In the treatise On Spirals, also dedicated to Dositheus, Archimedes studies a new curve, defined by a motion.

If a straight line in a plane revolves at a uniform rate about one of its extremities, which remains fixed, and returns to the position from which it started, and if, at the same time, a point moves at a uniform rate along the line, starting from the fixed extremity, the point will describe a spiral in the plane.

— Archimedes, On Spirals, definition

The equation

In polar coordinates, the distance $r$ from the point to the centre grows in the same proportion as the angle $\theta$ turned by the line:

$$r = a\,\theta.$$

With each full turn the radius grows by the same amount, $2\pi a$, so the successive turns are equally spaced, like the grooves of a vinyl record or a hose coiled on the ground.

Archimedes computed the area swept by the radius during the first turn. The technique is again exhaustion: divide the turn into many narrow circular sectors, some inside and some outside the curve, and add them up. In modern notation, a sector of angle $d\theta$ and radius $r$ has area $\frac{1}{2}r^2\,d\theta$, and so

$$A = \int_0^{2\pi} \frac{1}{2}(a\theta)^2\,d\theta = \frac{a^2}{2}\cdot\frac{(2\pi)^3}{3} = \frac{4}{3}\pi^3 a^2.$$

The circle whose radius is the final radius of the first turn, $R = 2\pi a$, has area $\pi R^2 = 4\pi^3 a^2$. The area of the spiral is exactly one third of it:

$$\frac{A}{\pi R^2} = \frac{\frac{4}{3}\pi^3 a^2}{4\pi^3 a^2} = \frac{1}{3}.$$

To get there, Archimedes needed a formula for the sum of squares $1^2 + 2^2 + \cdots + n^2$, which amounts to saying that this sum lies between $\frac{n^3}{3}$ and $\frac{(n+1)^3}{3}$. It is the same sum that, in calculus, gives $\int x^2\,dx = \frac{x^3}{3}$.

The tangent and the circumference

Archimedes also found the direction of the tangent to the spiral, a remarkable feat in an age without derivatives. The result has a surprising consequence: at the end of the first turn, the tangent cuts the line through the centre perpendicular to the radius at a point whose distance from the centre is exactly equal to the circumference of the circle of radius $R$. The spiral thus makes it possible to construct a segment the length of a circumference, and with it to square the circle. It is not a ruler-and-compass construction, because the spiral itself cannot be drawn with those instruments; but it shows that the problem can be solved with a curve only a little more complicated.

Exercise: the second turn

What is the area swept by the radius during the second turn, from $\theta = 2\pi$ to $\theta = 4\pi$? Compare it with that of the first turn.

$\int_{2\pi}^{4\pi}\frac{a^2\theta^2}{2}\,d\theta = \frac{a^2}{6}\left[(4\pi)^3 - (2\pi)^3\right] = \frac{a^2}{6}\cdot 56\pi^3 = \frac{28}{3}\pi^3 a^2$. The first turn had $\frac{4}{3}\pi^3 a^2$; the second sweeps seven times as much. (The area swept by the second turn includes the region already swept by the first; Archimedes also computes the area of the ring between the two turns, which is six times the first.)

Exercise: the sum of squares

Check Archimedes’s inequality for $n = 10$: $\frac{10^3}{3} < 1^2 + 2^2 + \cdots + 10^2 < \frac{11^3}{3}$.

The sum is $\frac{10 \cdot 11 \cdot 21}{6} = 385$. The bounds are $\frac{1000}{3} \approx 333.3$ and $\frac{1331}{3} \approx 443.7$. Indeed, $333.3 < 385 < 443.7$. As $n$ grows, the sum divided by $n^3$ approaches $\frac{1}{3}$, and that is where the spiral’s “one third” comes from.

Sources: Archimedes, On Spirals, definitions and propositions 10, 18–20 and 24 · Thomas L. Heath, The Works of Archimedes (1897) · E. J. Dijksterhuis, Archimedes (1956), ch. XI.

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