Prismoidal Formula

Problem

A prismatoid is a polyhedron whose vertices all lie in one or the other of two parallel planes. The perpendicular distance between the two planes is called the height or altitude of the prismatoid. The faces that lie in the parallel planes are called the bases of the prismatoid. The midsection is the polygon formed by cutting the prismatoid by a plane parallel to the bases halfway between them.

The cross-sections of a prismatoid follow a quadratic function along its height. The general formula for computing the volume of a prismoid is

$$V = \frac{h}{6}\left[A_T + 4A_M + A_B \right]$$

where $h$ is the height, $A_T$ is the area of the top base, $A_M$ is the area of the midsection, and $A_B$ is the area of the bottom base. Use integration to prove why this volumetric formula is exact.

Solution

Let $h = b-a$ be the height of the prismatoid. For a solid with cross-sections that follow a quadratic function along its height, $f(x) = Ax^2 + Bx + C$ and

$$V = \int_{a}^{b} f(x) \, dx.$$

Let $A_T = f(b)$, $A_M = f\left(\frac{a+b}{2}\right)$, and $A_B = f(a)$. Then it must be shown that

$$\int_{a}^{b} Ax^2 + Bx + C \, dx = \frac{b-a}{6}\left[f(b) + 4 \times f\left(\frac{a+b}{2}\right) + f(a) \right].$$

Working out the algebra on the right-hand side yields

$$\begin{align} \frac{b-a}{6}\left[f(b) + f\left(\frac{a+b}{2}\right) + f(a) \right] &= \frac{b-a}{6}\left[Ab^2 + Bb + C + 4 \times \left[A\left(\frac{a+b}{2}\right)^2 + B\left(\frac{a+b}{2}\right) + C \right] + Aa^2 + Ba + C \right] \\[5 pt] &= \frac{b-a}{6}\left[2A(b^2+ab+a^2) + 3B(b+a) + 6C \right]. \end{align}$$

The algebra is simple but tedious.

Now for the left-hand side (which requires more tricks).

$$\begin{align} \int_{a}^{b} Ax^2 + Bx + C \, dx &= \frac{Ax^3}{3} + \frac{Bx^2}{2} + Cx \Biggr|_a^b \\[5 pt] &= \frac{A}{3}(b^3-a^3) + \frac{B}{2}(b^2-a^2) + C(b-a) \end{align}$$

Use the following factorizations:

1. $b^2 - a^2 = (b-a)(b+a)$
2. $b^3 - a^3 = (b-a)(b^2 + ab + a^2)$

$$\begin{align} \int_{a}^{b} Ax^2 + Bx + C \,dx &= (b-a)\left[\frac{A}{3}(b^2+ab+a^2) + \frac{B}{2}(b+a) + C \right] \\[5 pt] &= \frac{b-a}{6}\left[2A(b^2+ab+a^2) + 3B(b+a) + 6C \right]. \end{align}$$

Since the left-hand and right-hand sides for evaluating the volume are equal,

$$V = \frac{h}{6}\left[A_T + 4A_M + A_B \right].$$