Area of a Disc

How much paint do you need for a circular wall? The question reduces to: what is the area enclosed by a circle of a given radius? Archimedes answered it twenty-three centuries ago by squeezing polygons in and out of the circle until they met, a method-of-exhaustion argument that prefigured calculus by two millennia. The answer is shockingly simple: $A = \pi r^2$.

First, define $\pi$

The number $\pi$ is defined as the ratio of a circle's circumference to its diameter:

$\pi = \frac{C}{d}$

This ratio is the same for every circle, large or small, which is one of the deep facts of Euclidean geometry. Numerically $\pi \approx 3.14159\ldots$; it is irrational (Lambert, 1761) and transcendental (Lindemann, 1882).

The area formula

For a disc of radius $r$:

$A = \pi r^2$

If you know the diameter $d = 2r$ instead, substitute: $A = \pi (d/2)^2 = \pi d^2 / 4$.

• Engineering: Pipe-flow rates depend on cross-sectional area $\pi r^2$; doubling the pipe radius quadruples the flow capacity, which is why municipal water mains are large even though pressure stays modest.
• Astronomy: Telescope light-gathering power scales as the disc area $\pi r^2$ of the primary mirror; a 10-metre mirror collects 4x the light of a 5-metre, which is why bigger telescopes resolve fainter objects.
• Statistics: The standard normal density has its mass concentrated within a disc-like region around the mean; computing "probability within radius $r$" in 2D Gaussian samples uses $\pi r^2$ weighted by the density.

Slide the number of rings $n$ from low to high. The disc on the left is partitioned into $n$ concentric annuli; on the right those rings are unrolled into rectangles stacked into a near-triangle. As $n$ grows, the rectangles fill out the triangle of base $2\pi r$ and height $r$, whose area is $\tfrac{1}{2}(2\pi r) \cdot r = \pi r^2$.

Where does $\pi r^2$ come from?

Inscribe a regular $n$-gon inside the circle. Slice it into $n$ thin isosceles triangles meeting at the centre. Each triangle has base $b_n$