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.
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$
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.
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'.
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 $.
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.
Hence
\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
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 $$ V_{cone} = \frac{\pi}{3} r^2 h$$
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.
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.
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 $$\color{blue}{\text{Volume of cone}=\frac{1}{3}(\text{Volume of cylinder})}$$
$\def\Vol{\text{Vol}}$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.
I found an answer with an infinite series. The answer feels satisfying because you derive the prefactor, after some calculation, from the simple equation $1-1/2-1/6 = 1/3$.
I think the Greek knew about infinite series. The solution anyway doesn't require complex mathematical tools.
You can make the cone and cylinder with the required dimensions (base radius $R$, height $R$) by revolving a triangle and a rectangle respectively.
We know the volume of the cylinder, so to find that of the cone, we need to subtract some parts. You can do this by subtracting cones, which are rectangles before revolution. You can of course also build a cone by addition of cylinders like this.
You can rearrange this infinite series to
$1 - S_1 - S_2$
where
$S_1 = \frac14 + \frac18 + \frac1{16} + \cdots$
$S_2 = \frac18 + \frac1{32} + \frac1{128} + \cdots$
By noting that $S_1 - 1/4 = S_1 /2$ and something similar for $S_2$ you find the numerical value:
$1/3 = 1 - 1/2 - 1/6$
I also like how this suggests a way to construct an approximation of $1/3$ if you only know how to divide something in two.
A rather empirical answer is that if you slice the cone in $n$ disks of equal thickness, the volume is proportional to the sum of the squares of the integers.
The square pyramidal numbers are $P_n=\frac{n^3}3+\frac{n^2}2+\frac n6\sim\frac{n^3}3$. This is where the divisor $3$ comes from.
By the way, this works with any section shape (e.g. pyramids), provided that the area varies quadratically.
By the same reasoning, the volume of the hemisphere is proportional to the sum of $n^2-k^2$, which is asymptotic to $(1-\frac13)n^3$.
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
Non-calculus proof:
Let $C_n$ be the volume of an $n$-dimensional cone with height 1 and base area 1 (note that scaling either the height or the base area scales the volume by the same factor, by Cavalieri's principle applied to the scaled cross sections).
For any $0 < h < 1$, we can cut this cone down to a height $h$ base area $h^{n-1}$ cone (where $0 < h < 1$) of volume $C_nh^n$, by removing a frustum of height $1 - h$ and volume between $(1 - h)h^{n-1}$ and $1 - h$.
Thus, $(1 - h)h^{n-1} < C_n(1 - h^n) < 1 - h$,
which factors to: $(1 - h)h^{n-1} < C_n(1 - h)(h^{n-1} + h^{n - 2} + \cdots + h + 1) < 1 - h$,
which simplifies to $h^{n-1} < C_n(h^{n-1} + h^{n - 2} + \cdots + h + 1) < 1$.
Now let $k = (nC_n)^{1/(n-1)}$ so that $C_n = \dfrac{k^{n-1}}{n}$. If $k > 1$ then taking $h = \dfrac{1}{k}$ yields a contradiction on the upper bound, and if $k < 1$ then taking $h = k$ yields a contradiction on the lower bound. So $k = 1$ and $C_n = \dfrac{1}{n}$.
Suppose the cone has volume $V$, base area $B$, and height $h$.
We can use Cavalieri's principle and completing the square to calculate the following volume:
The cross-section at height $0 \le y \le h$ has area
$$B\left(\dfrac{y}{h}\right)^2 + B\left(\dfrac{h - y}{h}\right)^2 = \frac{B}{2}\left(\frac{2y - h}{h}\right)^2 + \frac{B}{2}$$ which is the cross-section of the following:
where each cone has base area $\frac{B}{2}$ and height $\frac{h}{2}$, and the cylinder has base area $\frac{B}{2}$ and height $h$.
Thus, $$V + V = \frac{V}{2} + \frac{Bh}{2}$$ so $V = \frac{Bh}{3}$.
