Archimedes' Spherical Bands, math that surprises you lesson

Cut a slice one centimeter thick from the middle of an orange and another slice one centimeter thick from near its top. The middle slice goes around a much wider circle, so its ring of peel looks far bigger. For a perfectly round orange it is not, because any two bands of the same height on a sphere have exactly the same area, wherever they sit. This page explains how the shrinking circles near the top are made up for by the way the surface leans over there, finds the area $2\pi R h$ of a band of height $h$ on a sphere of radius $R$ , and gets the area of the whole sphere from it.

Key terms

Archimedes’ hat-box theorem

Definition Place a sphere of radius $R$ with its center at height $0$ on a vertical axis, so that it reaches from the South Pole at height $-R$ to the North Pole at height $R$ . Its circumscribed cylinder is the cylinder of radius $R$ around the same axis, from height $-R$ to height $R$ , which touches the sphere all around the equator. Cut both by two horizontal planes at heights $z_1$ and $z_2$ , where $-R \le z_1 < z_2 \le R$ , and let $h = z_2 - z_1$ be the vertical distance between the planes. The hat-box theorem says that the band of the sphere between the two planes and the band of the cylinder’s curved side between the same planes have the same area,

A = 2\pi R h .

The area depends only on the radius and the height of the band, not on where the band sits, so bands of equal height on one sphere have equal areas. The height is measured straight along the axis, not along the curved surface, and the comparison is with the curved side of the cylinder only, not with its two flat ends. A stronger form of the theorem compares any regions. Move each point of the sphere other than the two poles horizontally, straight away from the axis and keeping its height, until it reaches the cylinder. This horizontal projection gives every region of the sphere an image on the cylinder with the same area, although it changes lengths and shapes. Archimedes found the area of the sphere and of its caps in his book On the Sphere and Cylinder, written in the third century BC.

In plain words A band near a pole goes around a smaller circle than a band near the equator. But near a pole the surface leans over toward horizontal, so climbing the same height takes a longer walk along the surface, and the band is wider. The two effects cancel exactly, and every band has the area of the matching band on the cylinder, which unrolls into a plain rectangle.

Example Take $R = 2$ and two bands of height $1$ . The first runs from height $-\tfrac12$ to height $\tfrac12$ around the equator, and its two edge circles have radius $\sqrt{15}/2 \approx 1.94$ . The second runs from height $1$ to the North Pole at height $2$ . It is a cap whose edge circle has radius $\sqrt{3} \approx 1.73$ and which closes up to a single point at the top. Both have area

2\pi \cdot 2 \cdot 1 = 4\pi \approx 12.57

square units. The matching bands of the cylinder unroll into rectangles of width $2\pi \cdot 2 = 4\pi$ and height $1$ , with the same area $4\pi$ .

Spherical zone

Definition A spherical zone is the part of the surface of a sphere that lies between two parallel planes which both meet the sphere. Its height $h$ is the distance between the two planes. When one of the planes only touches the sphere, at a single point, the zone is a spherical cap, and when both planes only touch it, the zone is the whole sphere. A zone is a curved band of surface with no thickness. On a sphere of radius $R$ its area is $2\pi R h$ , wherever the planes are. To see why, turn the sphere so that the planes are horizontal, with its center at height $0$ . At a height $z$ with $-R < z < R$ the sphere’s horizontal circle has radius $r = \sqrt{R^2 - z^2}$ . Take a thin strip of the sphere between heights $z$ and $z + \Delta z$ and look at a vertical cut through the axis. There a short step along the sphere is perpendicular to the radius from the center, so the small right triangle made by the step and its vertical rise $\Delta z$ has the same shape as the triangle made by that radius, of length $R$ , and its horizontal part $r$ , with the step matching the radius and the rise matching $r$ . The step is therefore about $\tfrac{R}{r}\,\Delta z$ long, and the strip has area about

2\pi r \cdot \frac{R}{r}\,\Delta z = 2\pi R\,\Delta z .

A single thin strip is only approximately a band of this width, but the approximation gets better as the strips get thinner, and adding up the strips in the limit is the integral

A = \int_{z_1}^{z_2} 2\pi R \, dz = 2\pi R h .

At a pole $r = 0$ and the factor $R/r$ makes no sense, but the strip area $2\pi R\,\Delta z$ stays the same, and the area of a cap follows as a limit.

In plain words A zone is a ring of peel cut off by two parallel slices. Its area is the radius of the ball times its height times $2\pi$ , so a zone twice as tall has twice the area, and its position on the ball makes no difference.

Example On a sphere of radius $2$ , compare a thin strip of vertical thickness $0.01$ just above the equator with one just above height $1.5$ . At the equator $r = 2$ , the surface is vertical, and the strip has area about $2\pi \cdot 2 \cdot 0.01 = 0.04\pi$ . At height $1.5$ the circle has radius $r = \sqrt{4 - 2.25} = \sqrt{1.75} \approx 1.32$ and circumference about $8.31$ , much less than $4\pi \approx 12.57$ . The slant factor there is $2/\sqrt{1.75} \approx 1.51$ , so the strip is about $0.0151$ wide along the surface, and the product is

\begin{aligned} &2\pi\sqrt{1.75} \cdot \frac{2}{\sqrt{1.75}} \cdot 0.01 \\[0.4em] &\quad = 0.04\pi , \end{aligned}

because $\sqrt{1.75}$ cancels. The zone formula confirms that both thin zones have area exactly $2\pi \cdot 2 \cdot 0.01 = 0.04\pi \approx 0.126$ .

Circumscribed cylinder

Definition The circumscribed cylinder of a sphere of radius $R$ is the cylinder of radius $R$ and height $2R$ whose axis passes through the center of the sphere. Its curved side touches the sphere along the equator, halfway up, and its two flat ends touch the sphere at the poles. The curved side unrolls into a rectangle of width $2\pi R$ and height $2R$ , so its area is $4\pi R^2$ , and the two ends add $2\pi R^2$ , for a total of $6\pi R^2$ . Archimedes also showed that the ball fills two thirds of the cylinder’s volume, $\tfrac43 \pi R^3$ out of $\pi R^2 \cdot 2R = 2\pi R^3$ . The hat-box theorem compares the sphere with this cylinder by horizontal projection. A point $(x, y, z)$ of the sphere other than a pole lies at distance $r = \sqrt{x^2 + y^2} > 0$ from the axis, and the projection moves it straight away from the axis, at the same height, onto the cylinder,

(x, y, z) \mapsto \left( \frac{R x}{r}, \frac{R y}{r}, z \right).

Near that point, lengths around the axis are stretched by the factor $R/r$ , since the circle of radius $r$ through the point is carried onto a circle of radius $R$ . Lengths along the meridian, the north and south direction on the sphere, are shrunk by the factor $r/R$ , since a short step along the meridian is carried onto a vertical segment as long as its rise. These two directions are at right angles on both surfaces, so a small patch has its area multiplied by $\tfrac{R}{r} \cdot \tfrac{r}{R} = 1$ , and every region keeps its area. This is not projection from the center of the sphere, which would send a point at height $z$ to the height $Rz/r$ and would not keep areas.

In plain words Picture the sphere inside a can that fits it exactly, and push every point of the sphere straight out sideways onto the wall of the can. Near the poles a patch gets spread out around the can and squashed from top to bottom by the same factor, so it keeps its area but not its shape.

Example Take $R = 1$ and the region of the sphere from the equator up to the North Pole between longitudes $0^\circ$ and $90^\circ$ , one eighth of the sphere, with area $4\pi/8 = \pi/2$ . Its image is the part of the cylinder’s curved side over a quarter of its circle, from height $0$ up to height $1$ , which unrolls into a rectangle of width $\pi/2$ and height $1$ , again of area $\pi/2$ . Lengths do not survive. The edge along the equator keeps its length $\pi/2$ , but each edge along a meridian, of length $\pi/2$ on the sphere, becomes a vertical segment of length $1$ . Shapes do not survive either, because the region narrows to a point at the pole while its image keeps its full width all the way up to the top edge.

Surface area of a sphere

Definition The surface area of a sphere of radius $R$ is

A = 4\pi R^2 .

It follows from the hat-box theorem by taking the whole sphere as one band, from height $-R$ to height $R$ , so that $h = 2R$ and $A = 2\pi R \cdot 2R$ . This is the area of the curved side of the circumscribed cylinder, four times the area $\pi R^2$ of a flat disk of the same radius, and two thirds of the cylinder’s total surface area $6\pi R^2$ .

In plain words The surface of a ball has the same area as four flat disks of the ball’s radius. Doubling the radius multiplies the area by four.

Example Model the Earth as a sphere of radius $6371$ kilometers. Its surface area is then $4\pi \cdot 6371^2 \approx 5.10 \times 10^8$ square kilometers, about 510 million. The circle of latitude $30^\circ$ north lies $R \sin 30^\circ = R/2$ above the plane of the equator, and the circle of latitude $30^\circ$ south lies the same distance below it. The zone between them has height $R$ and area $2\pi R \cdot R = 2\pi R^2$ , exactly half of $4\pi R^2$ . So half of the surface of a spherical Earth lies within 30 degrees of the equator, even though that band covers only 60 of the 180 degrees of latitude.

Check yourself

A sphere has radius $5$ . Find the area of the band between the horizontal planes $1$ and $3$ units above its center, and the area of the cap above the plane at height $3$ . What fraction of the whole sphere is each?

Show the answer

Both have height $2$ , so each has area

2\pi \cdot 5 \cdot 2 = 20\pi \approx 62.8

square units. The whole sphere has area $4\pi \cdot 5^2 = 100\pi$ , so each covers one fifth of it. The band has edge circles of radius $\sqrt{24} \approx 4.90$ and $4$ , while the cap has an edge circle of radius $4$ and closes up to a point, yet the areas agree. In general a zone of height $h$ covers the fraction $h/(2R)$ of the sphere, here $2/10$ .

Change the surface. Stand a cone with base radius $3$ and height $4$ on its base, so that its slanted side has length $5$ . Cut it with a horizontal plane halfway up, at height $2$ , and compare the areas of the slanted surface below and above the cut. Use the fact that the slanted surface of a cone with base radius $r$ and slant length $\ell$ has area $\pi r \ell$ , because it unrolls into a sector of a disk of radius $\ell$ with arc length $2\pi r$ .

Show the answer

The part above the cut is a smaller cone of the same shape, with base radius $1.5$ and slant length $2.5$ , so its slanted surface has area $\pi \cdot 1.5 \cdot 2.5 = 3.75\pi$ . The whole slanted surface has area $\pi \cdot 3 \cdot 5 = 15\pi$ , so the band below the cut has

15\pi - 3.75\pi = 11.25\pi ,

three times as much as the part above. On a cone, equal heights do not give equal areas. Its slant is the same at every height, with each unit of height costing $5/4$ units along the surface, while its circles shrink steadily toward the tip, so nothing makes up for the smaller circumference. On a sphere the slant factor $R/r$ grows in exactly the proportion that cancels the shrinking circumference $2\pi r$ , and that is what the hat-box theorem needs.

Extend the pattern. Two meridians, each a half circle of radius $R$ running along the sphere from the North Pole to the South Pole, meet at an angle $\alpha$ at each pole, measured in radians, and cut out a region shaped like a segment of an orange, called a lune. Use the horizontal projection onto the circumscribed cylinder to find its area, and check the answer for $\alpha = 2\pi$ and $\alpha = \pi/2$ .

Show the answer

Describe each point by its height and its longitude, the angle around the axis. The lune is the set of points whose longitude lies in an interval of length $\alpha$ . The horizontal projection keeps both height and longitude, so it sends the lune, without its two poles, onto the part of the cylinder’s curved side over an arc of angle $\alpha$ of its circle, at heights between $-R$ and $R$ . Unrolled, that part is a rectangle of width $R\alpha$ and height $2R$ . Leaving out two points does not change an area, so by the stronger form of the hat-box theorem the lune has area

A = R\alpha \cdot 2R = 2\alpha R^2 .

For $\alpha = 2\pi$ this gives $4\pi R^2$ , the whole sphere. For $\alpha = \pi/2$ it gives $\pi R^2$ , a quarter of the sphere, as it should, since four such lunes fill it. The half of that lune north of the equator has area $\pi R^2 / 2$ , one eighth of the sphere, and it is the triangle with three right angles.