Proving that the volume of any cone is $\frac{1}{3}BH$ using fubini's theorem. $B$-base, $H$-height.

Question

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?

Answer 1

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.

Answer 2

If $B\subset \Bbb{R}^2$ is the base, we may describe the cone as: $$C = \{ (x,y,z) : (x,y) \in \frac{z}{H} B, 0 \leq z \leq H\}$$ (there are more complicated versions of cones). Note that ${\rm Leb}_2(zB)=z^2 {\rm Leb}_2(B)$ (2-dimensional Lebesgue) and use Fubini to calculate $$ \int_C dx\;dy\;dz $$ by first doing the $x,y$ integration for fixed $z$.