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**

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

**Attribution***Source : Link , Question Author : bryn , Answer Author : Glorfindel*