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

Question

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.

Answer 1

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$

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

$\def\Vol{\text{Vol}}$Here is one more approach.

Take a cone with radius $r$ and height $h$ and chop it into four pieces of height $h/4$; call these pieces $F_1$, $F_2$, $F_3$ and $F_4$ from narrowest to widest. Let $C$ be a cylinder of height $h/4$ and radius $r/2$. We claim that $$\Vol(F_1) + \Vol(F_4) = \Vol(F_2) + \Vol(F_3) + \Vol(C). \quad (\ast)$$ To see this, slice through the five solids at height $y$ from the top. The areas of the slices through $F_1$, $F_2$, $F_3$ and $F_4$ are $\pi r^2 (y^2/h^2)$, $\pi r^2 ((y+h/4)^2/h^2)$, $\pi r^2 ((y+2h/4)^2/h^2)$ and $\pi r^2 ((y+3h/4)^2/h^2)$ respectively, and the area of the slice through $C$ is $\pi (r/2)^2 = \pi r^2 (h^2/4)/h^2$. We check that $$(y + 3 h/4)^2 + y^2 = (y + 2 h/4)^2 + (y + h/4)^2 + h^2/4.$$ Multiplying by $\pi r^2/h^2$, we deduce the claim by Cavalieri's principle.

Now, let $K$ be the original cone. The volume of a cone must be proportional to $r^2 h$, so we have $$\Vol(F_1) = (1/64) \Vol(K)$$ $$\Vol(F_2) = (8/64-1/64)\Vol(K) = (7/64) \Vol(K)$$ $$\Vol(F_3) = (27/64-8/64)\Vol(K) = (19/64) \Vol(K)$$ $$\Vol(F_4) = (64/64-27/64)\Vol(K) = (37/64) \Vol(K).$$

Plugging this into $(\ast)$, we have $$\tfrac{1+37}{64} \Vol(K) = \tfrac{7+19}{64} \Vol(K) + \Vol(C)$$ $$\tfrac{3}{16} \Vol(K) = \Vol(C)$$ and $\Vol(K) = \tfrac{16}{3} \Vol(C)$. Now, remember that $C$ had radius $r/2$ and height $h/4$. So the volume of $K$ is $1/3$ the volume of a cylinder of radius $r$ and height $h$, as desired.