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

The volume of a cone with height $$hh$$ and radius $$rr$$ is $$\frac{1}{3} \pi r^2 h\frac{1}{3} \pi r^2 h$$, 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 $$r\sqrt\pi r\sqrt\pi$$. Such a pyramid has volume $$\frac13 \cdot h \cdot \pi \cdot r^2. \frac13 \cdot h \cdot \pi \cdot r^2.$$
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 $$a/h \times ra/h \times r$$. The side length of the square pyramid’s cross section will be $$\frac ah \cdot r\sqrt\pi.\frac ah \cdot r\sqrt\pi.$$
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:$$\frac13\cdot h \cdot \pi \cdot r^2 \frac13\cdot h \cdot \pi \cdot r^2$$