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Why is the volume of a cone one third of the volume of a cylinder? (2010)
- eigenvalue 3y agoWithout using calculus, it’s actually pretty mysterious where that 1/3 factor comes from. I found this somewhat shocking—- usually there is some easy (but perhaps clever or elegant) framing whereby the answer just “falls out,” but not in this case!
- jaclaz 3y agoThere is the demonstration analyzing a frustum and then extending its smaller radius until it is the same as the base one (i.e. stretching the frustum until it becomes a cylinder): https://web.maths.unsw.edu.au/~mikeh/webpapers/paper47.pdf https://web.maths.unsw.edu.au/~mikeh/webpapers/paper47.pdf This does not use calculus at all but still needs some time to digest, the idea is that there must be a coefficient c that multiplied by the base area of the cone A and by its height gives the volume (V=cAh) and that if we express the volume of the cylynder with the same approach we get V=3cAh.
- pharrington 3y agoYou said that demo doesn't use calculus at all, and yet I see a sum of infinitesimal divisions of a surface area. I know an integral when I see one.
- jaclaz 3y agoLook only at page 1, where the volume of the cone is calculated. The sum of surfaces is on page 2 about the surface of a sphere.
- pharrington 3y agoMy bad!
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- Solvency 3y agoI love how this document just randomly throws in "e" which mysteriously means "extra" with absolutely zero further explanation or rationale. This is why mathematicians drive me insane. There is zero boilerplate for explaining where these variables come from and why. But, cool. Let's just roll with this "extra".
- consilient 3y agoWhat explanation is there to give? It's just the ordinary meaning of the word: when you cut out a shape you end up with some extra material left over.
- zem 3y agoa frustum is the base of a cone when you slice off a conical cap, so the remaining height of the cone from which the frustum was sliced is the "extra".
- ColinWright 3y agoI wrote about this years ago, and submitted it here on HN where I think it sank without trace. It's on this page: https://www.solipsys.co.uk/new/ConsideringASphere.html?wg01hn https://www.solipsys.co.uk/new/ConsideringASphere.html?wg01h... A slightly different derivation is here: https://www.solipsys.co.uk/new/VolumeOfASphere.html?wg01hn https://www.solipsys.co.uk/new/VolumeOfASphere.html?wg01hn
- defrost 3y agoMany moons ago, twixt primary school and high school, I enjoyed coming up with two derivations of the volume of a sphere remaining after removing the volume of a hole drilled from pole to pole (cylinder plus end caps). These things can get a little addictive :-)
- recursive 3y agoThose end caps sound difficult.
- defrost 3y agoThere's a "surprising twist" on analysis. The question above is typically posed as (say) a hole of 2 units in length. That's all you get, the lip to lip length of the hole. Not the radius of the sphere, not the diameter of the hole - just the hole length. Nothing else (save the implication that it's solvable). There's a lengthy formula laden proof approach, and there's an Aha!! lighbulb moment solution.
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- zem 3y agothat's very nicely done, as was the linked proof of archimedes' hatbox theorem
- RicoElectrico 3y agoHot take: do we really need intuitive explanations for anything? I love 3b1b videos for example, but I am more in the "mechanical application of rules" camp rather than "a-ha moment" camp. That's why I hated integrals at the calculus course. It's all weird tricks all the way down - which a mortal couldn't come up with.
- eigenvalue 3y agoFor many people, it’s deeply dissatisfying to mechanically apply magical incantations that you don’t understand and could never arrive at yourself. Once you intuitively understand something, it also makes it trivial to remember it, and also helps connect it to other information in your brain.
- nostromo 3y agoIntuition seems to me to be the point at which you really understand something - and it allows you to make progress and predictions about other phenomena. For example, I don’t think anyone has an intuitive understanding of quantum mechanics yet, and it seems to be limiting progress. String theory was an attempt at an intuitive understanding that has so far been fruitless. Spacetime is a counter example, where a difficult concept was made graspable by some intuitive concepts and analogies, and progress very quickly seemed to follow.
- sethammons 3y agoDef encourage you to read Lockhart's Mathematician's Lament: https://www.maa.org/external_archive/devlin/LockhartsLament.pdf https://www.maa.org/external_archive/devlin/LockhartsLament.... The way we were introduced to integration, the different rules (tricks as you put it) made sense. If you are using rote memorization and cannot derive the "tricks," I imagine all of math is harder to approach. As a former inner-city high school math teacher, recovering, students who tried to memorize mechanical approaches could be easily spotted when they forgot or recalled incorrectly. Simple example: they may not remember X^1 and/or X^0 and they lack the understanding to figure it out.
- bowsamic 3y agoOne of the reasons I became a quantum physicist is because I always sucked at intuition
- javajosh 3y agoAh yes, but what interesting, related shape has the volume of 3 times that of a cylinder of a certain base and height, I wonder? Perhaps the coefficient here is just coincidentally interesting and there are actually infinitely interesting shapes connected in this manner.
- lisper 3y agoThe general result is that any volume consisting of a stack of cross sections whose area goes does as the square of the distance from the base will have a volume of 1/3 the area of the base times the height. The usual special cases are squares (pyramids) and circles (cones) but it works for any shape. It doesn't even have to be convex.
- lisper 3y ago> whose area goes does as the square of the distance from the base I tried to edit this earlier but apparently it didn't work, and now it's too late. That should read: "whose area goes up as the square of the distance from the tip" but I guess most people were able to figure out what I meant. It's not a particularly difficult concept.
- valbaca 3y agoIsn’t it just the surface area of a circle as the circle shrinks? (Limit as Radius goes to zero) That’s what a cone is
- blueblimp 3y agoI found the visual justification for partitioning a cube into three pyramids to be a bit confusing. For me, an easier way to think of it is that there are three coordinates (x,y,z) and each pyramid represents a region where a particular coordinate is largest. e.g. One such pyramid is {(x,y,z) | x = max{x,y,z}}.
- zeteo 3y agoPick a corner of the cube. There are three faces that go through the opposite corner, right? The pyramids are formed by the first corner and those three faces. (They're obviously equal.) Now imagine looking at the inside of the cube through that first corner. There is no angle where you don't see one of the three faces in the background! So the decomposition is complete.
- crdrost 3y agoThat is a very clean way to get this result for sure!
- patrickthebold 3y agoMy preference is to take the point in the center of the cube and make pyramids to each of the 6 faces. There are twice as many pyramids but they are half as high.
- codeflo 3y agoDoesn't the first answer attempt to give one? That's also the way many people learn it middle school, at least I think I remember I did. To spark a constructive discussion, it would be interesting to know what doesn't satisfy you about this answer.
- OJFord 3y agoThe submitter isn't asking us, they're just sharing the question and answers.
- eigenvalue 3y agoYes, it does attempt to give one-- but I didn't find it particularly satisfying. But some of the replies in this HN thread have been better in that respect, for me at least. The idea that it's just an extension of why the area of a triangle is 1/2 the enclosing rectangle to 3 dimensions to me is the "oh, that's why" framing for what otherwise seemed sort of magical to me.
- micheljansen 3y agoThey key ask being "without using calculus". I think my brain has been ruined by calculus in that I now consider approximating it as either an infinite stack of thin discs or as a pyramid with infinite sides to be intuitive explanations.
- Perenti 3y agoIs "circular pyramid" intuitive, or is it calculus based? It works for me.
- codeflo 3y agoIs Cavalieri's principle (two volumes are equal if every cross section has equal area) considered calculus or not?
- filmor 3y agoI'd say that this is an application of Fubini's theorem which is very clearly calculus (or analysis, not sure where the distinction is drawn in English).
- consilient 3y agocalculus = introductory real analysis aimed at engineering and science students. Same material, different emphasis.
- dmurray 3y agoI would say "circular pyramid" is intuitive, but "an n-sided pyramid works just the same as a square pyramid" is not. If instead of just a square pyramid, you demonstrated it for regular polygons with sides 3, 4, 5 - I wouldn't need any convincing that the relationship continued to infinity.
- pharrington 3y agoIt's both.
- sublinear 3y agoI'm going to say something that sounds silly to most people, but isn't it the same deeper reason why a triangle is half of the base times the height? I don't give a rat's ass if there's a proof or not connecting the two. I'm not a mathematician, but it makes sense from a calculus perspective that the number of dimensions ends up as a divisor.
- ozyschmozy 3y agoNow I'm curious if that holds for 4d shapes. Maybe someone with a better understanding of calculus or 4d geometry can answer Edit: apparently it does for pyramids, so it should for cones as well. General formula for volume of n-dimensional pyramid is A*h/n where A is the volume of the base.
- sethammons 3y agoIt intuitively holds for 1, 2, and 3 dimensions. I don't know how to visualize the 4th dimension, but if it holds, it should take up a quarter of the "space" (point -> area -> volume -> space time?) If the 4th dimension is time, does that mean a quarter of the "space" is spread out over time? What does that even mean? Anyone familiar with n-dimensional space able to weigh in?
- zoky 3y agoIt's probably better to think of time as the "zeroth" dimension, in that everything that exists in any other dimension has to exist for some duration of time first. Of course, we can still consider theoretical objects that exist outside of time, just like we can easily consider two-dimensional objects despite being in an existence with three spatial and one temporal dimension, but for an object to physically exist it must at least exist in time. When you look at time as the zeroth dimension rather than the fourth, it should be obvious that the fourth dimension is simply another spatial dimension. Think about how a two-dimensional plane is a cross section of a three-dimensional object, and line would be a cross section of a plane, and a (zero-dimensional) point is a cross section of a line. So too can our three-dimensional universe be looked as a cross section of a theoretical four-dimensional existence. A three-dimensional object that moves in the fourth dimension would simply cease to exist in our universe, and a four-dimensional object that moves in the fourth dimension would change which 3D cross section is visible to us.
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- albert_e 3y agoWhat would be the shape of a rotationally-symmetric (nearly conical) shape that has 1/2 of the volume of a cylinder with same base?
- crdrost 3y agoThat question does not have a unique answer. For instance one possibility is to take a beaker as your cylinder, fill it halfway with clay as your shape, scrape a thin layer off the top, form it into a little straight toothpick of clay the height of the beaker, and then insert the toothpick into the very center of the clay in the beaker so that it reaches to the top. This shape is rotationally symmetric, it has a volume of half the beaker, and the beaker is the smallest cylinder that encloses it. However you could do this toothpick idea to just about any shape that you could throw out of that same volume of clay on a potter's wheel, and you might not even need to. However one of the easiest algebraic ways to do it is to have the radius go like the square root of the distance from the top of the beaker, then the cross-sectional area goes linearly with that distance and when you integrate the integration gives you a clean factor of ½ rather than one-third.
- xeonmc 3y agoWe could add constraints to make it meaningful. For example, restricting the profile to only an Lp norm superellipse [0], what is the exact norm value that results in 1/2 volume? A 1-norm superellipse results in 1/3 volume (cone), 2-norm results in 2/3 volume (hemisphere), so it must be a norm between 1 and 2. A quick totally non-rigorous calculation on Desmos yields a power of 1.389857035315, you can see the shape in [1] [0] https://en.wikipedia.org/wiki/Lp_space https://en.wikipedia.org/wiki/Lp_space [1] https://www.desmos.com/calculator/x8y5evt0tc https://www.desmos.com/calculator/x8y5evt0tc
- zoky 3y agoTechnically correct, but I think it can be assumed from the question that it's referring specifically to a conical frustum. Of course, the answer is then that the shape of such an object would be, well, a conical frustum, which renders the question tautological. I believe that the intent of the question is something like “What are the dimensions of a conical frustum having 1/2 the volume of a cylinder, with the same base and height?”, or, put another way, what would the ratio of the diameter of the top of such a frustum be to its base? The answer being somewhere between 0 (as it would be for a cone) and 1 (for the cylinder). I'm sure there's a formula to calculate this value for a given volume, but I don't really care to figure it out at the moment. Although, it might be interesting to see if such a formula scales to values for the relative volume above 1, creating an “inverted” conical frustum…
- hamandcheese 3y agoA triangle is 1/2 the area of the surrounding rectangle, a cone is 1/3 the area of a cylinder... does this pattern extend into the 4th dimension?
- NovaDudely 3y agoContinue that logic, infinite dimensions and the cone would infinitely smaller? Im sure I have that wrong but cool to think about.
- tgv 3y agoWhen the number of dimensions increases, there's more and more space around the cone, so its volume would go to zero if the volume of the hypercube stays constant. It would become a needle. But spaces with infinite dimensions are difficult. They are usually required to have a finite norm for all points. Idk how that would affect volume.
- themoonisachees 3y agoWouldn't this be just continuation of the fact that 2-dimentional shapes have 0 volume? Or are you talking about hypercones ?
- yaakov34 3y agoI explained above what happens when the dimension grows - spheres and cones do indeed take up a smaller and smaller portion of their unit cube, eventually having negligible volume. This is important in the context of high-dimensional statistics and so on. If you want to actually have infinite-dimensional volumes, you can't just assign finite values to them in a simple way, or you will have contradictions such as a certain volume being completely covered by a union of things which have 0 volume. In infinite dimensions, you instead have various measures like the Gaussian measure. Feynman's path integrals are a kind of way to assign a value - called amplitude - to an infinite-dimensional manifold (a kind of "volume") of paths. But that takes us well to the side of the idea of the ratio between cube and inscribed figure volumes.
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- revskill 3y agoHope hackernews supports Katex soon.
- DonHopkins 3y agoBecause it used to be 1/PI until they legislated PI down to 3? ;)
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- OscarCunningham 3y agoVolume is defined in terms of limits, so there's no hope to prove this without taking a limit of some form. Whether or not you count that as calculus is up to you.
- walnutclosefarm 3y agoI don't know that limit is the key factor here, but I agree on the calculus part. The arguments that rely on Cavalieri's principle to relate the volume of the cone to the volume of a slice-by-slice identical pyramid, for example, don't really involve a limit per se, but they are clearly motivational for integral calculus. Basically every intuitional argument for this, and other volume formulas, are somewhere on the road to integral calculus.
- edent 3y agoIs it because Pi is 3?
- Xen9 3y agoCan someone give me this problem without any spoilers?
- boomlinde 3y agoHere: why is the volume of a cone one third of the volume of a cylinder?
- xept 3y agoPaul Lockhart wrote a great book called Measurement, about this and similar questions. I highly recommend it if you're interested in the topic.
- gglitch 3y agoThe math here is interesting on its own, but don’t miss the discussion of the animation creator’s reason for doubling the animation being so it can be seen in “stereo” as 3d if you defocus a bit.
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- justin_ 3y agoIf you're looking for where the 1/3 "comes from", I think using Pappus' second theorem[0] is the way to go. The theorem states that the volume of a solid of revolution formed by a revolving shape is the area of the original shape multiplied by the distance traveled by its _centroid_. The centroid of a triangle is determined based on distances from the three sides, and this is where thirds come from. For example, the centroid of a a right triangle with points (0,0), (0,1), and (1,0) sits at (0.33(3), 0.33(3)). That's a third! The OP alludes to this with the mention of a solid of revolution in the post. And someone mentions Pappus in the thread there. How did Pappus figure out the centroid was distance to consider? I think that's another mystery, since the proofs of the theorem seem to depend on calculus, which brings us back to the original question. Anyway, I was looking into the exact question of the OP a few months and this was the most satisfying answer I could find for where the 1/3 comes from geometrically. [0] https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem#The_second_theorem https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem#Th...
- Izkata 3y ago> The theorem states that the volume of a solid of revolution formed by a revolving shape is the area of the original shape multiplied by the distance traveled by the _centroid_. The centroid of a triangle is determined based on distances from the three sides, and this is where thirds come from. For example, the centroid of a a right triangle with points (0,0), (0,1), and (1,0) sits at (0.33(3), 0.33(3)). That's a third! Isn't the centroid of an equivalent rectangle rotated to make a cylinder at (0.5, 0.5)? 0.33 is not 1/3 of 0.5
- justin_ 3y agoTo make an enclosing cylinder that bounds the cone in my example, we can use a 1x1 square with centroid (0.5, 0.5). The cylinder has volume π. And that works out from Pappus' theorem too: 1 (the square area) * 0.5*2π (the distance traveled by the centroid)
- squirtlebonflow 3y agoNo, but it's 2/3. Then you divide by 2 because the area of the triangle is half the area of the rectangle.
- ineedasername 3y agoThe answers there all give a mathematical answer. The title made me think it was being asked in an almost philosophical sense (my own flawed reading of the question) like: “Why does the universe impose spatial constraints such that a a cone’s volume just happens to work out this way?” But on reflection I’m not sure that’s even a meaningful question.
- version_five 3y agoIf a triangle is 1/2 a rectangle, a cone is 1/3 a, cylinder, is there some kind of n-d hyperpyramid that is 1/n of its bounding hypercube? As in, does the 1/3 have to do with the dimensionality?
- Chinjut 3y agoYes, that's exactly what happens. Think of a point in an n-dimensional hypercube and the pyramids made by connecting it to the n many opposite faces. These pyramids each have volume 1/n times the hypercube's volume. And in the same way, in n dimensions, any tapering figure has volume given by its height times its base divided by n.
- tylerneylon 3y agoYes; one way to see this is with a little calculus. It all boils down to the fact that the indefinite integral of x^(n-1) is x^n / n. (That's where 1/n comes in.) Suppose we have an n-dimensional pyramid. It has an (n-1)-dim'l base with volume B, and that base tapers to a point; for simplicity, it has height 1. Take cross-sections as we travel from the base to the height. How big is a cross section at height y? Each cross section is a (n-1)-dim'l shape, and its volume is B * y^(n-1) because this is how volumes scale in dimension n-1. [Examples: In 2D if you x2 the sides of a shape, it's "volume" (area) is x4. In 3D if you x2 the sides of a shape, it's volume is x8; in 4D it would be x16, etc.] Now take the integral to add all of these cross-sections: integral(from 0 to 1 of B * y^(n-1)) = B * y^n / n evaluated from 0 to 1 = B / n. If the shape is scaled along the height dimension by h, then we get the more general formula: volume = h * B / n.
- renewiltord 3y agoThe stereoscopic animation is really something. That's pretty cool.
- nojs 3y agoIn case anyone missed it the gif in the first answer is actually in stereo, if you relax your eyes it’s 3D.
- UlisesAC4 3y agoI always thought it had to do with the number of dimensions. A triangle is half of the rectangle and when I learned that a cone was a third part I was like yeah it makes sense.