Volume of the pyramid

A pyramid with base of area $B$ and height $H$ is “sliced” into horizontal layers of thickness $dz$. At height $z$ (measured from the apex), the cross-section is similar to the base with linear scale factor $z/H$: the area of the cross-section is therefore $A(z) = B\cdot(z/H)^2$.

The pyramid sliced: at height $z$ from the apex the cross-section is similar to the base with factor $z/H$.

The volume of a single layer is $dV = A(z)\,dz = B\cdot\dfrac{z^2}{H^2}\,dz$. We sum from $z=0$ (apex) to $z=H$ (base): $V = \int_0^H B\cdot\frac{z^2}{H^2}\,dz = \frac{B}{H^2}\int_0^H z^2\,dz = \frac{B}{H^2}\cdot\frac{H^3}{3} = \boxed{\dfrac{1}{3}B\cdot H}.$

Remark

The formula $V = \frac{1}{3}Bh$ holds for any pyramid (even a non-right one, even with an irregular base): the only thing that matters is that the cross-sections at height $z$ be similar to the base with a linear factor. It is the same principle as the cone (a “pyramid with a circular base”) and it explains why the factor $1/3$ appears in both.

Links

Topics: Integrale
Concepts: Integrale definito · Principio di cavalieri · Volume per sezioni
Skills: Calcolare · Integrare