# Why is the volume of a cone one third of the volume of a cylinder?

The volume of a cone with height $h$ and radius $r$ is $\frac{1}{3} \pi r^2 h$, which is exactly one third the volume of the smallest cylinder that it fits inside.

This can be proved easily by considering a cone as a solid of revolution, but I would like to know if it can be proved or at least visual demonstrated without using calculus.

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+1. I always believed that a rigorous proof required calculus, but I'd love to be shown otherwise. – Larry Wang Jul 24 '10 at 3:41
The Egyptians new how to calculated pyramids. Turned out the shape didn't matter just the base area. Democritus put it together en.wikipedia.org/wiki/Democritus#Mathematics – Jonathan Fischoff Jul 24 '10 at 4:04
another twist on this: The other well-known calculus way to find the volume of the cone is to integrate over the disks that make up the cross sectional areas, as you go from the base at 0 to the top at h. The same argument (though it uses calculus) can show that, if you take an arbitrary region in the plane, and first form a "cylinder" of height h from it by extruding it a distance h, then form a "cone" from it by extruding and then tapering linearly, the volume of the resulting "cone" is 1/3 the volume of the resulting "cylinder" (loosely, making solids pointy nicely gives 1/3 the volume). – Jamie Banks Jul 24 '10 at 4:13
@Katie: Well it only takes calculus to formally prove it. Imagining a cylinder as an "infinite-a-gon pyramid" is really good for intuitively understanding thas – Casebash Jul 24 '10 at 6:48

A visual demonstration for the case of a pyramid with a square base. As Grigory states, Cavalieri's principle can be used to get the formula for the volume of a cone. We just need the base of the square pyramid to have side length $r\sqrt\pi$. Such a pyramid has volume $\frac13 \cdot h \cdot \pi \cdot r^2.$

Then the area of the base is clearly the same. The cross-sectional area at distance a from the peak is a simple matter of similar triangles: The radius of the cone's cross section will be $a/h \times r$. The side length of the square pyramid's cross section will be $\frac ah \cdot r\sqrt\pi.$
Once again, we see that the areas must be equal. So by Cavalieri's principle, the cone and square pyramid must have the same volume:$\frac13\cdot h \cdot \pi \cdot r^2$

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Great animation. I wonder if anyone makes any good (toy) physical models of such... – Jesse Madnick Jan 16 '13 at 6:04
But why are there two? – TonyK Jul 28 '13 at 17:41

One can cut a cube into 3 pyramids with square bases -- so for such pyramids the volume is indeed 1/3 hS. And then one uses Cavalieri's principle to prove that the volume of any cone is 1/3 hS.

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It is because a triangle in a box that has the same height and length is 1/2 if the square because it is in the second dimension so if you move in to the third dimension it will change to 1/3 and so forth.

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This is not much of an argument. Could you provide some algebraic detail? – Simon Hayward Nov 28 '12 at 21:53
Yes, it's not clear what the "so forth" means -- what is the fourth dimensional counterpart? – bryn Dec 1 '12 at 9:30
Its not much of an argument, but it is true. An n dimensional pyramid is made from an n-1 dimensional figure where every point is joined to a single point. A 2D pyramid is a triangle. The proof is not really practical in a comment, but is a simple generalisation of the 3D case for the cone or pyramid, and ultimately the factor of 1/n derives from d(x^n)/dx = nx^(n-1) where x^n is the volume of the enclosing hypercube, just as the factor of 1/3 derives from this in the 3D case. – Peter Webb Oct 22 '14 at 11:36

I just did a demonstration with my class that took about 2 minutes. Granted it was just inductive reasoning but it satisfied the students for now. I had 2 pairs of students come up to the front of the class. One pair had a cone and a cylinder. One pair had a pyramid and a prism. Each pair had solids with a congruent base and height. The person with the cone had to see how many times they could fill the cone with water and fit it into the cylinder. Similarly the person with the pyramid had to see how many times they could fill the pyramid with water and fit it into the prism. Other than ensuring that the cone and the pyrmaid were not overfilled (taking into consideration that the water has a curved skin at the top) the experiment was simple and the demonstration made it easier for the students to remember the relationship. Hope this helps.

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I remember someone else posting somewhere that their teacher did the same with their class, and that they had never forgotten the formula since. S/he even remembered which students were called up. – Akiva Weinberger Apr 5 at 19:22
What do you mean the water has a curved skin at the top? – Kyle Delaney Nov 9 at 3:56

You can use Pappus's centroid theorem as in my answer here, but it does not provide much insight.

If instead of a cylinder and a cone, you consider a cube and a square-based pyramid where the "top" vertex of the pyramid (the one opposite the square base) is shifted to be directly above one vertex of the base, you can fit three such pyramids together to form the complete cube. (I've seen this as physical toy/puzzle with three pyramidal pieces and a cubic container.) This may give some insight into the 1/3 "pointy thing rule" (for pointy things with similar, linearly-related cross-sections) that Katie Banks discussed in her comment.

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Let $r$ & $h$ be respectively the radius & the normal height of a cone. Now place it with its geometrical axis coincident with the x-axis then the cone can be generated by rotating a straight line:$\color{blue}{y=\frac{r}{h}x}$, passing through the origin, about the x-axis. Hence, the volume of the cone $$\color{blue}{V_{cone}}=\int\pi y^2 dx=\int_{0}^h \pi\left(\frac{r}{h}x\right)^2 dx$$ $$=\frac{\pi r^2}{h^2}\int_{0}^h x^2 dx=\frac{\pi r^2}{h^2} \left[\frac{x^3}{3}\right]_{0}^h=\frac{\pi r^2}{h^2} \left[\frac{h^3}{3}\right]$$$$\color{blue}{=\frac{1}{3}\pi r^2h}$$

Similarly, the cylinder with a radius $r$ & normal height $h$ can be generated by rotating a straight line:$\color{blue}{y=r}$, parallel to the x-axis, about the x-axis. Hence, the volume of the cylinder $$\color{blue}{V_{cylinder}}=\int\pi y^2 dx=\int_{0}^h \pi\left(r\right)^2 dx$$ $$=\pi r^2\int_{0}^h dx=\pi r^2 \left[x\right]_{0}^h=\pi r^2 \left[h-0\right]\color{blue}{=\pi r^2h}$$ Thus. we find that $$\color{blue}{\text{Volume of cone}=\frac{1}{3}(\text{Volume of cylinder})}$$

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The OP was interested in a proof that didn't involve taking the integral of a volume of a surface of revolution. – jkabrg Jun 16 at 11:54
yes, you are right – Harish Chandra Rajpoot Jun 16 at 12:25

imagine a pyramid inside a cube 1 of the point of the pyramid is touching the top face of the cube That one point i just mentioned can go anywhere as long as its on the top face of the cube and still not change the volume. Imagine that the point i just mentioned went to the corner of the cube Cut the top half of that pyramid The top half u just cut would look exactly like the pyramid,except that the volume is exactly 1/8 of the original Now look at the lower half You will notice that u can cut a part of it to get the exact same shape as the top half. Cut it so you have 2 of those small pyramids. The remaining object will have a volume 1/4 of the cube the two small pyramids is 1/8 of the original.since u have 2 of them the two parts combined will be 1/4 of the original pyramid Which means the remaining bit is 3/4 of the orignal pyramid,which is 1/4 of the cube the 2 parts are a 1/3 of the remianing part so if we add them together (1+1/3)/4=1/3 sorry this is confusing but it works :P

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You'd have to use calculus to derive this formula in the correct manner. I have created a playlist that you can watch if you'd like to know exactly how this formula can be derived from first principles. The playlist is almost an hour long:

Finding The Formula For Volumes Of Cones

In this playlist you will learn what slopes are, how to differentiate, how to find areas underneath curves, how to find volumes using co-ordinate axes and then how to find the volumes of cones formula using calculus.

If you have any extra questions regarding these videos, I will reply to them below.

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I managed to find the volume of a cone without calculus using an observation that I made.

First, I put a cone on a cartesian plane, with the tip at the origin. Thus, an equation to describe the radius(x) would be the radius over the height times x. Then, I substitued this equation into pi r squared to get cross sectional area as function of x.

I then observed how the volume of the cone could be approximated by using disks, the width of each being the height of the cone divided by the number of disks. So, the volume as a function of x would be the area as a function of x times the height divided by n, or the number of disks. However, instead of using integration to sum the volumes of all the disks, I observed that if I moved along the height in increments equal to the width of each cylinder, that the volumes of the cylinders increased in a sequence of squares, the second disk being 4 times the volume of the first, the third being 9 times, the fourth being 16 times, and so on.

To me, this showed that the second disk can be broken up into 4 cylinders equal to the volume of the first disk, the third into 9, the fourth into 16, and so on. So, the volume of a cone is equal to the volume of the first disk times the sum of all the cylinders, which we can get using the summation of squares formula. So, I got the volume of the first cylinder by putting the width of one cylinder into the volume as a function of x formula, which got pi r squared times the height over n cubed. I then multiplied this by the summation of squares formula to get pi*r^2*h*(n(n+1)(2n+1))/(6n^3) Then, I let n go to infinity, which resulted in the volume of a cone being (pi*r^2*h)/3.

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First You need to understand how to derive the formula to find the volume of apyramid. Once you have this down then it's simple to see why it is $\frac13\pi r^2 h$ because a cone is no more then a circular pyramid.

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