A band of a sphere is a band of its cylinder

The slice that has to match found the volume of a sphere by comparing it, slice by slice, with a cylinder from which a cone had been removed. Along the way it noted a second comparison and set it aside: the sphere’s surface matches the side of its cylinder, band for band, with no cone removed and no correction at all. Archimedes counted the two results together as his finest work — the sphere is two thirds of its circumscribing cylinder in volume and in surface — and asked for a sphere inside a cylinder to be carved on his tomb.

The volume result compares slices. The surface result compares bands, and it is the stranger of the two, because the band of a sphere and the band of a cylinder look nothing alike.

Take a sphere of radius one and the cylinder that just contains it, of radius one and height two. Cut both with two horizontal planes $0.3$ apart. On the cylinder the band is a strip of height $0.3$ wrapped round a circle of circumference $2\pi$, with area $0.6\pi \approx 1.885$. On the sphere the band depends dramatically on where it is cut. At the equator it is a wide belt, standing nearly upright. Near the pole it is a small cap, lying nearly flat. And every one of them has area $1.885$.

A band whose width grows as its girth shrinks #

The equality is not a coincidence of the three bands drawn. It holds for every band, and the reason can be seen on one thin band at a time.

A thin band of the sphere at height $z$ is almost a truncated cone — a lampshade. Its area is its circumference times its slant width. The circumference is $2\pi\rho$, where $\rho = \sqrt{1 - z^2}$ is the sphere’s radius at that height, smaller than the cylinder’s radius $1$. The slant width $ds$ is the length of arc between the two planes, larger than their separation $dz$, because the sphere’s surface is tilted.

The two factors are related by similar triangles. The radius from the centre to the band, of length $1$, makes with the horizontal the same angle that the tangent to the sphere makes with the vertical — the radius and the tangent are perpendicular. So the small triangle with sides $dz$ and $ds$ is similar to the large triangle with sides $\rho$ and $1$, and

$$\frac{ds}{dz} = \frac{1}{\rho}, \qquad 2\pi\rho\, ds = 2\pi\, dz.$$

The circumference shrinks by exactly the factor the slant grows. Every thin band of the sphere has the area of the thin band of the cylinder at the same height, and adding bands adds areas. The figure does not assume it: it measures the band from $0.45$ to $0.70$ by summing thin rings along the sphere’s profile and compares the total with $2\pi \times 0.25$, and the two agree to the precision of the sum.

The surface, and the tomb #

The whole sphere is the band from height $-1$ to height $1$, so its area is the cylinder’s side, $2\pi \times 2 = 4\pi$ — four times the area of a great circle. With a radius $r$ in place of one, $4\pi r^2$.

That is one half of what Archimedes asked to have carved. The other half is the volume, $\tfrac43 \pi r^3$, which the slice that has to match derived; the cylinder holds $2\pi r^3$, so the sphere is two thirds of it. And the cylinder’s whole surface, side and two ends, is $4\pi r^2 + 2\pi r^2 = 6\pi r^2$, of which the sphere’s is again two thirds. Cicero, as a magistrate in Sicily in 75 BC, found the neglected tomb by looking for the sphere and cylinder among the monuments outside Syracuse, and had it cleared.

The two results are also tied to each other. A sphere of radius $r$ is a stack of thin spherical shells, and the volume grows by surface times thickness as $r$ grows, so the volume’s rate of change with the radius is the surface: the derivative of $\tfrac43\pi r^3$ is $4\pi r^2$. That is the same relation a circle unrolled found between a disc’s area and its circumference, one dimension up.

A cap is a flat disc in disguise #

Archimedes stated the theorem in a form that avoids the cylinder altogether, and it is the most surprising way to say it.

A polar cap of height $h$ on a sphere of radius $R$ has area $2\pi R h$, by the band argument. Now draw the straight line from the pole to any point on the cap’s rim. Its length $c$ is a chord of the great circle, and the right angle that every angle standing on a diameter makes gives $c^2 = 2R \cdot h$: the chord is the geometric mean of the diameter and the cap’s height. So

$$\text{area of the cap} = 2\pi R h = \pi c^2,$$

the area of a flat disc whose radius is the straight-line distance from the pole to the rim. A skullcap has exactly the area of the flat circle of cloth whose radius reaches from its crown to its edge, measured straight through the head. For a hemisphere the chord is $R\sqrt2$ and the disc has area $2\pi R^2$, half the sphere; for the whole sphere the chord is the diameter $2R$ and the disc has area $4\pi R^2$, the whole surface as one disc of radius $2R$.

That form of the statement is what makes a certain map possible: sending each point of the sphere to the point at the same bearing from the pole, at a distance equal to its straight-line distance from the pole, preserves area — every cap becomes a disc of the same area, and so does every sector of a cap. It is Lambert’s other equal-area map, the azimuthal one, and it is the hat-box theorem again, read round the pole instead of up the axis.

Archimedes’ route, without a limit #

Archimedes had no derivatives and no limits, and his proof is a squeeze of the kind pinned between two sequences used for the circle.

Inscribe a regular polygon in a great circle and spin it about a diameter through two of its corners. The result is a stack of cones and truncated cones, each with a surface that can be computed exactly: a truncated cone with end radii $r_1$ and $r_2$ and slant $s$ has side area $\pi (r_1 + r_2) s$. Adding them up, the stack’s surface comes out, for a polygon of $2k$ sides, as exactly $4\pi \cos(\pi/2k)$ — less than $4\pi$ for every $k$, and as close to it as desired for large $k$. A stack circumscribed about the sphere gives a matching sequence from above.

Archimedes then argued by double contradiction. If the sphere’s surface were less than $4\pi$, some inscribed stack would exceed it, which is impossible because the stack lies inside; if it were more, some circumscribed stack would fall short of it, which is impossible because that stack lies outside. So it is exactly $4\pi$. The figure’s percentages are the squeeze made visible: $92.4\%$, $98.1\%$, $99.5\%$ and on, a sequence nobody has to take to its limit to see where it is going.

How much of the Earth lies in each zone #

The hat-box theorem makes some questions about the globe answerable by subtraction.

The Arctic Circle lies at latitude $66.56°$, at height $\sin 66.56° \approx 0.9175$ above the equatorial plane on a sphere of radius one. The cap north of it spans a height of $0.0825$ out of the sphere’s $2$, so it holds $4.1\%$ of the Earth’s surface. The tropics lie at $\pm 23.44°$, heights $\pm 0.3978$, so the belt between them holds $0.7956/2 = 39.8\%$ — two fifths of the planet, from a strip a quarter of the way to each pole. Each temperate zone holds $26.0\%$. Areas on the sphere are proportional to heights, so any band’s share is its height divided by the diameter.

The latitudes do not suggest those proportions. The tropics span $47°$ of the $180°$ from pole to pole, about a quarter, and hold two fifths of the surface; the polar caps span $47°$ between them too, and hold a twelfth. Latitude is an angle, and equal angles near the equator correspond to more height, and so to more area, than equal angles near a pole.

The only fair way to pick a point on a sphere #

That distinction settles a practical question: how to choose a point at random on a sphere so that every region is equally likely.

The obvious method is to pick a latitude and a longitude, each evenly. It is wrong, and the right-hand sphere shows how: the points bunch at the poles, because equal ranges of latitude near the poles cover much less area than equal ranges near the equator. The cap above height $0.8$ gets about a fifth of the points while holding a tenth of the area.

The hat-box theorem gives the correct method in one line. Since equal heights carry equal areas, choosing the height evenly between $-1$ and $1$ and the longitude evenly spreads points evenly over the sphere. The left-hand sphere is drawn that way, and a tenth of its points lie above $0.8$. The rule is special to the ordinary sphere: on a sphere in many dimensions a single coordinate of a random point is not evenly spread but concentrated near nought, and the flat distribution in three dimensions is the hat-box theorem and nothing else.

A map on which every area is true #

Pressing the sphere sideways onto its cylinder and unrolling the cylinder makes a map.

Move every point of the sphere straight out to the cylinder, keeping its height, and unroll. Meridians become evenly spaced vertical lines; parallels become horizontal lines at heights $\sin(\text{latitude})$, crowding towards the top and bottom. A region near a pole is stretched east–west, because its circle of latitude has become as long as the equator, and squashed north–south by exactly the same factor, because its height is so much less than its latitude would suggest. The theorem says the two effects cancel, so every region of the map has the area it has on the sphere. Johann Heinrich Lambert published the projection in 1772, and it is the parent of every equal-area world map with straight parallels, including the ones used to show the true relative sizes of continents.

The price is shape. Greenland on this map is correctly sized and grotesquely flattened, and no map can avoid paying one price or the other: the sphere’s curvature forbids a map that preserves both areas and angles, as the triangle a globe gets wrong shows for distances. The stereographic map of the sphere the complex numbers live on keeps angles and distorts areas; Lambert’s keeps areas and distorts angles.

Why it works in three dimensions and nowhere else #

The cancellation is exact only for the ordinary sphere, and seeing why makes the theorem less of a coincidence and more of a count.

On a circle — the sphere in two dimensions — the analogue of a band is an arc between two horizontal lines. Its length is not proportional to their separation: near the top of the circle a small change of height is a long stretch of arc. The circle’s arc is the slant alone, with no circumference to compensate, so near the poles the length per unit of height grows without bound.

In four dimensions the band of the three-dimensional sphere between two heights has a cross-section that is itself an ordinary sphere of radius $\rho$, with area $4\pi\rho^2$, times the same slant factor $1/\rho$. The product is $4\pi\rho$, which shrinks towards the poles. In $n$ dimensions the factor is $\rho^{\,n-2}$ against the slant’s $1/\rho$, leaving $\rho^{\,n-3}$, and only in three dimensions is the exponent nought. In two the poles are favoured, in three every height is equal, and from four on the equator is favoured, more and more strongly — which is the concentration that a sphere that is nearly all equator measures in a thousand dimensions. The hat-box theorem is the single dimension in which that tendency changes sign.

A shadow of a deeper theorem #

The hat-box theorem turns out to be the first case of a result proved only in 1982, and the connection explains why the sphere is special.

Spinning the sphere about its axis is a symmetry, and the height $z$ is the quantity that the symmetry conserves — in mechanics, the height function is the moment map of the rotation. Hans Duistermaat and Gert Heckman proved that whenever a circle acts in this way on a space of the right kind, the areas of the regions between the levels of the conserved quantity are given by a formula that is piecewise polynomial — and for the sphere, piecewise linear, with one piece. That linear formula is Archimedes’ theorem. The same principle gives exact integrals over far more complicated spaces, and physicists use its consequences to compute averages that would otherwise need approximation.

So the cancellation of radius and slant is not an accident of the circle’s geometry. It is what a rotation does to area on a space that it moves rigidly, and Archimedes found its simplest instance two thousand years before the general statement.

What the figures leave to the argument #

Every band’s area is a numerical sum. The figures add up thin rings along the sphere’s profile with a rule whose error is far below the digits printed, and compare the total with the cylinder’s band. That confirms the theorem for the bands drawn and is not a proof; the proof is the similar-triangles argument, which the lampshade figure illustrates and does not replace.

The Earth is drawn as a sphere. It is flattened at the poles by about a third of a per cent, and the zone shares change in the third digit when that is taken into account.

The random points are a sample. Twelve hundred points confirm the shares to within the fluctuation a sample of that size allows, and the claim that the method is exactly right follows from the theorem, not from the sample.

Still open: dividing a sphere into equal areas with the least boundary #

The theorem makes it easy to divide a sphere into pieces of equal area: cut it into horizontal bands of equal height, or into equal wedges between meridians, or both. What it does not say is how to do so economically. Among all ways of dividing a sphere into $N$ regions of equal area, which has the shortest total boundary?

For two regions the answer is a great circle, and for three it is three half-meridians meeting at the poles at $120°$, proved by Joseph Masters in 1996. For four it is the pattern of a regular tetrahedron’s faces pushed out onto the sphere, proved by Max Engelstein in 2010, and for twelve it is the regular dodecahedron’s, proved by Thomas Hales in 2002 alongside his proof that hexagons are the most economical way to divide the plane. For every other number of regions the best division is not known. Soap films on a sphere find good candidates, and the equal-height bands the hat-box theorem makes so easy are among the worst — their boundaries are long circles, and the optimal divisions look like curved honeycombs instead.

Two effects that cancel #

The habit worth keeping is the similar triangles.

A band of the sphere differs from the cylinder’s in two ways, and each difference is large: near a pole the circumference is a small fraction of the cylinder’s and the slant is a large multiple of the rise. The theorem is that the two differences are reciprocal, and the proof is that a radius and a tangent are perpendicular. Two distortions that exactly undo each other are invisible in the product, and finding the pair is the whole of the work — after which the sphere’s area, the fair random point and the true-area map all follow in a line each.