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
$$ V_{cone} = \frac{\pi}{3} r^2 h$$