Visual Proof: Where the Volume of a Pyramid Formula Actually Comes From

Everything you need to know about Visual Proof: Where the Volume of a Pyramid Formula Actually Comes From, featuring critical updates.

The transition from physical models to mathematical proof reaches completion through formal calculus. By treating the pyramid as a stack of infinitesimally thin horizontal cross sections, integration demonstrates why three dimensions inevitably produce the coefficient $\frac{1}{3}$.

Anchor the apex of a pyramid at the coordinate origin ($y = 0$) and orient the central axis downward along the vertical axis toward the base at $y = h$. Each cross section at depth $y$ forms an infinitesimal slab of thickness $dy$.

Using the quadratic cross-sectional profile derived through basic geometric scaling, the differential volume slice $dV$ equals:

$$dV = A(y) \, dy = \left( \frac{B}{h^2} y^2 \right) dy$$

Integrating these differential slabs from the apex at $y = 0$ to the full base depth at $y = h$ yields the total enclosed volume:

$$V = \int_{0}^{h} \frac{B}{h^2} y^2 \, dy$$

Pull the geometric constants outside the integral sign:

$$V = \frac{B}{h^2} \int_{0}^{h} y^2 \, dy$$

Evaluating the definite integral through the fundamental power rule:

$$\int{0}^{h} y^2 \, dy = \left[ \frac{y^3}{3} \right]{0}^{h} = \frac{h^3}{3} - 0 = \frac{h^3}{3}$$

Substitute that result back into the volume expression:

$$V = \frac{B}{h^2} \left( \frac{h^3}{3} \right) = \frac{1}{3} B h$$

The integration exposes the core mechanic. The one-third fraction is not an empirical estimate or an arbitrary geometric constant. It is the direct consequence of integrating a second-degree polynomial across space. Because physical area expands quadratically over distance ($y^2$), integrating across the third dimension inevitably produces a factor of $\frac{1}{3}$.