. As soon as you go real world, you also have to deal with the Planck length, but the paint must be infinitely thin, or it couldn't fill the parts of the horn where 2y < Planck length. Then you end up with a situation where the paint "particles" have no dimension (or we would need an additional dy term to account for the paint application, or the paint wouldn't be able to fit into the infinitely small part of the horn) but they still have volume, or how else could they fill the horn...?
Of course you can paint it! Create a second horn with 2* the radius and fill the gap between the 2 horns with paint. You'll need 3*Pi paint ... Actually you can use as little paint as you want with the second horn of radius 1+epsilon with epsilon going to zero. Or just cover with a layer of paint that goes to zero thickness fast enough with x...
9:00 Wrong!!!! This is for A in the middle of the part not the small curvy side. And the proof is not that long: The Area of the difference of the discs of radius r+B/2 and r-B/2 is SD = π(r+B/2)^2 - π(r-B/2)^2 = ... = 2πBr But we only consider a part of length A of the perimeter of the circle of radius r. The ratio is A/2πr So S = SD * A/2πr = AB 😎
I'm a student of topography, I'm gonna be very simple, you are basically just stretching a solid with finite volume infinitely, so no matter how long it becomes it's always gonna have that finite volume, and for Gabriel's horn that's why it's width becomes infinitesimally small in order to compensate for that finite volume the shape got. Same goes for 2d shapes.
You are right, that didn't even cross my mind. Only difference is, that in this paradox the shape is finite. Which is the same paradox, a finite shape of infinite length. Just like pi. A finite value, yet it has an infinite number of decimals. That's the core of this paradox. But basically it's just Zeno's paradox in a different setting. As in your example. Half the width and double the length into infinity. (Thinking of a rectangle for simplicity). Same area, infinitely long circumference.
@@LOKOFORLOKI1 u can actually fill it if u puor the paint in it, the paint will start filling up at nearly infinite speed at the beginning and then as it reaches the upper wider end it will fill up a bit slower and it will take the same amount of time it would have taken to fill up a container with the same constant volume.
I like the paradox, thank you. Also, your enthusiasm is contagious 🙂 Using a cylinder for the surface calculation is much faster, but the detour via this weird shape, the name of which I can't even remember, was interesting too. Thanks so much!
Great video, thanks! Question though: When you do the volume, you go over dx because the slant of the surface doesn't matter, but then for the surface area you do use ds because now somehow the slant does matter. Can you elaborate on why the slant doesn't either matter for both, or for none? Thanks!
Actually, he *did* use that approximation, he just wasn't clear about it: The approximation used for the volume is ds = dx Apply to SA = 2*pi*(1/x)*ds -> 2pi(1/x)dx As shown above, the integral between 1 and infinity of 1/x is, itself, infinite. And since for all values of x, ds > dx, the true value of SA will be infinite.
The slant does not matter for volume because as you take shorter and shorter segments for the approximation, the outer ring's length approaches 0, thus removing the slant from the equation. Technically, there is an infinitesimal slant still present, but it's so small that it isn't worth consideration when doing this sort of problem. Surface area is different, in that you aren't subdividing to work it out, and intuitively the slant makes a huge difference because it makes the length of the horn infinite. The outer surface is unending, therefore no amount of paint will ever cover it. The inner volume, however, is bounded and reduces further along the horn, so much so that there is a number that is larger than the sum of it. Because of that, the volume is finite. I hope this helped clarify and didn't just add more confusion.
@@mavericksolomon395 Hi, I can understand the idea that the slant does not matter for volume as you said. But I don't get the sentence below: " Surface area is different, in that you aren't subdividing to work it out, and intuitively the slant makes a huge difference because it makes the length of the horn infinite. " As far as I know, the way to derive formulas for surface area and volume stems from the idea of "splitting" an interval into an infinite number of infinitesimal intervals and approximating each of them by an appropriate surface or shape. Therefore, I don't quite understand what you meant by not subdividing to work out the surface area? Thanks.
@@mavericksolomon395 Even if you estimated using a cylinder in this case you'd still get an infinite result. The surface area of each cylinder would be the circumference of the circle times the width, 2πR dx. Since R is 1/x, you'd wind up with 2π * Int(1, infinity) 1/x dx. But if you tried to do a calculation on a finite slice of the horn, your answer would be slightly off for the surface area. But not for the volume. It would have made more sense to do both of these with a conic frustum. Volume of a conic frustum is 1/3 π h (R1^2 + R1*R2 + R2^2) ... In this case h is dx and both R1 and R2 are 1/x. So Int(1, infinity) of 1/3 π dx (3 (1/x)^2) ... Do the arithmetic and it simplifies to the same answer as when using cylinders.
I remember once feeling like it made sense that you could fill it but couldn’t paint the outside, but now I can’t see how I felt that. It really is a paradox.
Not a paradox, as a given volume of mathematical paint can paint an infinite surface area due to having infinitely low viscosity. For illustration, if you poured the paint out on a perfectly flat plane it would continue to spread infinitely without ever changing volume.
"Gabriel must have an incredibly small mouth if he's blowing in the small end". End?
Beginning? Says the Gabriel with his small mouth on the other side
This is guy doing mathematic, what a world are we living in?
well after all, considering how many angels can dance on the tip if a pin, it makes sense the mouthpiece must approach zero. something like that.
Yeah
What he means is that there is no end as it's an infinite object, so there is no part of to put in your mouth and blow.
Yo, making an approximation of this horn and filling it up then measuring the volume would be a great new addition to Matt's weird ways of calculating π
@@c0ldc0ne ... calculated every March 14th.
Yeah, use approximation techniques to see how long your horn would have to be to be accurate out to...say like 10 decimal places, and then make a prototype.
Man, I want a 3D printed one of these now.
Doesn’t one have to use pi during the creation of the horn? I am a bit skeptical to the claim that pi comes out of nowhere
@@Nia-zq5jl If you 3D print it, each layer would be a circle, with radius 1/x, no pi here.
@@Nia-zq5jl Do you use pi when drawing circles using a compass?
“It’s always nice when π appears out of nowhere”
Starts pushing squares against each other on a frictionless surface.
What KIND of pie, though? Rhubarb or pumpkin? No thank you....
For any who don't know this reference, this is a reference to a 3Blue1Brown video where he calculated pi based on how many times blocks with different mass ratios (all multiples of 10) hit each other before the bigger block and smaller block diverged. It's an amazing video
@@rtpoe its prob a typo, he will change it
@@JGMeador444 but its squares not cubes
@@JGMeador444 not true, the Pi comes from taking logarithm of the mass ratio. CollisionCount = Log10 10^(I N Pi) = I N Pi, where I is imaginary.
These were extraordinarily intelligent questions asked. "If you fill the horn with paint, haven't you painted it too?" Excellent point.
I have a can of Banach Tarski math paint that covers infinite surface areas such as the Gabriel's horn. Works great.
Yeah, the issue is that at some moment the width of the horn is less than an atom, so you can't really "paint the inside". But this is a physisists' problem that math people can't be bothered by.
I think the real key to the paradox is that these are different units. Any finite amount of volume can theoretically spread into an infinite area if you can spread it infinitely thin, so you CAN paint the horn (e.g. the inside, by filling it) with a finite amount of paint.
Exactly. No matter how much 2D paint you use it will always add up to zero volume.
There is a limit to how thinly paint can be applied. 1 molecule thick is the minimum. So theoretically, Any finite amount of volume can not be spread over an infinite surface area. These type of problems are where pure math separates from applied maths. The paradox makes the same assumption you made here. The math isn't wrong, its the assumption that you can limit ds as dx goes to zero in the physical universe, which you can't.
If you're saying paint can only be applied at one molecule thick and not infinitely thin, than the diameter of the horn also has to have the limitation of a one molecule thick wall, and it can only get as small as one molecule at its point, making it finite. They are both infinite or both finite
@@keyonastring Exactly.
The serious problem consists of a minimal thickness of ink. So, from outside, you cannot paint it with a finite amount of ink (pi). Now, considering the minimal thickness, filling it from inside will not guarantee that the interior surface area will be all painted. There comes a point where the regional volume will not suffice to paint the surrounding surface area. It's simple. I don't think this has to do with the boundary between applied and pure math. Also, the volume is clearly finite and surface area infinite.
The animations on this channel just keep getting better.
:: Slaps horn :: "This baby can hold π paint"
😭😭😭😭
pi paint, otherwise known as gravy
nniice))))))))))))))
@Someone Who is but NONE what do u mean infinite? Pi certainly isn’t infinite
@@shivam5105 but it doesn't end, so the amount of paint would not end.
It’s amazing how intuitive his explanation of the volume of a solid of revolution is.
It's probably not the solid that you imagine ... it is not defined at the center. ←----------O-----------→ You can't rotate something around the very something that doesn't exist and expect it to represent the volume defined by the exterior surface area, because it is undefined at the axis. That undefined center has infinite volume. That is what he leaves out to produce the 'paradox' ... by using a slight of hands you are tricked into believing it. I don't know if it is by accident ... maybe he is tricking himself. But I suspect he is smart enough to know better.
@@ejrupp9555 I don't get you, can you be more specific ?
@@ejrupp9555 Disc integration is a completely legitimate method of computing the volume of a solid of revolution. The axis in the center doesn't matter; it doesn't contribute any amount of volume. If you want, you can use disc integration to calculate the volume of a finite solid, such as a cone, and you will get the correct result. You can even use disc integration to calculate the volume of a finite section of Gabriel's horn, which could theoretically be verified with a physical model, and as you take bigger and bigger sections, the volumes you get will asymptotically approach π.
@@ejrupp9555 you're speaking absolute nonsense
@@kameronpeterson3601 well you are speaking about being @infinity ... so ...
Interestingly, the word “frustum” is taken directly from the Latin which means “crumb, or morsel” so comparing it to a pizza crust was more accurate than he ever imagined!
but also, Conical Frustum = C(onical F)rust(um) = Crust (+ some detritus)
@@CaTastrophy427 Onical Fum
@@RuthlessDutchman oyfum
That man has the disheveled nature and wild look in his eyes of someone who has looked into infinite and saw it staring back.
He looked into the void. The void ran away.
i thought he looked like dantdm
Or some spell-slinging anime wizard boy.
He even has some sick pokeball ink.
He has the look of someone with gender disphoria.
"You can't argue with the maths." Well. I CAN argue with the maths. I'll lose. They kicked me out of the math club a long time ago and told me never to return.
I can argue too, I just can't win...
I got booted from debate club because my only argument was “well f@ck you too!” Lol
AN idiot will argue the math based on their own ignorance... perfect case in point... the fool that thinks one over infinity is equal to zero... mic drop.
@@slthbob I smiled ... seems we live in the same reality ... not many of us here.
Be thankful they didn't try to drown you
Thank you for bringing our attention to this bug. Our team is working on it and we hope to have it patched in the next update.
LOL
Hahaha 😂
damn, devs really do be taking all the fun out of the game.
"It's always nice when π appears out of nowhere" *sad revolution noises*
I know what you mean but in this case I think it came from the volume of the section and after intigration pi was all that was left.
exactly what i though lol. "we plug pi in the beginning and get it at the end. out of nowhere, incredible!"
@@captainoates7236 His point is that is not "out of nowhere". Pi is defined by the geometry of a circle. When there is a circle, pi usually appears some way or another. In some cases we can all agree that it is hard to see where there is a circle in the problem, but in this case it was really obvious. It is a figure you get by rotation along an axis. Pi was bound to show up
You actually always get pi when doing a revolution 😃
@@Sigmav0 you just took that whole thing to a new level haha
A nice way to think about this (and resolve the paradox), is that we are basically thinking about 2 different types of paint as if they are the same. When you fill the horn you are thinking of 3-dimensional paint particles, but when you are painting the horn's surface you are only thinking of 2-dimensional paint plates.
If we would try to fill the horn with 2D paint, we would never finish. In fact, we wouldn't even be able to fill a tiny bit of it, because our 2D paint plates have height zero. This matches our intuition that something that fills out a volume must be more than something that only covers the surface.
Contrary, if we would try to paint the surface of the horn (let's say the outside) with the 3D paint, we are making it unintentionally thicker everywhere as well (presumably by a constant epsilon > 0). If a covered surface is epsilon > 0 thicker than before, then the thickness of the horn doesn't converge to zero like explained at the beginning of the video, but converges to a constant > 0. Now, of course the paint we would need to cover the surface is still infinite, but the paint we would need to fill the entirety of the horn including the epsilon-thick surface is also infinite.
But the volume inside wouldn't change no matter how thick of a layer of paint is on the outside, wouldn't it?
But then again, it wouldn't be surprising anymore that you would need an infinite amount of 3D paint to paint the outside when the outside is always bigger than a constant > 0.
The paint we need to cover the entity of the horn is just pi not infinite even if we consider the tiniest of volume.
Yeah, Imagine if you would pour a finite amount of (“mathematical”) liquid on an infinite plane. If it spread across only a part of the plane it would have some thickness. You could then take the top half of the liquid and spread it across another part of the infinite plane. The liquid will now have twice amount of area but half the thickness. This can be repeated infinite times such that the thickness becomes infinitely small but the area become infinitely big.
Finite paint can paint any infinite area in that sense, and in that sense there is no paradox
I like the abstract theoretical dimensional analysis of paint, but volumes and surface areas aren't types of paint, theres no reason we can't have finite-volume-filling infinite-areas, finite-area-filling infinite-lengths, finite-length-filling points, etc., space filling curves and convergent series are just a part of calculus and without them there would be no calculus
Exactly. It's like how a coastline has infinite length but the area of the country is not infinite. This "paradox" may get you into Oxford but you would lose marks in grade school for not including the units!
"Welcome to Sherwin-Williams. How much paint do you want?"
"Pi."
There's your horn of Pi Paint Mr. Backes!
Careful though, it's infinitely long. But we do sell infinite limos in the infinity aisle, just outside this dimension. Have fun shopping!
You get Pi cubic-units of paint in a paint can that is 2 units wide and 1 unit high. Not standard paint can dimensions at Home Depot, but easy enough to imagine.
You don't need exactly pi units, just at least that much, so I'll just take 4 and have some left over.
@@QW3RTYUU just have the horn be set with the skinnier end touching the edge of an infinity pool. Easy peasy engineering solution.
Absolutely loved it where you skipped the pain and substituted for “bigger than 1” just to get to the proof. Something I didn’t really encounter in maths at school. Mind blown in lots of ways on this video
This was used in my Calc 2 class in things like the comparison test, where we can compare a very complicated series to something that is easy to compute to prove divergence or convergence. For example we know that 1/((n^2)+1) < 1/(n^2), and since we know an infinite series of 1/(n^2) converges, we then know the first converges as well. the same can also be applied to divergence
"Gabriel's horn has a finite volume"
Never have I so violently disagreed with such a logically sound and well explained argument/claim.
I remember going over this in college. It wasn’t called Gabriel’s Horn, we didn’t talk about paint, and it wasn’t called a paradox. We simply derived and noted the relationship between the two solutions as being counterintuitive. A surface of revolution with a finite volume and an infinite surface area. The infinite surface area is no problem at all. The horn is infinitely long, of course the surface area is infinite. It’s the volume that’s tricky. But this may help- remember that this is a bottomless pit. The volume approaches a limit, but never gets there. You can always add a fraction of what you just added, dividing it forever and never getting to the volume. So the sense of infinity is still present.
@@HollywoodF1 Indeed sir... Pi is a ratio not a definable numerical value...
@@Mcboogler Well said.... Pi is a ratio not a definable value
@@HollywoodF1 Wow, you seem very smart
@@Mcboogler Precisely- it is not that mysterious. There will always be a little bit more of Pi, just as there will always be a little bit more volume to fill. In that sense, the infinity is present
I am a Mechanical Engineering drop-out of 17 years, and I literally followed every single step of this. Thank you, Mr. Walton, Mr. Roy and whoever my calculus professor was in college.
No problem!
@@danielwalton0416 lol imagine if this was actually real
@@danielwalton0416 you being his professor I mean
@@mihailmilev9909 it is
@@jenm1 doubtful, but it could be considering this is a math video.
Anyone else wondering why he didn't just use cylinders again for the surface area?
It should work. Since we're doing surface area, we'd have the area of the cylinder, which is 2пr dx. With the limiting/integral process, this should become exactly the surface area. When you do the actual integral, you get 2п ln|x|, with the bounds from 1 to infinity. Plugging the bounds in yields infinity.
Especially since the area of the conical frustum is the same as the area of a rectangle, A * B.
(I realize now I was wrong, but he does end up approximating the area with a rectangle anyway, so the slant does not matter, since even the rectangle approximation diverges..)
This is the approach I take when I show this to A-Level students. But I also liked the approach here showing off a neat trick for proofs, whilst also avoiding having to go into integrals of arc lengths.
@@gabrielhermesson9926 yes both approaches yield infinity but they are not equal (e.g. if the problem were finite). For the volume it doesn't matter. I will try to explain this intuitively.
Let dV be the volume of the truncated cone and dV'=(1/x)^2пdx the volume of the (less accurate) cylinder. There is some factor v(x) that depends on x, such that dV=v(x)*dV'. For dx->0 it is clear that v(x)->1 and dV=dV'.
However, the situation is different for the lateral surface area. Let dA be the lateral surface area of the truncated cone and dA'=2п(1/x)dx the lateral surface area of the (less accurate) cylinder. There is some factor a(x) that depends on x, such that dA=a(x)*dA'. For dx->0 the factor is still a(x)->sqrt(1+(d(1/x)/dx)^2) and dA≠dA'.
As you can see, the factor that correlates the differentials is what matters and if you do the integration (=summation), the factor will still affect the whole sum, even if dx->0. (e.g. by factoring out the factor if it were constant, you get the idea)
@@markussteiner1105 That makes sense, though it feels a little hand wavy with the factors. Also, it feels intuitive to my math sense, but not my common sense.
I was ABOUT to argue that, due to that factor a(x) that the integral with cylinders would provide a smaller surface area--and by solving that you can then state that the actual surface area would be larger. But then I realized that is exactly what they did in the video around 14:50 (though they didn't state it as such with the cylinders).
I think you could paint the outside by placing the horn into a horn made by revikving 2/x and filling the larger horn which itself has finite volume
You have a very serious point here and because of displacement you'd only need whatever the volume of 2/x - pi is as well.
7:12 "Can't fit any more. Can't fit any less"
...you definitely could fit less though!
Probably just to make it more cool
Can't fit less if you're adding exactly pi.
does it count as fitting if it's less? I feel like if it doesn't fill the space it's a partial fit. You can put a person that's too sizes too small into a shirt, but there's extra space so it doesn't fit unless you get more person (I guess we usually think about that one the other way around).
@@CaptainOblivion looks like the problem here is that "fit" can mean either "is the right size for" and also "can be placed inside of". I was using the latter (e.g. "two units of paint would fit in the horn").
"Conical Frustum" sounds like an uncomfortable urological condition.
proctological
never trust a frustum
thrustum
Not to mention his formula seems off, it should be Area = mean({minor arc-length, major arc-length})×B
I have my frustum snipped off, but that must have been frustrating.
how 2 paint the horn:
fill the inside with paint
dump the paint out, this will result in a coating of paint inside the horn.
turn the horn inside out
refill the horn with paint and dump it out again
Wouldn't the shapes fit inside each other and stack? Fill one with paint then dip a second inside.
you would be infinitely trying to turn the horn inside out
WOW!
I think, in that case.
you'll dump out (Pi - infinty that stayed on the surface) paint.
booom
wanna try twice?
that's expensive
@@1988ryan1 Come back when you work out how to get an infinitely long object inside another.
@@Bosstastical Just like the concept of the horn is theoretical, i could easily say go create an infinity long object to begin with, and then take an infinity long amount of time to get the paint inside. So finding the end and doing my process would take no time at all because you already have to use up an infinite amount of time finding the end and getting the paint in there in the first place.
My point is just that the paint can in fact cover the surface area you just need an infinity thin layer of paint by squishing it.
as a chemist i was assuming the volume would be dependent on the size of a molecule of paint haha. great video i love your channel
I thought of that too, but then it occurred to me that the molecule size would affect the painting of the outside of the horn as well. An infinitely small horn would be smaller than a molucule. But then, what is the horn made of? Since you are the chemist you can say, what is bigger, a molecule of tin/copper or one of paint?
So if I make Lucifer's horn from a revolution of 2/X, put 3π of paint in it, put Gabriel's horn inside with π paint in it, I seemed to have accidentally painted Gabriel's horn inside and out. 😂
Mathematician: Rotate.
Pi: It's Free Real Estate.
π
Attn: Jim Boonie
Ro tat e
I think I can resolve the paradox. When we normally talk about "painting" a surface, we choose a thickness of paint, say 1 millionth of a meter. Then the amount of painted needed to paint the surface would be the surface area times the thickness. If the surface area is infinite, that's an infinite amount of paint, no matter how thin the layer of paint is. Does that mean you can't paint the surface?
No, not really. You can instead use a variable thickness of paint. Suppose you paint the region from x=1 to x=2 to a thickness of 1 micrometer. Then the region from x=2 to x=3 to a thickness of 0.5 micrometers, and the region from x=3 to x=4 to a thickness of 0.25 micrometers, etc. The region from x=n to x=n+1 is given a thickness of 1/n micrometers. Then this only requires a finite amount of paint.
That's what's happening when you fill Gabriel's horn with paint. You're painting the inside of the horn, but the thickness of paint gets less and less as you go.
Nope. That's contradicted by the video. Nowhere in the video do they talk about "normal" paint with a thickness. Only about *surface area*. The surface area of the outside of the horn is infinite! (The same is true of the inside, too, because the horn has no thickness, therefore the two surface areas are the same.) So even an infinite amount of zero-thickness paint is needed to coat the horn, never mind your "thinner and thinner" paint.
@@ModestJoke an infinite area of zero thickness paint can be painted by an arbitrary amount of paint.
@@ModestJoke The comment of Daryl only deals with the apparent paradox that arises when one imagines actually filling the horn and sees that by doing so paint will cover its whole inner surface. What is brought by Daryl's comment is that the mathematical area does not equal the amount of paint needed to cover it: the thickness of your coat of paint is a critical value to determine the amount of paint you will use. Therefore if you want to apply Gabriel's horn problem to a somewhat more realistic setting, you need to consider the thickness of your coat of pain, and if your thickness decreases as suggested by Daryl, the amount of paint needed to cover the horn will remain finite. Therefore the paradox is solved.
@@mcaelen2539 exactly if you fix the thickness of the paint to 0.01% of the diameter of the horn the amount of paint becomes finite.
@@ModestJoke You are a prat.
I feel like part of the reason that this is confusing is that it implies that it is reasonable to compare sizes of different dimensions. In some sense, any 3-dimensional size larger than zero is larger than any 2-dimensional size, even infinite. A single drop of paint that has no limit to how thinly it can be spread can cover an entire plain.
That's what I was going to say. What is the surface area of the interior of a volume? If you reduce it to molecules then it's no longer infinite, and neither is the surface of the horn. This is essentially Hilbert's Curve in another form.
Exactly this. The whole thing is only paradoxical if we assume that the paint layer has a fixed thickness. Of course, when painting the inner side of the horn this is not possible. The thickness of the paint layer must go to 0 as the diameter of the horn does.
Lowe’s left the paint mixer on overnight.
Yes, this is the intuitive resolution to the paradox. Real paint is made of small particles of finite size. If you had a paint made of infinitesimally small particles, then it would only take an arbitrarily small amount of it to paint Gabriel's horn. Since Pi is larger than many arbitrarily small numbers, you could use that volume of paint to cover the surface of Gabriel's horn and still have enough left over to fill it up like a cup.
This kind of rational explanation is refreshing to read. I'm getting _really_ tired of so called paradoxes that always seem to involve algebra with infinities. Once they even mention infinity, it should be obvious that all logic based on real-word physics flies out of the window.
It occurred to me that painting - in mathematical terms - involves a layer of paint that is infinitely thin. This would cancel out the infinitely large surface area if we were using even a very small amount of volume of paint. In other words, even the tiniest volume of paint can paint even the largest area - if the thickness of the paint is zero.
Indeed. In fact, this comparison of volume to surface area is nonsensical unless there is some defined thickness to the layer of paint. A mathematical "painting" of the surface with no thickness would literally require no paint at all in terms of volume (thickness times area).
@@googleyt2622 The whole thing with the paint is to visualise that the surface area is infinite
@@shadowcween7890 No, the surface area is infinite whether your imagine it painted or not. This "thought experiment" implies that painting that surface would require an infinite amount of paint. The "visualization" of the infinite surface is the horn of infinite length. None of these purely mathematical constructs can be related to a volume of paint (i.e., an amount of paint) except a measurement of volume. Even this is pure hypothetical in this situation since at some point the molecules of paint will no longer fit in the ever decreasing space between the walls of the horn. The horn is not real and could not be constructed physically and could never be filled with something that will not fit inside.
The whole thing with the paint is to force your mental concepts of purely mathematical constructs to be confounded by notions regarding actual physical objects in the real world.
This makes this make sense to me. The word problem simply doesn't work in the real world.
@@googleyt2622 Yeah but if the wall of the horn is infinitely thin then the internal surface area of the horn would be equal to the exterior surface of the horn. If the horn is filled with paint then clearly, there is more than enough paint to coat the entire internal surface _and_ the external surface too. In fact, you would have enough paint to cover the external surface in a layer of paint that is almost half as deep as the radius of the horn at the points being painted. Right?
I mean; okay, we can slice the horn to give us a way to calculate volume and surface area but we are missing the fact that each slice represents a disc of paint. And only the paint at the circumference of that disc is in contact with the internal surface of the horn. And of course, the amount of paint in contact with the internal surface of the horn is very small compared to the amount of paint contained in the rest of the disc. If you were to transfer all the paint _not_ in contact with the internal surface to the external surface, you would end up with a torus with almost twice the radius of the original disc, wouldn't you? And both the internal _and_ external surfaces would most definitely be coated with paint, right?
The real question is: how many infinite horns could be coated with paint, inside and out, using paint from just one horn?
Why cannot the area of a slice be 2*pi*(1/x)*dx ?
Because the error would be proportional, and not decreasing as dx goes towards 0. The error would be only be dependent on where you are along the horn
I agree. This is more simple and area is directly infinite.
he said 10:03 slant is important. it's because of something to do with the Taylor expansion
I was wondering the same thing. Integral of 2*pi*(1/x)dx would have diverged as well.
"Let's go!"
"GRAMMAR?!"
I was thinking about Brady’s question about being able to fill up the volume with a finite amount of paint but needing an infinite number of paint for the surface. I think the idea is this: in ANY finite amount of VOLUME of paint, we can cover an infinite surface area. If you put your finite volume of paint in a can, there are an infinite number of cross sections we can take of the can of paint, so we can cover an infinite surface area.
Thank you for this. I was approaching a similar answer, but hadn't quite gotten there yet, and in the meantime I was going a bit crazy.
You can put the thing differently like this. If the horn is filled with paint, each cross section has different thickness of paint. As the horn narrows down to nothing as we reach infinite, the cross section of paint will narrown down to zero. The further down you go, the less paint you need. No matter how large surface area you cover with it, the less paint you need. The area goes to infinite, but the amount of surface area a volume of paint covers goes to infinte too. As you divide those two infinities with each other, they will cancel each other. The total volume of paint needed to paint the whole inner surface is π.
@@timokokkonen5285 you cannot divide and cancel two infinities
@@pulkitmohta8964 he literally just did.
@@pulkitmohta8964 You can if the two rates at which the infinities are being approached are the same (i.e. L'Hopital's Rule)
"If you are bigger than infinity, you are infinity." -Dr. Tom Crawford
I remember when they covered this trick in a module on proof, my mind was blown. Such a simple idea. Similar to the monotone convergence theorem, just with divergence instead.
@CHARLEE SANTOS infinity is not 2^1024, it's infinity.
@CHARLEE SANTOS infinity is a concept not a number dude.
@CHARLEE SANTOS 2^1024 is a finite number. Meaning it is a measurable static value. Infinity is the concept of being infinite ie not finite, where there is no countable number to describe the situation, and it has no “end” or “value” so to speak
@CHARLEE SANTOS I’m letting it sink in and I’m still confused
for those who can not wrap their head around this - imagine a pancake. The thinner you make it the bigger pane you need. As the thickness will approach 0, the area covered by the pancake will approach infinity. However, the volume is still the same.
While calculating the volume he ignored the curved edge of the infinitesimal slice, but while calculating the surface area he took into account the curve of the slice edge. Wondering why is that
Thanks that made it a lot more easy to understand, also reminded me that we're not talling about real physical objects
I’m learning from Numberphile! I was like “one series will converge, but the other will diverge”. All thanks to Numberphile and other similarly awesome channels.
this^^ :)
@@TomRocksMaths it’s the guy from the video!!! Hello!
@@romerobryan83 Hello!
CORRECTION: At 8:55, Area of "Net of Conical Frustum" = A*B, but "A"arc should be taken at mid section of Frustum and not at the uppermost portion as shown in the video.
Thanks, I was wondering why my calculations didn't work! ^^'
Yeah. It was coming as AB + B^2(θ/2).
Yeah, I also didn't get why he chose not to derive it. Just have the area of the ring with radii a and b: pi(b^2-a^2) and multiply it by the proportion that the angle makes to 2pi, so pi(b^2-a^2)*theta/(2pi). Cancel the pi, expand the difference of squares and note that a*theta is the length of the inner curve, b*theta - outer curve.
Thank you. His statement didn't seem to match any kind of stretching I could imagine to transform the shaded shape into a rectangle of equivalent area.
Area = 0.5 * (sector angle in radians) * (outer_radius^2 - inner_radius^2)
Area = 0.5 * (sector angle in radians) * (outer_radius + inner_radius) * (outer_radius - inner_radius)
Area = slant_height * (sector angle in radians) * (outer_radius + inner_radius) / 2
Area = slant_height * (outer_arc_length + inner_arc_length) / 2
Area = slant_height * average_arc_length
As (slant height) approaches zero, the difference between (the outer arc length) and (the inner arc length) approaches zero.
So the ( *_average_* of the outer and inner arc lengths) is approximately equal to either (the inner arc length) or (the outer arc length).
He chose the inner arc length as the approximation he carried forward because that is where he was calculating 'y'.
I went on a web search because I didn't want to do the "four pages of algebra" he described. Nothing I found seemed to suggest that the area of a similar shape of finite size would be the (inner arc length) * (slant height).
“The surface area of a conical frustum is just AB”
No talk to me I angy
I tried to repeat the calculation and got AB(1+B/(2R)). Where R is the distance to the center. Since B is dx, and goes to 0, this doesn't impact the calculation, but the statement that its just AB seems wrong.
@@yoelcalev2763 Definitely. If you test A = R, AB as shown would give the area of any circle to be 0. My hunch is he sloppily labelled the wrong side of the shape and B should be the outside curve, but I haven't done the math.
I think ab+(theta b^2)/2 is equivalent to AB(1+B/(2R)) because:
ab(1+b/(2R)) = ab+ab^2/(2r)
a/r = theta
ab+ab^2/(2r) = ab+ (theta b^2)/2
@@yoelcalev2763 a) You need to expand again so you'll get [Adx +(dx)^2]/(2R) the (dx)^2 is sloppily said nothing compared to dx (which is already infinitesimally small). Provided that your formula was derived properly.
b) If you go by arc lengt sector formula you'll find also a cure to the paradox: let C be the arc length of the outer circle so area is the difference between [C*(R+B)]/2 and [A*R]/2. Integration over both, individually culminates in a difference infinity minus infinity. Something quite often measured "finite" by physicist but highly frowned upon by mathematicians.
c)Try to calculate {(A+C)/2}*B ;-)
@@yoelcalev2763 I got the same thing. Perhaps the shape is incorrect and it's not the area of a big sector minus the area of a small sector, but a rectangle in which the ends are shifted up into a smile, so those lengths are still vertical instead of slanted like in the video
Thanks!
I love Tom's enthusiasm and he's great at explaining things. Very nice!
"... we will need to do an integral."
Ok. That's my stop. Have a nice weekend!
Why bro integrals are fun
You've been doing integrals ever since you had to find the area or volume of a simple shape, such as a square, cube, circle, or cylinder.
@@harleyspeedthrust4013
To me it's like watching a French video when you don't speak a word of French. Just more headache-inducing because you get a liiitle bit of it and your brain automatically tries to understand the rest (and fails spectacularly, every single time).
@@mesa176750 That's quite a stretch. Collegial integrals are nothing like that.
He does walk through the integrals really nicely
so this is the 3D version of the “mathematicians walk into a bar” joke, where the bartender just serves them 2 pints
Their bellies may be full of beer, but their throats remain dry.
...when all they really needed was some pi to be full
They should have known their limits.
I think you could actually paint the horn. The paradox comes from assuming that a finite volume of paint can't cover an infinite area but the horn proves that you can't assume that. Simply fill the horn with paint, freeze it, and take it back out of the horn. The paint is now a solid copy of the horn and thus has finite volume but infinite surface area, therefor a finite volume of paint can have and cover an infinite area.
I agree with your comment that Gabriel's Horn actually shows that a finite amount of paint can cover an infinite volume.
Construct the horn out of a very gradually permeable paper (like a kind of felt) and fill it with paint.
You are not painting it, you are just filling it
@@BrazilianImperialist True but in each case the entire surface is in contact with a substance.
@@BrazilianImperialistWouldn’t its volume always be infinity minus a lesser degree of infinity though? The horn’s external boundaries would still have to be infinitesimally slightly larger than its interior, no?
I adore how excited he is about this problem. Absolutely captivating!
We gotta give props to Brady for asking such a wonderful question, 16:04.
That was a brilliant question. Currently looking for an explanation in the comments...
Explanation is that you CAN paint the horn. By pouring the paint into the horn you've painted it, though the coat of paint gets thinner and thinner and thinner as you go down the horn. This is one example of how you can paint an infinite surface with a finite amount of paint.
Why can't we just fill the horn with paint and let it dry?
I think to answer this we need to ask what does it mean to paint a surface?
One answer is: Painting a surface is to cover the suface with paint such that the paint forms a 1mm thick coating on top. (1mm is not important and can be any non-zero amount).
By this meaning, to paint a surface with surface area S we need S*0.1mm amount of paint. Thus we need infinity*0.1mm=infinity amount of paint to paint the surface of the horn.
Now let us try painting by pouring pi amount of paint into the horn. Because the horn becomes thiner and thiner as we go, after a point the horn will become thinner than 1mm and the paint inside it cannot form a 1mm coating required for 'painting the surface'.
To sum up: If we try to paint the horn by pouring paint into it, as we go further near the mouth piece the coating of paint becomes smaller and smaller and the paint becomes fainter and less visible as we go and thus pouring the paint will not 'paint' the horn in the narrower regions.
Think of it this way: What is the surface area inside the paint in the bucket? It's a volume so it's like a larger order of infinite because it has an additional dimension to it. Take a peek at Hilbert's Curve (infinite in length but folded into a finite area), or see that the probability of selecting any single point (1D) from a continuous number line (2D) is exactly 0% even though you obviously did pick one, to get a sense of how moving between dimensions makes them not really comparable. You could use any ridiculously small 3D volume and cover the horn completely. The paradox here is setting up the expectation that you could never paint a 2D surface. The paint only runs out if you are considering it as molecules of paint bonding to an infinite brass surface, but if it's just a 3D, infinitely divisible volume, then it could be spread to nearly ( but still >0) thickness and cover any infinite surface. In fact, 0% of the volume makes contact with the interior of the horn, it's all still leftover to fill the horn to the rim.
I've used this a few times as an Oxford Maths admissions interview question, but I guess I can't anymore now you all know the answer... :p
Hehe... BPRP also did a video about this a while back, so yeah.
Well, I won't be aplying, but I do feel bad that my professors were nothing like you when explaining. Yes, I can function in life without the higher math knowledge, but after finding your and Numberphile videos, my hunger increases with each one.
Better that than my Oxford interview where they made me sketch tons of different trig functions. Always hated sketching graphs freehand.
What I found puzzling is why the slant was taken into account for the area but not for the volume. Will the result be the same if the slant is used even for the volume?
Yes this bothers me as well - there's approximation going on in a problem that deals with infinitesimal measurements. Is that not relevant?
Thank you! That is exactly what has been bothering me from the start !
So, someone like Matt Parker could, for instance, construct a long enough Gabriels horn, in order to fill it all the way up such that he gets an approximation for Pi? 🤔
No. This is pure maths and assumes that the horn has no width to its surface. If you were to construct something like it it wouldnt be infinite and the surface width would make it never come near a pi aproxymation.
@@egggge4752 yeah but for the real world wouldn't it be close enough? like anything after 3.14 for most things is just an academic exercise. Like if you make Gabriel's horn 100 units long your height is 0.0100, adding another 100 units only changes that height to 0.0050 units, another 100 height is 0.0033 units. The returns are diminishing. You've effectively already converged on the volume within a precision level of 3-4 sig figs.
It would depend on how close you get the interior of your horn to the ideal 1/x cross section and, if you get it decently narrow at its end, upon the viscosity and adhesion of your paint.
The error bars are going to be pretty big when constructing it, but that's never stopped him before
For those curious -- If a horn was constructed with a mouth radius of 1cm, it would need to be roughly 5 meters long (or longer) to achieve 3.14~ ml precision.
It's a trick. The thickness of the coating of paint we're applying is zero, which is why it doesn't "use up" any volume of paint at all!
We painting nothing on the outside
Yeah, any finite volume of paint, if you can spread it infinitely thin, can cover an infinite area. The paradox is the intuitive thinking that an infinite area requires infinite paint.
But if the coat of paint has a thickness of 0, then you haven't put any paint on. Zero paint means unpainted. In order for a surface to be painted, it has to have some positive thickness of paint on it. Thus, Gabriel's Horn takes an infinite amount of paint to paint. But since the interior becomes smaller and smaller, it eventually gets thinner than any thickness of the coat of paint you care to define. At some point, the "thickness" of the paint filling it up, will be less than the thickness of the coat of paint you want to apply to the outside. And it will get thinner and thinner, in a way that converges to a finite number. So the volume is finite, but the surface is infinite.
You can reach a similar paradox with fractals. The Koch Snowflake has a finite area, but an infinite perimeter. Which means you can paint over the whole thing, but you can't trace its outline.
The interesting thing about Gabriel's Horn is that you can create this paradox without having to make it a fractal. It's a very simple shape.
@@PhilBagels Best explanation
@@PhilBagels I think this is purely about how you interpret words that have no everyday meaning in the situation you are trying to describe here, and that's the full source of the paradox.
Let's say that if I fill a volume completely bounded by surfaces with ideal liquid - there are no places where the liquid is not touching the surface(s), otherwise there would be void to fill. If I interpret that liquid as paint, I have definitely painted all inner surfaces that bound that volume for a very reasonable definition of painiting of an inside of anything, I don't have to reflect an arbitrary number for thickness of paint, if it's full, it is painted from inside (and for usual objects it's more than just painted, I could pour something out). Then for Gabriel's horn I have just painted it's infinite area (from inside) with finite amount of paint (and also could pour some out). Well, that is exactly what mathematically happens here. Its just that for infinitely thin and long horn neither filling the inside nor painting from outside is something you can do with physical paint, and that's where the "intuition" breaks and paradox emerges. All the horns pictured in the video are infinitely shorter and (on average) infinitely thicker than Gabriel's horn (the mathematical one, not the physical one), so they also don't help with building any reasonable intuition about Gabriel's horn's properties.
Mathematically, it's just properties of the geometry, that are (invented/discovered to be) very different from everyday objects. And same goes for the fractals you mention.
Surface area and volume is two different units, it's like comparing apples to oranges. To know if we have enough paint to cover the area we need to state the amount of paint as an area. Thus if the area of the horn is infinite m2 then we need to get the, for arguments sake, pi m3 volume to a surface area. How do we do this? Let's spread the paint out onto a surface, to get the volume to become a surface we need to devide by the thickness of the paint on the surface that we are covering. Since this is a purely mathematical construct and we are not dealing with phisics we can say that the thickness will tend towards 0. Thus lim x->0 (pi/x) and since the thickness will always be positive we can say that the area that the paint can cover tends towards infinity.
Yes that answers Brady's question
Yeah, the paradox runs on the assumption that the amount of surface a perfect mathematical paint can cover is dependant on it's volume, when it's not. I'm surprised the video didn't mention this...
I like how the paradox is making people assume you cannot paint an infinite surface. Yes, you can, just have to make the pain infinitely thin
... or have an infinite amount of paint.
Well, it’s more about how you would never be able to paint it in a finite amount of time
@@fun_nuggets2514 Sure, but you'd never be able to make that much paint in finite time either.
Exactly what I was thinking.
Exactly, they make a horrible mistake claiming that you can never paint the outside, or even inside because the surface area is infinite. If course you can, in fact it requires no paint at all (an infinitesimal fraction of PI)
But.. By filling the horn, the paint literally touches the horn.. aAaAA infinity melts my brain every time
But only because the paint has surface tension
Edited on July 23 2021. This was as an example of bringing molecular density into it.
@@ADHD9009 The calculation didn't account for surface tension, it just gave the volume of the horn, which is finite, meaning it can be completely filled up using paint leaving no part of the horn untouched. But at the same time the surface area is infinite which means it shouldn't be possible to touch one entire side of the horn.. Because that would require an infinite amount of paint. But it is possible because the volume is finite
@@rysea9855 It does actually account for surface tension. Just in an abstract way. At some point x the diameter will be to narrow for paint or even protons to fit into. But it will still continue on towards infinity.
@@rysea9855 you can fill an infinite surface area only with paint that you can spread infinitely thin.
@@cerealdude890 What you're saying is that ∞/∞
This is an awesome production - very clean and extremely well explained by Mr. Crawford. The derivation of dS I would never come up with on my own. It's elegant to see the various tools of mathematics applied to a problem like this. The animations on this channel are strictly top drawer.
I guess this is similar to cutting a square in half infinitely, with the formula 1/(2^n). The total area is 1, but you can always add total perimeter length when you cut the next square in half.
A non-paradoxical paradox: the reason there seems to be a paradox here is because of an unstated (and absurd) assumption, that the thickness of the paint is uniform throughout the length of the trumpet! If the paint thickness tapers off as 1/x, adding another 1/x term to the integral, the volume of surface paint will be finite. This is exactly why the interior surface is coated with a finite volume of paint: that paint must taper off at least as fast as 1/x to fit inside the trumpet.. I didn’t read all 3,904 comments to see if anyone else mentioned this and apologize if they did. And I find it disappointing that the author of the post didn’t realize this. It is not at all unique to have a finite volume(surface) enclosed with an infinite surface(boundary), e.g., the Mandelbrot set is a finite surface area but its boundary is (hideously) infinitely long.
Really appreciated the walkthrough of the calculus, I always enjoy explanations of calculus that really focus on logical thought and physical characteristics and don't just go "this is the answer"
I'm guessing he doesn't have a tattoo of Gabriel's Horn because it'd take infinite ink?
Sounds like a challenge...
Not if, for example, he had the tattoo wrap around his wrist. That would make the smallest part touch the largest part (because it's wrapping around), so he could just stop there and say that it keeps going.
@@vibaj16 or it could be facing outwards / inwards
@@BlackKillerGamer If you mean viewed along the x axis, yeah, finite ink then. It might be hard to know what one's looking at, might just look like a filled circle depending on the rendering choices, but then the story.
Ooh! Formula for crossection! NO! Formula shadow!
This is exactly what we want. *The real maths*
Either human brain should be wrong or mathematics
I'm glad you approve :)
Newton time traveling to the future to learn calculus from Numberphile, so he can go back in time and pretend to invent it.
I prefer complex maths
Is the first Numberphile video to calculate in full an integral using antiderivatives? After the video from a couple weeks ago with ‘e’ where they did a derivative. Real calculus is so rare on Numberphile!
Assuming the horn is of negligible thickness, this claim asserts that the finite volume of paint that fills the horn will not coat the horn. However, if the horn is of infinitesimal thickness, the surface area of the inside and the outside are the same. Since the inside is holding the finite volume of paint, that volume of paint is touching the entire surface area. CLEARLY that volume of paint can also completely cover the surface of the horn.
I came to the same conclusion that the volume includes the surface. This requires a proper answer.
It's the same paradox. You just painted an infinitely large surface with a finite amount of paint. There are many versions of this, Zeno's paradox, Koch's paradox.
The Gabriel's horn in the ear was a classy touch, well played
ngl it's kind of the opposite of classy
I'm glad you noticed
@@ogg5 "well, that's just like... Your opinion man"
@@TomRocksMaths may I ask you a really tough question?
@@K.E.L-117 sure
the inside and outside surface areas are the same so just fill it and then invert it to fully paint the outside.
The thing is, paint needs a thickness to be seen. It doesn't matter how arbitrarily thin you coat of paint is, there will be an infinite amount of horn where the diameter is too small to fit that much paint inside, so it'll look unpainted.
@@DanielHarveyDyer if you want to get into actual physical limitations of paint then at some point the diameter of the horn is smaller than a paint molecule leaving an unfillable void.
Yeah, imagine if you would pour a finite amount of (“mathematical”) liquid on an infinite plane. If it spread across only a part of the plane it would have some thickness. You could then take the top half of the liquid and spread it across another part of the infinite plane. The liquid will now have twice amount of area but half the thickness. This can be repeated infinite times such that the thickness becomes infinitely small but the area become infinitely big.
I claim there is no paradox in that sense. You can pain ANY infinite area with finite paint
@@Nia-zq5jl Makes sense, I agree. In other words, there are infinitely many areas in a finite volume
Jo a You are dealing with two infinities so you have to show that they are the same size of infinity. There are infinite kinds of infinity and they are not equal to each other.
Loving the vibe of this guy. Pokemon tats, bright eyed when talking math, and lip piercing is rarely seen in the same person.
I'm also like him , I like both piercing and maths 😂😂
Yeah, none of the other guys with Pokémon tats that talk about math have lip piercings.
I'm just kidding btw, it's a joke
@@jjackandbrian5624 lol
@@TomRocksMaths DAMN ITS YOU WOAHHHHH
7:23 can't we still assume the small limiting part is still cylinder, use 2πrdx and then just integrate it from 1 to ∞?
S.A=integral of(2πdx/x) from 1 to ∞ would give
S.A=2π(ln∞) which would still tend toward infinity, but an easier calculation
My thought exactly. Especially because B should go to 0 as it is equal to dx. I stopped watching after that thought. Thx.
Instead of getting stick bugged in math the equivalent is getting Pi'd where Pi just pops out of nowhere lol
Except here it's not really out of nowhere, since the entire thing is circular. The general difference between calculating a plain old area under a curve and this kind of rotational volume calculation is literally just a pi in front of the integral. Same paradox arises with jsut the plain intagration, no 3D needed, and there's no pi. But this version showcases the bizzareness more, I guess.
Next video: Gabriel's wedding cake, the discrete version of Gabriel's horn
Vsauce subscriber, I see
Hey Vsauce, Michael here
@@jimi02468 Or is it?
you can make the sponge cake, but never completely ice it :)
@@mondolee yes you can: pour icing over it until every part is iced
For anybody bothered by the lack of intuition in the result, maybe this will help. What we have here is a finite volume contained within an infinite surface area. Consider the fractal coastline problem, wherein the coast length of a piece of land increases towards infinity as you account for smaller and smaller bumps, but the area of the land contained within the coastline stays approximately the same. This is an analogous situation where we have a finite area contained within a boarder of infinite length. Just increase everything by one dimension to get back to Gabriel's horn. If you were comfortable with the fractal coastline problem, now maybe you're comfortable with Gabriel's horn. If you aren't comfortable with the fractal coastline, then I've just doubled your anxieties; you're welcome.
How about the reverse? Can we have the infinite measure of (n+1) dimensional figure enclosed in finite measure of (n) dimensional figure?
Seems obvious, but how do one goes about proving it? Or just definition of higher measure relies on finite measure of lower one?
The trouble comes when thinking of the paint as a surface. It isn't, but it's tempting to think of it that way as if it's equivalent to the surface area. But the paint has some volume, even if we make it infinitely thin, otherwise we couldn't fill the horn.
Think of the paint being between your original horn and a slightly larger horn, and the paint is filling the gap between the two horns. We know the paint can't have infinite volume, or it wouldn't fit in the horn. Note that making the outer horn a little bigger does not make its volume infinite. We can make the outer horn as big or as small as we want as long as it's less than infinity, so the paint can be as thick or thin as we want it. But as long as the paint is between the gap between the two horns, it must be a finite amount, since it takes up finite volume. If we took off the horn when the paint is dry, it would be clear we still don't have an infinite amount of paint.
And yes, the paint would still have infinite surface area, but still finite volume, just like the horn.
This is not a paradox; any finite amount of volume of paint can be painted on an infinite surface, as long the thickness of the paint is infinitely thin....
Yep. This. It's a mixing of two limits (numbers approaching ends). One goes to infinity, and one goes to a smaller number (possibly still infinity, but a smaller one ;) ).
My man in the video literally took you step-by-step and yet you still missed it
Can’t be thinner than a certain limit which varies based on the material
the "paint" is a metaphor. the paradox is that if this object existed it would have a finite volume but an infinite surface area. don't think of it as paint if that confuses you, but it's there to make the problem easier to think about
Yeah, it's possible for a finite amount of paint to have an infinite surface area as long as it takes the right shape (e.g. it fills the horn).
The area for the side of the conical frustum is incorrect!! The "A" should be the length of the arc measured in the middle of the frustum, kind of like its midriff.
That seems correct! I tried the algebra and got an extra +AB^2/R for A being the inner arc
I was thinking the same thing when I saw it but didn't bother to math it out. It would make sense that if you're "averaging" the slope you would also "average" the distance from the center point, so to speak, intuitively. That makes perfect sense.
I was looking for this comment. I did the derivation because he said it was 4 pages of algebra. I did it on a sticky note but got an extra term of angle*A²/2, which accounts for that extra length if you moved B to the middle. I guess it still works out in the video because as the slices get infinitesimally small, the inner length approaches the middle length.
Thank you so much for pointing this out. I did the algebra 3 times using A times B and get getting an extra term. I thought I was going nuts till you pointed this out. Works out perfectly using the "midriff" times the length.
I haven't worked this out but I suspect that if B is very small relative to A then the area approximates to AB. And that approximation gets better and better as B gets smaller which it does as B is delta x which is taken to the limit, ie towards zero.
One really cool detail about this: Early on, the voice behind the camera points out that the volume of the horn appears infinite because you can always keep adding a little bit more at the end, infinitely, which is technically true. So the fact that the answer resolves to Pi, a number that famously has an infinite number of digits, really fits this perfectly. You can always keep adding smaller and smaller amounts at the end, and yet it is a finite constant. The volume is finite, but indefinite.
Why filling the horn with paint doesn't mean you can paint it:
Although it isn't mentioned here, it has to be assumed that painting the horn means giving it a coat of paint with non-zero thickness (otherwise you don't need any paint to do it). Let's call the thickness t. Since the area of paint held by a section is pi r^2 dx and the paint required to paint that section is 2pi r t dx, there r
I think it makes sense somehow.
In fact, with real world paint, made out of atoms of finite size, it would not be possible to fill the horn either. At some point, as x gets bigger and the diameter shrinks, it would get stuck (and probably before that because of surface tension anyway). That's the same reason why it would be impossible to paint the outside area: a finite volume of paint cannot be infinitely spread as particles cannot be spread.
However, if one considers a mathematically ideal paint, made out of infinitely small particles, then the horn could be both filled and painted with it !
In real world paint, the smallest radius would be the radius of a paint molecule. That's big enough to make the horn extremely short compared to how much paint you need to fill it. For example, if the horn opening has a radius of 1m, and for argument's sake I'll say a paint molecule is about 0.1 micrometer in radius, then the horn is an impressive 1000km long. However, the surface area is just 2*pi*13.816 square meters = approximately 87m^2 (roughly the walls of an apartment) and you have 3142L (or 830 gal.) of paint to do the job. You can easily give it a few coating (thousands of atoms thick) and have 3000L of paint left.
No, because the parts of the horn beyond the blockage, where the last paint molecule gets stuck, would remain unpainted, as no paint gets there.
Holy cow. Good job on the animations!
But I'd like to ask how one does mathematically "paint" the surface of something? When we suppose that "painting" is a surjection from the set of points of a 3-d object (e.g. a bucket of paint) to the set of points of a 2-d surface, it is surely possible to "paint" the infinite AREA with finite VOLUME of paint. This answers the question from 16:02 too.
I agree, that a 3d coating of paint on the horn would have finite volume. However, again that paint would have infinite area. I find it perplexing, but basically my intuition of infinity is not applicable. Comparing area to volume doesn’t really make sense anyway. Volume provably finite, area provably infinite.
@Dalton Long There is no essential difference between 3D paint (which fills the horn) and 2D paint (which covers the surface); the whole horn and its surface are both subset (also infinite set) of R^3, and there exists a surjection from the horn to its surface. It means it possible to cover all the surface (infinite area) with the points from the horn (finite volume, but infinite points). That was my point.
@@rc5989 I like how you describe the paint as a 3d coating. It might be a better representation of real-life paint to consider the coating of paint has non-zero thickness. In this case, you are right; after we painting the surface of the horn, we will end of making another weird 3d object with finite volume and infinite surface area. Anyway, I was talking about what if we apply a paint with zero thickness as an absolute 2d object.
It is possible to paint the infinite area with any volume of mathematical (zero thickness) paint. However, if you do this at a finite rate (m^2 painted per hour), it still takes you infinite time.
Then again, if you pour paint into an object at a finite speed, it still takes infinite time for the paint to reach the (un)end of the object, so you can't fill it either... ;-)
@@junyoungshin4719 a useful equivalence between the horn and its surface being a bijection sounss wrong to me. pure set bijections can take random paints and map them to random points just based on the cardinality of both these being 2^alef0. a useful idea of equivalence would have to at least be a homotopical equivalence, maybe even a homeomorphism, so that you can spread your paint continuously. neither of those equivalences happen tho.
I initially learned about Gabriel's horn from a book I'm reading called "The Math Book" and came to this video to learn more about it. I watched this video twice now and just noticed the book I'm reading is on the top shelf in the video!
Great explanation by the way!
Explanation of the paradox for anyone who is curious:
The paradox is the result of using two seperate models for how it is treating the paint in the two different scenarios.
In the "Fill" scenario we treat the paint as a pure mathematical volume; which can be compressed and shrunken infinitely. This is not how paint actually exists in the real world. Which is to say, in the real world paint is made up of a finite number of paint particles with their own, immutable volume.
However, in the "Paint the surface" scenario he switces from treating the paint as the mathematical idea of volume to treating it like actual paint, which can only be spread over a finite area because it is made up of a finite number of paint particles, and once they run out, you are out of paint.
So now we can choose between two different paint models:
If we treat paint as it actually is (as a finite collection of particles with finite, immutable volume) then you can neither paint the surface area nor can you completely fill the horn. You cannot paint the surface because you will eventually run out of particles, and you cannot fill the horn completely because eventually the horn will be too thin for the paint particles to fit.
However, if we stick to the "Fill" example's model and treat paint like a pure mathematical volume. Then you can easily both fill the inside and also paint the surface area. In fact, when treating paint like this, *Any* amount of "paint" would be sufficient to paint the surface, both inside and out trivially.
This is because you can (as demonstrated in his explanation) morph the "paint" into a shape wherein it maintains it's volume, but has infinite surface area and is wrapped around the horn.
Interestingly enough, while treating paint consistently in this manner, you could not both fill the horn completely And paint the outside if you only had π units of paint.
However if you had any amount of paint more, say π+0.0000000000000001 units of paint, you could both fill the horn completely and paint the outside completely as well.
Very useful, thanks
thanks for thisd explanantion
Thank you that was amazing
pi units of paint... a pi-nt?
"Technically this is a derivative, I kinda treated them as fractions, it's fine for now"
Is this physics or what?
Physics is maths anyway so 🤷♀️
Be aware that's just two steps away from engineering where Pi becomes 3,14 and everything has to have finite values.
You can treat them as fractions for the most part and you'll be gucci. If you ever solve differential equations you basically abuse the differentials and beat them into submission with math hacks
@@harleyspeedthrust4013 "And now dx cancels out dx and we have a perfectly fine Integral in respect to dy"
@@Fubinii thats why i love fizicks
The solution of the paradox is simply that there are pi volumetric units, but infinite surface units. However, since there are infinite surfaces in a volume, you can "paint" infinite surfaces with any finite volume of paint.
This guy teaches surface area and volume of revolution in 20 minutes better than my professor does in 2 weeks.
Or it is just you paying more atention because u chose to watch this video lol
@@pedroaleb omg comment was best but reply was savage 😀
@@pedroaleb i'll argue that since my calc professor completely skipped over the 'why' of a lot of things (for example, he never talked about ds or the area of a conical fruntum, simply gave the formulas as needed with no explanation) and as a result I barely retained the info. This guy managed to actually explain it in a 17 min video which I applaud
@@reilandeubank i agree. he does it very well. but there is certainly a difference between school, wich you have to attend even if you dont want to, and youtube videos and other content that you actively go after and chose to watch
Learning Calc 3 right now. Crazy that I know what he is talking about. A year ago I would have had no idea.
Same here. Did solids of rotation just 2 days ago in class
5 or 6 months ago i would've had no idea
"Some other big horns" in the middle of this mathy explanation nearly slayed me
It could have been *so* different...
Regardless of how many angels can dance on the head of the pin, the music they're dancing to is definitely played by an angel blowing on the infinitesimally small end of an infinitely long trumpet
12:20 The cold rage of a thousand hells is entering my heart
Quite clearly, you just reduce the thickness of the coat as you go towards the end of the trumpet.
"Horn is completely full of paint. We need more paint."
If it assumed that the Horn has no thickness then the inside surface area and the outside area are the same therefore the horn is already painted inside and out.
Also if you have 1 unit of thickness for the paint and slice the Horn to this thickness then the maths for a cylinder of this height reduces to :- Paint volume inside r^2 , Paint required outside 2r.
When we counted the volume we sliced the horn into thin cylinders, and when we counted the surface we sliced it into less approximated shapes (that have bigger volume). That doesn't feel right. If you ask me, we counted volume for different solid and surface for a different one. I know it was done most surely right, but I don't get it why we didn't slice the horn for same shapes counting both volume and surface. Would love an answer or a link for further reading. Thanks.
Exactly my thought! I am not questioning that it was done right, just asking why it was done differently for the two calculations
The idea is that in the limit as dx goes to 0, the approximation of the volume gets arbitrarily close to the actual volume of the slices, since the volume not counted looks like π(1/x^2-(1/x+dx)^2)dx ~ dx^2 -> 0. On the other hand, if we were to calculate the surface area of the slice approximated as a cylinder, it would not approach the correct surface area, since the error is proportional to sqrt(1+(dy/dx)^2). Incidentally, since we throw away that term in the end, approximating the surface area with cylinders is sufficient for this proof, since it is an underestimate of the true surface area that still diverges.
That's actually not important, because he gets rid of the square root anyways. The point is that the volume is proportional to 1/(x^2) and the surface is proportional to 1/x. Now if you integrate you get 1/x for the volume and ln(x) for the surface. If x gets infinite 1/x goes to zero but ln(x) goes to infinity. Therefore the volume is finite and the surface is infinite.
(copied from my reply to another comment): As I mentioned in the video, I did this calculation (if only to convince myself) which was several pages of algebra, so we decided to exclude it from the video. If you want to try it for yourself then use the slanted edge for the volume also, and then when taking the limit as dx tends to zero (ie. doing the integral) you'll find that we ignore the terms in dx^2 at leading order - and these are exactly the terms that come in when using a slanted edge for the volume as opposed to a straight edge. For the surface area, the 'slanted edge' terms are order dx and so we do need them.
@@TomRocksMaths I think the question is, though, why bother calculating the exact integral when you could instead calculate a lower bound using the innermost cylinder at each approximation? In other words you know the length of ds at each slant has to be greater than if you simply cut horizontally straight across at the inner base with width dx. So the surface area has to be greater than 2pi I(1/x) dx - the same inequality you end up with in the video only without having to actually figure out the approximate length of ds.
"wouldn't filling the horn paint the inside"
We're defining painting as covering the surface with a layer of paint of finite thickness. When you fill the horn, the 'thickness' of the paint covering the inside tends to an infinitesimal
If you declare a finite thickness of paint, then the problem of painting the inside solves itself. Because once the horn gets too small, that layer of paint won’t fit inside.
Honestly that doesnt make any sense of it. Filling it would technically be a maximum thickness layer of paint on the entire inside. Filling any container with zero wall thickness and no wall touching areas would definitely mean you could paint the surface. Inside area is the same and you can cover the entire area with paint when filled.
Yes filling it would mean maximum thickness, but it still works because the maximum thickness is still infinitely small as it converges to zero.
@@noahtraylor525
I am unsure which side you are arguing. There isnt even a "thickness" paint in this thought experiment, just area. The op makes it sound as though filling cannot also be coating, which it definitely can be. If you fill a container, it would also be coated.
I’m just trying to say that it doesn’t seem that paradoxical to me. You can fill it, but you can’t cover it. Well this is because it is an infinite horn. So there is always another another length to be covered. But you can fill it because the volume converges to pi. Basically the surface area to volume ratio is infinite. It’s not that you don’t have enough paint, because any volume of paint can cover an infinite area. It’s just that you will never finish covering it. So I’m just trying to say that it doesn’t seem that paradoxical to me
Why do we ignore the _conial_ shape as illustraed in 7:41 when calculating the volume (3:28), but not when calculating the surface area?
I think they could have ignored the conial shape in the surface area calculation too since they got rid of the square root stuff
Exactly. I thought the area would be an integral of 2*PI(1/x)dx right a way, you are basically adding up circumferences of the slices with dx so small that it does not matter. Apparently, it does.
(copied from my reply to another comment): As I mentioned in the video, I did this calculation (if only to convince myself) which was several pages of algebra, so we decided to exclude it from the video. If you want to try it for yourself then use the slanted edge for the volume also, and then when taking the limit as dx tends to zero (ie. doing the integral) you'll find that we ignore the terms in dx^2 at leading order - and these are exactly the terms that come in when using a slanted edge for the volume as opposed to a straight edge. For the surface area, the 'slanted edge' terms are order dx and so we do need them.
The longer the horn gets, the higher its volume is, and the higher its surface area is. The paradox only arises if you conceive of a limit as being the actual measure of something. If you state it as "as the horn gets longer, the relative increase in its surface area diminishes at a slower rate than the relative increase in its volume, because surface area increases with r and volume increases with r squared," then it doesn't seem paradoxical at all.
I remember learning of this in calc 2, and I was fascinated. This recaptured that exact feeling for me
If Gabriel's horn is infinitely long, it can't be played, because there would be no end for the mouthpiece to be at.
That's why it's called a Gabriel's horn. I mean how many angels can fit on a pin's head right? Angels may very well be differential beings. A differential being wouldn't have any issues with travelling to infinity to blow a horn.
@@1urie1 When you start using magic and the supernatural to explain things, yeah literally anything is possible.
If people get too serious, they don't notice they take things far too literal. Gabriel is an angel. And angels can do miracles. In this case, he could fly an infinite length and blow you a horn partita by Johann Sebastian. And he even flies back to you, to say "And that's what you can learn in heaven, dude! And now I got to fly there again, I forgot my mouthpiece." And now you'll start protesting the music would never reach the bell of that horn. You're beginning to understand.
@@DeadlyDanDaMan How about quantum effects? Entanglement and two connected things happening at the same time? And the first quantum computers are running. Still not perfect, but it IS an application of theory, and it will explode processor performance in the near future. A quantum computer is entirely different from the digital ones. And you can call it supernatural, for it is hard to explain WHERE the calculus is happening. In parallel universes? Still, there are some WORKING small quantum computers around the globe.
If Gabriel's horn is played, you can rest assured that the end is actually very close.
I feel like this, much like the perimeter of a fractal, can be painted. You would need to cover an infinite surface area in this case, but that's entirely possible. You would have to dump the entire horn in paint and the outside would be painted. If you made another infinite horn say, twice as large at every point so your horn fits inside it, you could fill it with paint while your horn is there (wouldn't take much paint) and your horn would be painted.
This reminded me of like an infinitely long, higher dimensional fractal. You can easily have a fractal with an infinite perimeter and very finite area, and you could easily surround the whole thing with paint to paint the edges.
If you have paint that is so infinitely thin that it is able to run down into the tip you are also able to use a finite amount of it to spread an infinitesimally thin layer across the whole surface.
I’ve thought of a second paradox that arises when you try to fill the horn with paint. Assume we are in a frictionless, surface tension-less, atmosphere-less etc., etc. world but gravity is the same as earth’s. You start pouring the paint and the first bit is continually accelerating down the horn. However, because the horn is infinitely long, it will never reach the bottom in a finite amount of time. That will mean there is always space to be filled further down the horn, so you actually cannot empty your pi units of paint from its can. But, since the first paint poured will always be accelerating, it can’t “get in the way” of the paint you continue to pour. So you should be able to empty the can. But, of course that would mean there is still unfilled space inside the horn. Which would mean, actually you can’t empty the can and so on.
That's the problem with using real world analogies for mathematical concepts that approach limits. As soon as you go real world, you also have to deal with the Planck length, but the paint must be infinitely thin, or it couldn't fill the parts of the horn where 2y < Planck length. Then you end up with a situation where the paint "particles" have no dimension (or we would need an additional dy term to account for the paint application, or the paint wouldn't be able to fit into the infinitely small part of the horn) but they still have volume, or how else could they fill the horn...?
Of course you can paint it! Create a second horn with 2* the radius and fill the gap between the 2 horns with paint. You'll need 3*Pi paint ...
Actually you can use as little paint as you want with the second horn of radius 1+epsilon with epsilon going to zero.
Or just cover with a layer of paint that goes to zero thickness fast enough with x...
9:00 Wrong!!!!
This is for A in the middle of the part not the small curvy side. And the proof is not that long:
The Area of the difference of the discs of radius r+B/2 and r-B/2 is SD = π(r+B/2)^2 - π(r-B/2)^2 = ... = 2πBr
But we only consider a part of length A of the perimeter of the circle of radius r. The ratio is A/2πr
So S = SD * A/2πr = AB
😎
I was just about to comment this, but figured I'd check to see if someone already had 👍🏿
Finite volume, but infinite horniness 😳😩
I approve this message.
Thanks, i hate it.
then clop
@@mamoonblue you're welcome
@@SuviTuuliAllan ...
I'm a student of topography, I'm gonna be very simple, you are basically just stretching a solid with finite volume infinitely, so no matter how long it becomes it's always gonna have that finite volume, and for Gabriel's horn that's why it's width becomes infinitesimally small in order to compensate for that finite volume the shape got.
Same goes for 2d shapes.
You are right, that didn't even cross my mind. Only difference is, that in this paradox the shape is finite. Which is the same paradox, a finite shape of infinite length.
Just like pi. A finite value, yet it has an infinite number of decimals. That's the core of this paradox. But basically it's just Zeno's paradox in a different setting.
As in your example. Half the width and double the length into infinity. (Thinking of a rectangle for simplicity). Same area, infinitely long circumference.
Ok, but if we fill it first then stretch it infinitely there is always more paint to always fill the expanding surface area no?
@@LOKOFORLOKI1 u can actually fill it if u puor the paint in it, the paint will start filling up at nearly infinite speed at the beginning and then as it reaches the upper wider end it will fill up a bit slower and it will take the same amount of time it would have taken to fill up a container with the same constant volume.
I like the paradox, thank you.
Also, your enthusiasm is contagious 🙂
Using a cylinder for the surface calculation is much faster, but the detour via this weird shape, the name of which I can't even remember, was interesting too.
Thanks so much!
Great video, thanks! Question though: When you do the volume, you go over dx because the slant of the surface doesn't matter, but then for the surface area you do use ds because now somehow the slant does matter. Can you elaborate on why the slant doesn't either matter for both, or for none? Thanks!
Actually, he *did* use that approximation, he just wasn't clear about it:
The approximation used for the volume is ds = dx
Apply to SA = 2*pi*(1/x)*ds -> 2pi(1/x)dx
As shown above, the integral between 1 and infinity of 1/x is, itself, infinite. And since for all values of x, ds > dx, the true value of SA will be infinite.
I have the same question. They approximated volume with a cylinders but not surface area.
The slant does not matter for volume because as you take shorter and shorter segments for the approximation, the outer ring's length approaches 0, thus removing the slant from the equation. Technically, there is an infinitesimal slant still present, but it's so small that it isn't worth consideration when doing this sort of problem.
Surface area is different, in that you aren't subdividing to work it out, and intuitively the slant makes a huge difference because it makes the length of the horn infinite. The outer surface is unending, therefore no amount of paint will ever cover it. The inner volume, however, is bounded and reduces further along the horn, so much so that there is a number that is larger than the sum of it. Because of that, the volume is finite.
I hope this helped clarify and didn't just add more confusion.
@@mavericksolomon395 Hi, I can understand the idea that the slant does not matter for volume as you said. But I don't get the sentence below:
"
Surface area is different, in that you aren't subdividing to work it out, and intuitively the slant makes a huge difference because it makes the length of the horn infinite.
"
As far as I know, the way to derive formulas for surface area and volume stems from the idea of "splitting" an interval into an infinite number of infinitesimal intervals and approximating each of them by an appropriate surface or shape. Therefore, I don't quite understand what you meant by not subdividing to work out the surface area?
Thanks.
@@mavericksolomon395 Even if you estimated using a cylinder in this case you'd still get an infinite result. The surface area of each cylinder would be the circumference of the circle times the width, 2πR dx. Since R is 1/x, you'd wind up with 2π * Int(1, infinity) 1/x dx.
But if you tried to do a calculation on a finite slice of the horn, your answer would be slightly off for the surface area. But not for the volume.
It would have made more sense to do both of these with a conic frustum. Volume of a conic frustum is 1/3 π h (R1^2 + R1*R2 + R2^2) ... In this case h is dx and both R1 and R2 are 1/x. So Int(1, infinity) of 1/3 π dx (3 (1/x)^2) ... Do the arithmetic and it simplifies to the same answer as when using cylinders.
I remember once feeling like it made sense that you could fill it but couldn’t paint the outside, but now I can’t see how I felt that. It really is a paradox.
you just dunk it in the paint
Not a paradox, as a given volume of mathematical paint can paint an infinite surface area due to having infinitely low viscosity. For illustration, if you poured the paint out on a perfectly flat plane it would continue to spread infinitely without ever changing volume.
Why do you calculate the volume of a cylinder but the area of a cone? Either ds=dx for both or ds>dx for both. No?