Proving that the volume of any cone is $\frac{1}{3}BH$ using fubini's theorem. $B$-base, $H$-height. Proving that the volume of any cone is $\frac{1}{3}BH$ using fubini's theorem. $B$-base, $H$-height. This was on the exam, and thought to be one of the harder proofs. The base , i'm told can be either an elipse or circle, obviously the origin of the cone is irrelevant but should be taken at the coordinate begining. I tried a bunch of things but only kept spining my wheels, wielding nothing of general value. Has anyone done this before with experience with multivariable calculus, and integration?

It is enough to consider a plane parallel to the base, cutting the cone at a distance $x$ from the apex. By homothety, the area $A_x$ of such section equals $C x^2$, where $C$ is a constant for which $CH^2=B$ (the shape of the base does not matter). By Cavalieri's principle / Fubini's theorem the volume of the cone is given by $$ V=\int_{0}^{H} A_x\,dx = \frac{B}{H^2}\int_{0}^{H}x^2\,dx = \color{red}{\frac{1}{3}}BH.$$ A subtle variation: since the shape of the base/sections does not matter, it is enough to prove that the volume of a square pyramid is $\frac{1}{3}BH$. By homothety, it is $DBH$ for some constant $D$. Since we may dissect a unit cube in six square pyramids having height $\frac{1}{2}$ and unit base area, $$ 1^3 = 6\cdot\left(D\cdot 1\cdot\frac{1}{2}\right) $$ and $D=C=\frac{1}{3}$ follows.