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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    $\begingroup$ +1. I always believed that a rigorous proof required calculus, but I'd love to be shown otherwise. $\endgroup$
    – Larry Wang
    Jul 24 '10 at 3:41
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    $\begingroup$ 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 $\endgroup$ Jul 24 '10 at 4:04
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    $\begingroup$ 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). $\endgroup$ Jul 24 '10 at 4:13
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    $\begingroup$ @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 $\endgroup$
    – Casebash
    Jul 24 '10 at 6:48
  • $\begingroup$ Integrating the disk area by Simpson's rule is exact because Simpson's rule is exact for polynomials up to cubic. And the disk area for a cone or pyramid is only a quadratic. In fact, Simpson's rule works for many shapes, including spheres, cylinders laying on their side, and others. Simpson's rule is H*(B + 4*M + T)/6, with H height, B area of bottom, M area of middle, T area of top. You can forget several formulas if you can remember this one. $\endgroup$ Sep 8 '16 at 17:27

10 Answers 10


alt text
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. $
alt text
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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    $\begingroup$ Great animation. I wonder if anyone makes any good (toy) physical models of such... $\endgroup$ Jan 16 '13 at 6:04
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    $\begingroup$ But why are there two? $\endgroup$
    – TonyK
    Jul 28 '13 at 17:41
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    $\begingroup$ There are two so you can see the animation in 3D. Place a piece of paper between the two cubes so each of your eyes can only see one of the cubes. If you sort of cross your eyes you can bring the two cubes together and make a stereo image. It is actually pretty cool! $\endgroup$
    – Hucker
    Mar 7 '17 at 19:17

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.


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.

  • $\begingroup$ 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. $\endgroup$ Apr 5 '15 at 19:22
  • $\begingroup$ What do you mean the water has a curved skin at the top? $\endgroup$ Nov 9 '15 at 3:56
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    $\begingroup$ He's referring to the surface tension. When overfill a container with water, it doesn't spill immediately -- instead it clings to the top making a "curved skin" on top. $\endgroup$
    – Cyllindra
    Feb 16 '16 at 18:38
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    $\begingroup$ Of course this is a neat demonstration, but it isn't a mathematical answer at all. $\endgroup$ Aug 11 '17 at 9:26
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    $\begingroup$ @BrianChandler Good thing StackExchange allows multiple answers $\endgroup$
    – Miles Rout
    Nov 8 '17 at 21:12

Here is a derivation of the volume of a cone which does not use calculus, Cavalieri's principle, the method of exhaustion, or any other infinitesimal arguments.

[Edit There is a flaw in this argument, see below]

[Edit 2 The flaw has been fixed, by considering the ratio of the volume of a cone to its circumscribing cylinder under different scalings]

We can split a cone horizontally into two pieces, so that the upper part forms another cone with a smaller base, and the lower part is no longer a cone but an object called a 'frustum'.

                            cone split into frustum and upper cone

For a cone with base radius $r$ and height $h$, we can use a parameter $b$ with $0 \lt b \lt 1$ to define the height of the frustum as $b h$. Because the whole cone and the upper cone form similar triangles in the vertical cross section, the upper cone with height $ (1-b) h $ will have a base radius of $ (1 - b) r $.

                                        cross-section of cone

The volume of the frustum will be equal to the volume of the original cone, less the volume of the upper cone. We don't yet know what form the function representing the volume of a cone will take, so for now we will just write $V_{cone} = V_{cone}(r,h)$ to remind us that it will be some function of the height and base radius. So the volume of the frustum is $$V_{frustum} = V_{cone}(r,h) - V_{cone}((1 - b)r,(1 - b)h)$$

At this point we make the observation that the ratio of the volume of a cone to the volume of it's circumscribing cylinder must be invariant under a scaling on the coordinates (the ratio is homogeneous of degree 0).

$$\frac{V_{cone}(r,h)}{\pi r^2 h} = \frac{V_{cone}(sr,sh)}{\pi (sr)^2 sh}$$

for all $s>0$. If we write $V_{cone} = \hat{Q}\,F(r,h)\, r^2 h$ where $F(r,h)$ is some as yet unknown function and $\hat{Q}$ is a constant, then

$$F(r,h) = F(sr,sh)$$

so $F(r,h)$ is also homogeneous of degree 0.


\begin{array}{l@{}l} V_{frustum} &{}= V_{cone}(r,h) - V_{cone}((1 - b)r,(1 - b)h) \\ &{}= \hat{Q} \, F(r,h) \,r^2 h - \hat{Q} \, F((1-b)r, (1-b)h) \, (1-b)^2r^2 (1-b)h \\ &{}= \hat{Q} \, F(r,h) \, r^2 h ( 1 - (1-b)^3) \\ &{}= Q \, r^2 h (3 b - 3 b ^2 +b^3) \end{array}

where $Q = \hat{Q} \, F(r,h)$

Now consider the following figure

                                        cone with inscribed and circumscribed cylinders of height bh

It is clear that the volume of the frustum of height $b h$ must be bigger than the inner cylinder of radius $ (1-b) r$ and height $b h$ and it must also be less than the volume of the outer cylinder with radius $ r $ and height $b h$.

$$\pi (1-b)^2r^2 b h \lt V_{frustum} \lt \pi r^2 b h$$

Substituting the expression for $V_{frustum}$ from above and dividing everything through by $ b \pi r^2 h $

$$ (1-b)^2 \lt \frac{Q (3 - 3b + b^2)}{\pi} \lt 1$$

This must hold for all $0 \lt b \lt 1$.

At this point, we could use the familiar argument about limits - in particular, as $b$ gets closer to zero, the lower bound approaches the upper bound of $1$, so $\frac{Q 3}{\pi} = 1$ or $Q = \frac{\pi}{3}$.

However, it is possible to find the value of $Q$ in a different way, that does not involve some limit process.

First, observe that the value of $Q$ has bounds placed on it by the geometry of the problem $0 \lt Q \lt \pi$ since the cone must have some volume, and that volume must be less than the volume of a cylinder with radius $r$ and height $h$. What we are going to show is that for all values of $Q$ in this range, with just one exception, there is a choice of $b$ with $0 \lt b \lt 1$ that causes the above inequality not to hold. In the spirit of Sherlock Holmes, '..when you have eliminated the impossible, whatever remains, however improbable [or in our case, expected], must be the truth'.

We split the problem up into two parts. The upper bound of the inequality does not hold when

$$\frac{Q (3 - 3b + b^2)}{\pi} = 1$$

Solving for $b$

$$b = \frac{3}{2}-\sqrt{\frac{\pi}{Q}-\frac{3}{4}}$$

Now introduce a parameter $\alpha$ and write $Q= \pi / (1+\alpha+\alpha^2) $. Then for $0 \lt \alpha \lt 1$ we have $\pi/3 \lt Q\lt\pi$ and the above equation reduces to $b=1-\alpha$, so $0 \lt b \lt 1$.

The lower bound of the inequality does not hold when

$$(1-b)^2 = \frac{Q (3 - 3b + b^2)}{\pi}$$

Solving for $b$

$$b = 1 - \frac{ (\frac{1}{2} + \sqrt{\frac{\pi}{Q}-\frac{3}{4}})}{\frac{\pi}{Q}-1}$$

Introduce a parameter $\alpha$ as before, but this time write $Q=\pi \alpha^2 / (1+\alpha+\alpha^2) $. Then for $0 \lt \alpha \lt 1$ we have $0 \lt Q \lt \pi/3$ and the above equation again reduces to $b=1-\alpha$, so $0 \lt b \lt 1$.

Therefore we have $0 \lt Q \lt \pi$ by the geometry of the problem, but whenever $0 \lt Q \lt \frac{\pi}{3}$ or $\frac{\pi}{3} \lt Q \lt \pi$ there exists at least one value for $b$ with $0 \lt b \lt 1$ for which the inequality does not hold. The only remaining possibility on the interval $0 \lt Q \lt \pi$ is $Q=\frac{\pi}{3}$ (for all $r,h > 0$), and so $$ V_{cone} = \frac{\pi}{3} r^2 h$$

  • $\begingroup$ Your fundamental inequality about volume of frustum of a cone assumes that $r, h$ are constant and the volume is a function of $b$. Thus your argument after that inequality does not hold. $\endgroup$ Aug 11 '17 at 8:33
  • $\begingroup$ I understand your first sentence, it is correct. I don't see how your second sentence follows from it? $\endgroup$
    – Will
    Aug 11 '17 at 9:31
  • $\begingroup$ @ParamanandSingh Sorry, I'm new here, didn't realise how the comments work. The $r$ and $h$ are the dimensions of the full cone, and $b$ is the proportion of the height at which the cone is cut. I don't see how your second sentence follows from that? $\endgroup$
    – Will
    Aug 11 '17 at 9:39
  • $\begingroup$ It should clear that the volume of the conical part also depends on $b$ and as $b$ varies from $0$ to $1$ it varies from $0$ to the volume of full cone. Let this function be called $C(b) $ and you have set $C(0)$ as volume of full cone. Then volume of frustum say $F(b) $ is given by $F(b) =C(0)-C(b)$. Now your inequality is $(1-b)^{2}b<F(b)/k<b$ where $k$ is a constant $\pi r^{2}h$. From this how can you figure out anything about $C(b)$ or $C(0)$? $\endgroup$ Aug 11 '17 at 17:53
  • $\begingroup$ In fact you assume that the volume of a cone is a constant multiple of product of square of radius and height and on the basis of that assumption and the inequality you try to show that the constant must be $\pi/3$. This reasoning is correct. But there is no proof that the volume of cone must be a constant multiple of $r^{2}h$. This fact does not follow from the inequality satisfied by volume of frustum of a cone. $\endgroup$ Aug 11 '17 at 17:53

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 substituted 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 square's 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.

  • $\begingroup$ How did the ancient Greeks do it? They did not have approximation or limit theory, nor could they write and manipulate a formula like pi*r^2. For us it is easy. For them, it was formidable. $\endgroup$ Sep 14 '19 at 22:08

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.


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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    $\begingroup$ This is not much of an argument. Could you provide some algebraic detail? $\endgroup$ Nov 28 '12 at 21:53
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    $\begingroup$ Yes, it's not clear what the "so forth" means -- what is the fourth dimensional counterpart? $\endgroup$
    – bryn
    Dec 1 '12 at 9:30
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    $\begingroup$ 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. $\endgroup$
    – Peter Webb
    Oct 22 '14 at 11:36

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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    $\begingroup$ The OP was interested in a proof that didn't involve taking the integral of a volume of a surface of revolution. $\endgroup$
    – ogogmad
    Jun 16 '15 at 11:54
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    $\begingroup$ yes, you are right $\endgroup$ Jun 16 '15 at 12:25

Imagine a pyramid inside a cube; One of the point of the pyramid is touching the top face of the cube, the point can be anywhere as long as it is 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, it would look exactly like the pyramid, except that the volume would be exactly $\frac {1}{8}$ of the original.

Now let's look at the lower half, you would probably notice that you can cut a part of it to get the exact same shape as the top half. Cutting it so you have $2$ of those small pyramids. The remaining object will have a volume $\frac {1}{4}$ of the cube, the two small pyramids is $\frac {1}{8}$ of the original.since you have 2 of them. The two parts combined will be $\frac {1}{4}$ of the original pyramid, Which means the remaining bit is 3/4 of the original pyramid,which is 1/4 of the cube the 2 parts are a 1/3 of the remaining part so if we add them together $\frac{1+1/3}{4}$=$\frac 1{3}$

This might be a little confusing But it works :P

  • $\begingroup$ You didn't deal with cones here at all. $\endgroup$
    – Allawonder
    Jul 8 '20 at 23:37

It is easy to imagine that the volume of a Cone is equal to the volume of a square pyramid as stated in other answers.

Now we only have to show that the volume of square pyramid is 1/3 b h.

For that, consider this cube made of 6 square pyramids: enter image description here

Let the height of the pyramid be h.

So the side length of the cube is 2h.

The volume of the cube

= base $\times$height of the cube

= base $\times$ $2h$

From the figure,

The volume of the pyramid

= $\frac{1}{6}$ $\times$ volume of the Cube

= $\frac{1}{6}$ $\times$ base $\times$ $2h$

= $\frac{1}{3}$ $\times$ base $\times$ $h$


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