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Why is the volume of a cone one third of the volume of a cylinder? (2010) (math.stackexchange.com)
233 points by eigenvalue on July 2, 2023 | hide | past | favorite | 112 comments

If 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#Th...

The centroid makes sense intuitively when you think about it as a center of mass. Since it's the average position, for every portion that's father away there is, in a sense, a portion that's closer - the portion being equal either insofar as it is larger and closer to the center or smaller and father away, and since the size/volume of a ring increase linearly with distance in an intuitive manner, you can justify to yourself that you should be using the centroid without any formal calculus, just by considering the "weight" of a ring.

This is an excellent explanation. If we imagine the triangle being swept about in a circle, the points in the "outer" section (forming the circle at edge of the base our our cone) will cover more distance the points in the "inner" section (the points near the center line of our cone that we're revolving around). The distance moved by the centroid, then, is the average distance a point travels.

> 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

No, but it's 2/3. Then you divide by 2 because the area of the triangle is half the area of the rectangle.

To 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)

I'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.

The downvotes are ignorant. See Lockhart's Mathematician's Lament, where he laments the state of math education comparing it to learning music through sheet music and not pickings up an actual instrument.

He asks how to visualize the area of a triangle by bounding it in a rectangle, where it becomes obvious that the triangle takes up half the area.

He then poses the question about the volume of a pyramid where you can use a similar technique.


Here's an attempt at making sense of that intuition. Let's compute the volume of a pyramid that's inside a cube. Imagine that you're inside a cube room (this will make it easier to identify the faces). You are sitting in a corner.

From your corner draw lines to the four ceiling corners -- a cube diagonal, two face diagonals and a cube edge. This is the pyramid. Now draw lines to the corners of one of the walls opposite you -- it's the same pyramid, but on its side. And lastly draw lines to the other wall opposite you -- again the same pyramid. You have now covered the entire cube with three identical pyramids, so the volume of the pyramid is a third of the volume of the cube. The interesting part is that the bases of these pyramids are on the three dimensions, giving the intuition that the /3 is due to the dimension. This proof might even generalize in higher dimensions.

Edit: I think this is also the gist of the top answer in TFA, but it's weirdly formulated imo.

That's a beautiful proof! And yeah, it generalizes to higher dimensions. If you're in the corner of an N-dimensional hypercube, then the hypercube has 2n faces of dimension n-1. You're touching half of these faces, leaving n faces to form pyramids out of. And just like in the 3D case, they're arranged symmetrically so they must all have the same volume.

EDIT: got around to reading TFA and it's also the top answer, so there's a nice visualization.

Diving deeper into the proof, say the hypercube is a unit hypercube, spanning from (0, ..n, 0) to (1, ..n, 1) -- where "..n" means "a sequence of length n". Then:

Each corner is a point (b_1, ..n, b_n) where each b_i is 0 or 1.

You're located at (0, ..n, 0).

The hypercube has 2n faces, which can be represented as a pair (i, b) where 0<=i<n and b is 0 or 1. The face (i, b) touches the 2(n-1) corners whose i'th coordinate is equal to b.

You touch half of those faces: specifically the faces (i, 0) for each 0<=i<n.

You can draw a pyramid to the other half of the faces, since you don't touch them. These pyramids must all have the same volume, by the symmetry of coordinate permutations, which is an operation that preserves volume.

TFA? :-/

TFA - What does TFA stand for? The Free Dictionary


Very cool! Thanks for sharing.

>I don't give a rat's ass if there's a proof or not connecting the two.

Life is too short for proofs. - Gilbert Strang (in one of his lectures)

In the same vein:

Beware of bugs in the above code; I have only proved it correct, not tried it.

Donald Knuth

The only effective way to raise the confidence level of a program significantly is to give a convincing proof of its correctness. [0]

Edsger W. Dijkstra


Argument three is based on the constructive approach to the problem of program correctness. Today a usual technique is to make a program and then to test it. But: program testing can be a very effective way to show the presence of bugs, but is hopelessly inadequate for showing their absence. The only effective way to raise the confidence level of a program significantly is to give a convincing proof of its correctness. But one should not first make the program and then prove its correctness, because then the requirement of providing the proof would only increase the poor programmer’s burden. On the contrary: the programmer should let correctness proof and program grow hand in hand. Argument three is essentially based on the following observation. If one first asks oneself what the structure of a convincing proof would be and, having found this, then constructs a program satisfying this proof’s requirements, then these correctness concerns turn out to be a very effective heuristic guidance. By definition this approach is only applicable when we restrict ourselves to intellectually manageable programs, but it provides us with effective means for finding a satisfactory one among these.

[0] Edsger Dijkstra - Turing Award Lecture - The Humble Programmer - 1972


It is exactly that and you even can use the same proof for a triangle, a cone or higher dimensional analogies. You always get 1/dimension.

Proof: Take height 1 for simplicity of notation. Then we get the volume as the integral over crosssections (Fubini). Each cross section is rescaled by a factor (1-z), where z is the height. Rescaling changes the cross sections measure by (1-z)^(n-1). Integrate that to get 1/n. Done.

Now 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.

All of these are examples of how the integral of x^n is 1/(n+1) x^(n+1). Combined with the fact that rescaling a n dimensional object by a factor x makes it x^n times bigger and that the (n+1)-volume can be expressed as an integral of n-volumes.

It will generalize to any shape consisting of stacked rescaled slices of some other shape where the scaling factor increases linearly.

It 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?

It'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.

No, not really, and I don't think the other reply about it being a 0th dimension makes much sense either. The key word is `spacial-dimensions`, whenever math talks about just spacial dimensions, there is no time dimension. In other words, a 0-dimensional 'world' would just be a point, 1 dimensional a line, etc. In something like flatland, where there are living beings, of course there would be a time dimension. But if you are just talking about the volume of shapes, all the dimensions are spacial. So when we go from 3D -> 4D in this scenario, we are adding an extra spacial dimension perpendicular to all the other 3. The idea that 4th D is time is more of something from physics, although the fact that we call it the 4th dimension doesn't mean much, it could be called 1st 2nd 3rd dimension and then a spacial one gets moved to a different number.

I found this video explaining quaternions to actually be a great way to explain (and visualize) higher spatial dimensions: https://youtu.be/d4EgbgTm0Bg

When you think about the nth dimension as a projection of the n+1th dimension, it begins to make a lot more sense. That doesn’t mean you can necessarily visualize the 4th spatial dimension directly, but you can at least visualize how the 3rd dimension can be a projection of the 4th, just like how we can easily see that the 2nd dimension as a projection of the third.

There was a paper that showed “bee dances” were very similar to some 6th dimensional shapes projected into 3d. IIRC, it was retracted due to people not taking it seriously and making it into a joke (maybe it was, but I wouldn’t know). But anyway, people who can visualize more than 3 dimensions should be considered magical wizards.

Time is just one possible meaning you can assign to a dimension if it's convenient for you to do so in order to model a problem.

It makes sense that the number of dimensions ends up somewhere, but I think there's no particularly obvious a priori reason why it should be a denominator. For example, the length of the long diagonals of a unit cube in n dimensions depends on the dimension, but it's \sqrt n, not anything divided by n (well, it's n^{3/2} divided by n, but that's not much use).

Of course one can fill in why in this case it happens to be the dimension in the denominator—several sibling posts, particularly gniv's (https://news.ycombinator.com/item?id=36562093), have done so nicely—but I was responding specifically to the disclaimer:

> 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.

which seemed to suggest that it should be obvious without mathematical reasoning.

While I agree about proofs it's not intuitive to me. A triangle can clearly be visualized as half a square. Look at any videogame square made of 2 polygons.

A 2D projection of a cone isn't intuitively 1/3 of a cylinder with the same dimensions.

They 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.

I think the calculus-based explanation is the simplest one, and also easy to extend to an arbitrary number of dimensions.

If we look at the direction from the tip of the cone to the base, the volume is an integral of the area of the circles that are the cone sections. The radius of those circles grows linearly when the point is moving from the tip of the cone to its base. The area grows proportionally to a square of the distance from the tip. An integral of x^2 is x^3/3. Hence 1/3.

(x^3)/3 is so much faster for me to read and far harder to read incorrectly. Yes there's the memory trick for order of operations, but division symbols are so easy to get confused around that the extra clarity of the parenthesis should be standard for all non-trivial (one item per side) cases.

x^(3/3) would just be x though

I think this is an elegant explanation

It's my first thought too, but I never developed an intuitive understanding of why the integral of x^2 is (x^3)/3 (even though I know it is by heart).

Do you have intuitive understanding of the derivative rules? For example:

d/dx of x^3 = lim h->0 of ((x+h)^3 - x^3)/(h)

= lim h->0 of (x^3 + 3hx^2 + 3xh^2 + h^3 - x^3)/(h)

= lim h->0 of 3x^2 + 3xh + h^2

= 3x^2

So then if we started with (x^3)/3 the 3s would cancel and we’d get x^2. This tells us that the antiderivative of x^2 is (x^3)/3.

Or did you mean some other intuition? Such as why the fundamental theorem of calculus (that integration and differentiation are inverses of one another) is true?

Graphing y=x^0 -> y=x^1 and y=x^1 -> y=(1/2)(x^2) and doing math on the shapes (just simple squares and rectangles) makes those pretty intuitive. I just see it as an extension of that.

I'm the opposite, I never really understood calculus until I realised that integration is low pass filtering and differentiation is high pass filtering, and since I was in high school I've only ever converted calculus into R, C, Fin, and Fc, and solved the filter.

Differentiation is high pass filtering? How so? Is low pass filtering followed by high pass filtering the same as leaving a signal unchanged?

The answer to your question requires a lot of math but in my opinion, thinking about derivatives and integrals in terms of signal responses is not at all intuitive since your have to first interpret the signal in terms of its fourier transform (so a sum of different magnitude and frequency sinusoids) and then you have to think of the operations as phase shifts that are only applied to a certain set of frequencies (cosines and sines are 90 degrees out of phase and are the derivates and ontegrals of eachother). The integrator and differentiator will apply the phase shift to different sets of frequencies. What complicates things are that the magnitudes are also changed based on the frequency of the sinusoid. I think the person you are replying to happened to learn this stuff before actually doing calculus and to him it seemed intuitive. If you would like to learn more, any control systems textbook will be able to do the work a lot more justice than I can.

In image processing an edge detector like Sobel works (IIRC) by forward differencing. This is not quite differentiation but may help connect the dots a little bit. My guess is that some people may find this more intuitive.

In simple terms, every sin(wt) or cos(wt) term becomes wcos(wt) and -wsin(wt) when differentiated, so the higher the frequency, the higher the multiplier.

This is not what most people would call high pass filter, as there's no cutoff frequency etc, and the phases get out of whack quickly, but you can think of it in these terms if it helps.

If you call multiplying the Fourier transform by a linear function "high-pass filtering" and dividing by a linear function "low-pass filtering", then I understand what you are saying. But I'm not accustomed to calling those high-pass filters and low-pass filters. I think of high- and low-pass filters as multiplying the Fourier transform by a step function (ideally, or some close approximation of such).

I spent freshman year taking calculus and physics and just being mad at all of the science classes I’d taken so far.

All this 1/2 bullshit is just integration?

My first intuition is integrating a "triangle" using cylindrical coordinates.

Is "circular pyramid" intuitive, or is it calculus based? It works for me.

Is Cavalieri's principle (two volumes are equal if every cross section has equal area) considered calculus or not?

I'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).

calculus = introductory real analysis aimed at engineering and science students. Same material, different emphasis.

I 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.

It's both.

I wrote about this years ago, and submitted it here on HN where I think it sank without trace. It's on this page:


A slightly different derivation is here:


Many 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 :-)

Those end caps sound difficult.

There'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.

that's very nicely done, as was the linked proof of archimedes' hatbox theorem

I 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}}.

Pick 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.

My 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.

That is a very clean way to get this result for sure!

If 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?

Yes, 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.

Yes; 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.

The 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.

A 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?

Continue that logic, infinite dimensions and the cone would infinitely smaller? Im sure I have that wrong but cool to think about.

Actually, you have that exactly right, and it's a very important fact in mathematics and statistics. A unit sphere takes up a smaller and smaller part of a unit cube as the dimension grows (and a unit cone is similar). In other words, a unit circle fills up most of the unit square (~3.14 out of 4), a unit sphere fills a little over half of the unit cube (~4.2 out of 8), and as the dimension grows, the fraction becomes negligible.

Imagine that you have something which depends on many variables (hundreds), and you're trying to predict its behavior based on your previous experience. There is a high chance that the next combination of variable values that you see will be in one of the corners of the many-dimensional cube, because that's where the volume is (the central part of the cube has negligible volume, as we said above). This means that every measurement is in effect an outlier along several dimensions, making predictions very difficult. This is part of the "curse of dimensionality" in statistics. I have seen some people with excellent understanding of mathematics trip themselves up in this area.

Thanks, that is a highly intuitive explanation

When 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.

I 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.

Wouldn't this be just continuation of the fact that 2-dimentional shapes have 0 volume? Or are you talking about hypercones ?

Yes, the SO link posted has a reply which explains by cutting a cube into 6 equal pyramids. This is easily extended to a higher dimensional case, we can cut a n dimensional hypercube into 2n hyperpyramids in a similar way


Without 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!

There 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):


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.

I 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".

What 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.

a 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".

You 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.

Look 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.

My bad!

The 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.

Ah 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.

The 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.

> 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.

Doesn'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.

The submitter isn't asking us, they're just sharing the question and answers.

When I posted this, the HN submission headline was along the lines of “There’s no intuitive explanation for the volume of a cone”, clearly a comment by the submitter that the SE answers aren’t sufficient for them.

Ah ok, sorry.

Yes, 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.

Paul Lockhart wrote a great book called Measurement, about this and similar questions. I highly recommend it if you're interested in the topic.

Isn’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

In case anyone missed it the gif in the first answer is actually in stereo, if you relax your eyes it’s 3D.

Hope hackernews supports Katex soon.

The stereoscopic animation is really something. That's pretty cool.

I 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.

Volume 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.

I 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.

What would be the shape of a rotationally-symmetric (nearly conical) shape that has 1/2 of the volume of a cylinder with same base?

That 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.

Technically 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…

Perhaps he's also asking for some interpolating curve, see my sibling comment. You can step through the numerical program to see how the shape converges:


We 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

[1] https://www.desmos.com/calculator/x8y5evt0tc

Can someone give me this problem without any spoilers?

Here: why is the volume of a cone one third of the volume of a cylinder?

Is it because Pi is 3?

Because it used to be 1/PI until they legislated PI down to 3? ;)

Hot 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.

For 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.

Intuition 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.

Def encourage you to read Lockhart's Mathematician's Lament: 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.

Interesting. I disliked calculus for exactly the opposite reason. I mean, it also struck me as "all weird tricks, all the way down". But it seemed like a list of tricks one needed to memorize, lacking an intuition.

One of the reasons I became a quantum physicist is because I always sucked at intuition

I was a TA for the EE course that all the ME's had to take. They kept trying to understand Kirchhoff's laws and do circuit analysis using intuition, using water through pipe analogies, stuff like that. The problem is, electricity just isn't intuitive, the analogies are not perfect, nothing in the mechanical world that we understand is quite like it. In fact, we don't really even know why electricity works like it does. I finally told them to stop trying to understand and just do the math and then they had a much easier time.

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