Area and volume

The last chapter found the area of a disc by a trick: the disc was sliced into thin sectors, and they were put together into a shape almost indistinguishable from a rectangle with sides $\pi r$ and $r$. The trick worked, but it hides two questions. Why doesn’t area change when a figure is cut up and rearranged? And what is area anyway: a number we assign to a figure, or something that’s “contained” in it?

In this chapter a workshop is open. There are two tools: scissors that cut only in straight lines, and a table on which pieces can be slid, rotated and flipped, but not stretched or piled on top of one another. At first the workshop can handle any order: a parallelogram, a triangle, a trapezium, an arbitrary polygon, even a triangle that has to be turned into a square. Then two orders come in that it can’t fill, a disc and a cube, and it becomes clear what scissors are missing.

The workshop rules

The school definition, “area is how many little squares fit in a figure”, is fine for a rectangle. For a triangle the squares along the edges have to be cut, and for a disc they don’t fit at all. So mathematicians do something else: instead of explaining what area “really” is, they list its properties. Everything else is derived from them.

The fourth rule is an agreement about the unit: a centimetre, a metre or a square of an exercise book. But the famous “length times width” isn’t among the rules. It has to be proved.

The plan: fill the rectangle with equal little squares, and where they don’t fit exactly, squeeze the area between two approximations.

First let the sides be whole numbers, say $a = 5$ and $b = 3$. The grid lines cut the rectangle into $3$ rows of $5$ unit squares. Each square has area $1$ by rules 2 and 4, and the whole rectangle, by rule 3, has area $5 \cdot 3 = 15$. Cut the unit square into $n \times n$ equal little squares of side $\frac1n$. They’re congruent, there are $n^2$ of them, and together they have area $1$. So each has area $\frac{1}{n^2}$. If the sides are fractions, bring them to a common denominator: $a = \frac pn$, $b = \frac qn$. The rectangle is made up of $p \cdot q$ little squares of side $\frac1n$, and its area is $\frac{pq}{n^2} = \frac pn \cdot \frac qn = ab$. If a side is irrational, say $a = 3\sqrt2$, take its decimal approximations from below and above: $4.2 < 3\sqrt2 < 4.3$. The rectangle with the smaller side fits inside ours, and ours inside the rectangle with the larger side. If one figure lies inside another, its area is no larger: the remainder is also a polygon, and its area is non-negative (rules 1 and 3). So the area $S$ is squeezed between two products, and they come arbitrarily close to $ab$ if you take more digits. A number that, at any precision, lies between two numbers converging to $ab$ is itself $ab$. So $S = ab$ for any positive $a$ and $b$.

The rules also give the workshop’s main property: if a figure is cut up and another is assembled from the pieces, the area doesn’t change. The pieces in the second figure are the same, only slid or rotated, so their areas are the same (rule 2), and each of the two figures has area equal to the sum of the areas of the pieces (rule 3). There are words for such pairs.

Equidecomposable polygons, as we’ve just seen, are equal in area. The workshop’s main question is the converse: if two figures are equal in area, can one always be re-cut into the other? For polygons the answer will be “yes”, and that’s not obvious at all. For a disc and a square it’s “no”, if you cut with scissors. And with volumes, scissors fail even for polyhedra.

Three re-cuts

The first orders are the figures whose areas are taught at school. Each formula can be derived with one or two cuts, and each time the figure turns into another whose area we already know.

The trick is to cut a triangle off the parallelogram and attach it on the other side.

Let $ABCD$ be a parallelogram with base $AB = a$. Drop from $D$ the perpendicular $DH$ to the line $AB$; its length is the height $h$. First suppose the point $H$ landed on the segment $AB$. Cut along $DH$: we get the right triangle $AHD$ and the trapezium $HBCD$. Slide the triangle $AHD$ along the base by the distance $a$, in the direction from $A$ to $B$. The point $A$ lands on $B$, and $D$ on $C$: opposite sides of a parallelogram are equal and parallel, so the path from $D$ to $C$ is the same shift as from $A$ to $B$. The point $H$ moves onto the extension of the base, to a point $H'$, and the triangle fits snugly against the side $BC$. We get the quadrilateral $HH'CD$. The angles at $H$ and $H'$ are right (a shift doesn’t change angles), $HH' = a$, and the sides $DH$ and $CH'$ equal $h$. It’s a rectangle with sides $a$ and $h$. The rectangle’s area is $ah$, and the parallelogram consists of the same pieces. So $S = ah$.

The proof relied on luck: the point $H$ landed on the base. In a very slanted parallelogram the height falls outside, and the triangle we cut off no longer fits into place. A different cutting pattern comes to the rescue. Every horizontal line crosses the parallelogram in a segment of length exactly $a$. Cut the figure with vertical lines every $a$ units and slide the strips to the left by whole multiples of $a$. All the strips land in the same $a \times h$ rectangle, and the horizontal segments at each height join end to end, with no gaps and no overlaps. There are more pieces, but the method works for any slant.

The side chosen as the base. Any side will do. The height to that side: the distance between the lines $AB$ and $CD$, not the length of a slanting side. Example: $a = 5$, $h = 3.5$, so $S = 5 \cdot 3.5 = 17.5$. With base $AD$ we’d have to take the height to $AD$, and the product would come out the same.

A parallelogram has sides $5$ and $3$. What is its area?

$15$ $7.5$ Not enough information $8$

The area is the base times the height, and the height depends on the angle between the sides. Squash the parallelogram flat and the height tends to zero; straighten it into a rectangle and the height grows to $3$. The area can be any number from $0$ to $15$, and exactly $15$ happens only for a rectangle.

The triangle. In chapter 17 the formula $S = \frac12 ah$ came from the fact that two identical triangles make a parallelogram. The workshop can do more: re-cut the triangle itself, without doubling it.

Cut the top off along the midline and rotate it about the midpoint of a side.

Let $M$ and $N$ be the midpoints of sides $AC$ and $BC$. The segment $MN$ is a midline: it’s parallel to $AB$. Triangle $MNC$ is similar to $ABC$ with scale factor $\frac12$, so its height from $C$ is $\frac h2$, and the midline runs exactly at half the height. Cut along $MN$ and rotate the triangle $MNC$ by $180^\circ$ about the point $N$. Since $N$ is the midpoint of $BC$, vertex $C$ lands on $B$. A $180^\circ$ rotation takes any line through the centre of rotation to itself, so $M$ ends up on the extension of $MN$ beyond $N$, at a point $M'$ with $NM' = NM$. A $180^\circ$ rotation takes any line to a line parallel to it. The line $CM$ went to the line $BM'$, so $BM' \parallel AM$. Besides, $MM'$ lies on the midline, which is parallel to $AB$. The quadrilateral $ABM'M$ has both pairs of opposite sides parallel: it’s a parallelogram. The base of the parallelogram $ABM'M$ is the segment $AB = a$, and its height is the distance from $AB$ to the midline, that is, $\frac h2$. Its area is $a \cdot \frac h2$, and so is the triangle’s. The method works for any triangle, obtuse ones included.

The trapezium. The same trick, a $180^\circ$ rotation about the midpoint of a side, turns it into a triangle.

Turn the trapezium into a triangle of the same height whose base is the sum of the trapezium’s parallel sides.

Let $ABCD$ be a trapezium with parallel sides $AB = a$ and $DC = b$, and $M$ the midpoint of the slanting side $BC$. Cut the trapezium along the segment $DM$ and rotate the triangle $DCM$ by $180^\circ$ about $M$. Vertex $C$ goes to $B$, because $M$ is the midpoint of $BC$. Vertex $D$ goes to a point $E$ on the extension of $DM$, with $ME = MD$. Under the rotation the line $DC$ went to the parallel line through the point $B$. Only one line through $B$ is parallel to $DC$, namely $AB$: that’s what the fifth postulate says. So $E$ lies on the extension of $AB$, and $BE = DC = b$. The trapezium has turned into the triangle $AED$ with base $AE = a + b$ and the same height $h$. Its area is $\frac12 (a + b)\,h$. The lower parallel side. The upper parallel side. With $b = 0$ the trapezium degenerates into a triangle, and the formula becomes $\frac12 ah$; with $b = a$ we get a parallelogram and $S = ah$. The height: the distance between the parallel sides. Example: $a = 7$, $b = 3$, $h = 4$, so $S = \frac{7 + 3}{2} \cdot 4 = 20$. The fraction $\frac{a + b}{2}$ is the length of the trapezium’s midline: the trapezium has the same area as a rectangle with sides “midline” and “height”.

A plot of land is a trapezium: its parallel sides are $11$ m and $5$ m, and the distance between them is $6$ m. Find the area of the plot in square metres.

$S = \frac{11 + 5}{2} \cdot 6 = 8 \cdot 6 = 48$ m². Check by re-cutting: a triangle with base $11 + 5 = 16$ and height $6$ has area $\frac12 \cdot 16 \cdot 6 = 48$.

Squared paper

Draw on squared paper a polygon with its vertices at grid points. Its area can be found with no cutting at all, just by counting points: the grid points strictly inside and the grid points on the boundary (vertices count too). The formula was found in 1899 by the Austrian mathematician Georg Pick.

The number of grid points strictly inside the polygon. The number of grid points on the boundary, vertices included. A boundary point is “half inside”, so it gets weight $\frac12$. A correction for the corners. A vertex of the polygon takes less than half from its grid point, and together these shortfalls add up to exactly one. Example: the “house” with vertices $(2, 1)$, $(7, 1)$, $(7, 4)$, $(4, 6)$, $(2, 4)$ has $14$ points inside and $14$ on the boundary, so $S = 14 + 7 - 1 = 20$. Check by cutting: a $5 \times 3$ rectangle and a roof triangle with base $5$ and height $2$ give $15 + 5 = 20$.

Let’s show that the quantity $f = I + \frac B2 - 1$ behaves like area: it adds up under cutting and coincides with area for the simplest figures.

Cut the polygon into two parts with a segment whose ends are grid points; suppose that apart from the ends there are $k$ more grid points on the cut. These $k$ points are on the boundary of the parts and inside the whole: $I = I_1 + I_2 + k$. The ends of the cut and these $k$ points are counted on the boundaries of both parts: $B = B_1 + B_2 - 2k - 2$. Substitute: $I + \frac B2 - 1 = I_1 + I_2 + k + \frac{B_1 + B_2}{2} - k - 1 - 1 = \left(I_1 + \frac{B_1}{2} - 1\right) + \left(I_2 + \frac{B_2}{2} - 1\right)$. The quantity $f$ adds up, like area. An $m \times n$ rectangle with sides along the grid lines. There are $(m - 1)(n - 1)$ points inside and $2m + 2n$ on the boundary. Then $f = mn - m - n + 1 + m + n - 1 = mn$: exactly the area. A right triangle with legs $m$ and $n$ along the grid lines is half of such a rectangle cut by a diagonal. The two halves go into each other under a $180^\circ$ rotation about the centre of the rectangle, and this rotation takes grid points to grid points, so $f$ is the same for both halves. By step 1 their sum is $mn$, so for the triangle $f = \frac{mn}{2}$: the area again. An arbitrary triangle with vertices at grid points. Put it inside the smallest rectangle with sides along the grid lines. The rest of the rectangle falls into at most three right triangles from step 3 and possibly a rectangle. For each piece $f$ equals the area, and by step 1 the values of $f$ add up just as areas do. Subtract the pieces from the rectangle: for the triangle too, $f$ equals the area. Any polygon can be cut by diagonals into triangles with vertices at the same grid points. For each triangle $f$ equals the area (step 4), and when they’re glued together $f$ adds up (step 1). So for the whole polygon $I + \frac B2 - 1 = S$.

Why any polygon can be cut into triangles by diagonals

It’s enough to show that any polygon with more than three vertices has a diagonal lying entirely inside it: it cuts the polygon into two polygons with fewer vertices, and the process can be repeated. Take the leftmost vertex $V$ and its neighbours $U$ and $W$. The angle at $V$ is less than $180^\circ$: there’s nothing to the left of $V$. If triangle $UVW$ contains no other vertices of the polygon, the segment $UW$ is the diagonal we want. If it does, take the one farthest from the line $UW$; the segment from it to $V$ crosses no sides and will do too.

The formula collapses as soon as the conditions are broken. If a vertex isn’t at a grid point, the points no longer count what’s needed: try it in the widget. If the polygon has holes, one is added to the right side for each. And in space there’s no analogue at all: in 1957 the British mathematician John Reeve constructed tetrahedra with vertices at the points of a cubic lattice that have no lattice points inside or on the faces apart from the vertices, yet have arbitrarily large volume.

A triangle has vertices at the grid points $(0, 0)$, $(6, 0)$ and $(2, 4)$. How many grid points lie strictly inside it? Hint: the area is easy to find, and the grid points on the sides can be counted with greatest common divisors.

The area is $\frac12 \cdot 6 \cdot 4 = 12$. The side from $(0, 0)$ to $(6, 0)$ has $7$ grid points. The side from $(6, 0)$ to $(2, 4)$ goes with the shift $(-4, 4)$ and passes through $\gcd(4, 4) + 1 = 5$ grid points, and the side from $(2, 4)$ to $(0, 0)$ with the shift $(-2, -4)$ through $\gcd(2, 4) + 1 = 3$. We’ve counted each vertex twice, so $B = 7 + 5 + 3 - 3 = 12$. By Pick’s theorem $I = S - \frac B2 + 1 = 12 - 6 + 1 = 7$.

Any polygon into any other

So far the workshop has re-cut figures into ones whose area we already knew. The main question was broader: can any two polygons of equal area be re-cut into each other, say an L-shape into a regular pentagon of the same area? In the first third of the nineteenth century several people proved this independently. The theorem is named after two of them: the Hungarian mathematician Farkas Bolyai, father of János Bolyai, one of the creators of non-Euclidean geometry, and the German Paul Gerwien, who published a proof in 1833.

The plan: learn to turn any polygon into a rectangular strip of width $1$. If we can do this for both polygons, the strips will come out identical, since the length of each equals the area, and all that’s left is to lay one cutting pattern over the other.

The first step is to cut the polygon into triangles by diagonals, as in the proof of Pick’s theorem. Re-cut each triangle into a parallelogram by rotating its top about the midpoint of a side, and each parallelogram into a rectangle by strips. Now the polygon is cut into pieces from which several rectangles can be assembled. The main lemma: an $a \times b$ rectangle can be re-cut into a rectangle with a side of $1$. If $b > 1$, cut it across in half and put the halves side by side; repeat until the height is at most $1$. Now build a parallelogram with the same base $a$ and the same height $b$, but with a slanting side of length exactly $1$; it can be tilted like that because $1 \ge b$. The rectangle and this parallelogram are equidecomposable: they share the base and the height, and the strips work for any slant. Look at the same parallelogram differently: let the slanting side of length $1$ be the base. The area hasn’t changed, it’s $ab$, so the height to the new base is $ab$. Re-cut the parallelogram by strips once more, into a $1 \times ab$ rectangle. Stack all the rectangles of width $1$ on top of one another: we get a $1 \times S$ strip, where $S$ is the area of the polygon. Do the same with the second polygon and get the same strip. The strip now has two cutting patterns on it: the lines of the first and the lines of the second. Together they cut it into small pieces, and each small piece lies entirely in some piece of the first pattern and in some piece of the second. Cut the first polygon into these small pieces, assemble the strip from them, and from the strip the second polygon. All the cuts are straight, and there are finitely many pieces. The polygons are equidecomposable.

The proof gives a recipe, but a wasteful one: there are many pieces, and the less alike the figures, the more pieces. Finding the most economical dissections is a sport of its own, and its champion lived in England more than a hundred years ago.

The haberdasher and four pieces

Henry Dudeney (1857–1930) was perhaps the most inventive puzzle setter of his time. In 1902, in a newspaper column, he set the haberdasher’s problem: cut a piece of cloth shaped like an equilateral triangle into four parts that can be assembled into a square. In 1907 the problem went into his book The Canterbury Puzzles. The solution turned out more beautiful than anyone could have hoped: the four pieces can be hinged into a chain that folds now into the triangle, now into the square. Dudeney showed a mahogany model with brass hinges at a meeting of the Royal Society of London in 1905.

The cuts are found with compass and straightedge. A triangle with side $2$ has area $\sqrt3$, so the side of the square we need is $\sqrt{\sqrt3} = \sqrt[4]{3} \approx 1.316$. Such a segment is constructed as a geometric mean. Let $E$ be the midpoint of side $BC$; the median $AE$ has length $\sqrt3$. Extend it beyond $E$ by $EF = 1$ and draw a semicircle on $AF$ as diameter. The perpendicular to $AF$ at the point $E$ meets it at a point $H$, and $EH^2 = AE \cdot EF = \sqrt3$: triangle $AHF$ is right-angled because it stands on a diameter, and the altitude of a right triangle is the geometric mean of the pieces of the hypotenuse (this follows from the similarity of the two triangles the altitude divides it into). Then a circle of radius $EH$ centred at $E$ marks a point $J$ on side $AC$, the segment $JK = 1$ is laid off from it, and perpendiculars are dropped onto $EJ$ from $K$ and from the midpoint $D$ of side $AB$.

For more than a century nobody doubted that four pieces was the limit, but nobody could prove it either. Only in 2024 did Erik Demaine, Tonan Kamata and Ryuhei Uehara prove that a square can’t be made from an equilateral triangle in three pieces.

A disc that won’t be cut

The workshop has coped with polygons. Now the order it was really opened for: big question 2. A disc and a square of the same area: can the disc be cut into pieces and a square assembled from them? With scissors cutting in straight lines, no, and the reason can be seen with the naked eye.

The point is that every piece’s arcs bulge outwards, and for a bulging arc to have somewhere to go, its neighbour needs a hollow one.

The boundary of any piece consists of segments of cuts and arcs of the original circle. The piece lies inside the disc, that is, on the hollow side of its arc: for the piece, the arc always bulges outwards. Sliding, rotating and flipping don’t change this. Suppose the pieces made a square. Take a piece $Q$ with an arc and a point $p$ on the arc that isn’t anyone’s vertex and doesn’t lie on the sides of the square. Such a point exists: there are finitely many vertices, and the circle meets each of the four sides in at most two points. So around $p$ the square is filled with pieces on all sides. On the bulging side of the arc, next to $p$, there are points of the square not belonging to $Q$. Another piece $Q'$ occupies them, and its boundary near $p$ must run along the same arc: otherwise there’d be a gap between the pieces or they’d overlap. But for $Q'$ this arc curves inwards: $Q'$ lies on its bulging side. By step 1 no piece has a hollow arc. A contradiction: the square can’t be assembled.

It’s not only a matter of the cuts being straight: in 1963 the American mathematicians Lester Dubins, Morris Hirsch and Jack Karush proved that scissors cutting along any curves won’t help either. A square of the right area can’t be constructed with compass and straightedge either: that follows from the transcendence of $\pi$, proved by Lindemann in 1882 (chapter 6). And yet in 1990 the Hungarian mathematician Miklós Laczkovich proved that a disc can be cut into finitely many pieces and reassembled into a square, and the pieces don’t even need rotating, only sliding. It takes around $10^{50}$ pieces, and they aren’t shapes but bizarre clouds of points that can be neither drawn nor cut out. How this is possible, and why it doesn’t violate area, is explained in chapter 54: that’s where big question 2 will be closed.

The cube and the tetrahedron

Let’s move into space. Volume, like area, is given by rules.

The volume of an $a \times b \times c$ box is $abc$: the proof repeats the argument about the rectangle word for word, only with little cubes instead of little squares and three edges instead of two sides. It would be natural to expect a three-dimensional analogue of the Bolyai–Gerwien theorem too: any two polyhedra of the same volume can be cut into each other. Then the volume of any pyramid could be found by re-cutting it into a box.

But school textbooks derive the volume of a pyramid differently: through an infinite sequence of ever finer cuts, as Euclid did in Book XII of the Elements. Gauss, in letters to his student Christian Gerling, regretted that the volume of a pyramid couldn’t be found without such a limiting process. In 1900, at the International Congress of Mathematicians in Paris, David Hilbert named 23 problems for the new century, and the third was exactly this: find two tetrahedra with equal bases and heights that can’t be re-cut into each other. In the same year it was solved by Hilbert’s student Max Dehn. It was the first problem on the list to be solved.

The conclusion for the workshop is discouraging: for volumes scissors alone aren’t enough, and a limiting process is needed, slicing into infinitely thin layers. Its handiest version fits into a stack of coins.

A stack of coins

Stack up some coins and push the stack from the side. It will lean and bend, but it won’t gain or lose any metal: the coins are the same. If the coins are infinitely thin, shifting the layers turns an upright cylinder into a slanted one or into any curved “snake”, and the volume stays the same. In 1635 the Italian monk and mathematician Bonaventura Cavalieri, a pupil of Galileo, made a method out of this observation in his book Geometry, Developed by a New Method Through the Indivisibles of the Continua.

Cavalieri had a predecessor at the other end of the world. In China, around the turn of the fifth and sixth centuries, Zu Geng, son of the astronomer Zu Chongzhi (the one who found the fraction $\frac{355}{113}$ for $\pi$), stated the same rule: if the areas of the cross-sections at every height are equal, the volumes can’t differ. With it he found the volume of a ball, a problem at which the commentator on the Nine Chapters, Liu Hui, had stopped more than two centuries earlier.

A pyramid is a third of a prism

That a pyramid takes up a third of the prism with the same base and height was known long before any proofs. The Moscow Mathematical Papyrus, written around 1850 BC and kept in the Pushkin Museum of Fine Arts in Moscow, finds in problem 14 the volume of a truncated pyramid with square bases $4$ and $2$ and height $6$ by a rule equivalent to the formula $V = \frac h3\left(a^2 + ab + b^2\right)$: it comes out as $2 \cdot (16 + 8 + 4) = 56$. According to Archimedes, the statement about the third was first made by Democritus and proved by Eudoxus.

It’s clearest of all on a cube: a cube can be cut into three identical pyramids with a common apex.

Each of the three pyramids has an $a \times a$ square as its base and the edge $a$ as its height, and its volume is a third of the cube: $\frac13 a^3 = \frac13 \cdot a^2 \cdot a$. For the general case we’ll need Cavalieri’s principle and one lemma.

Compare the cross-sections at the same height and apply Cavalieri’s principle.

Stand the pyramids on the same plane: both have height $h$ and base area $S$. Draw a plane parallel to the bases at height $z$. The cross-section of a pyramid is a scaled-down copy of the base. The plane of the cross-section is parallel to the base, so it divides the height and every segment from the apex to a point of the base in the same ratio, just as parallel lines do in the intercept theorem. From the apex to the cross-section is a fraction $k = \frac{h - z}{h}$ of each such segment. So the cross-section is similar to the base with scale factor $k$, and the area of a similar figure is multiplied by $k^2$ (it can be cut into triangles, and in each the base and height are multiplied by $k$). So for both pyramids the area of the cross-section at height $z$ is $S\left(\frac{h - z}{h}\right)^2$: the same number, whatever the shape of the base. For a cone it’s the same: its cross-section is a disc scaled down in the same ratio. The cross-sections are equal at every height, so by Cavalieri’s principle the volumes are equal.

Euclid’s idea (proposition XII.7): a triangular prism is cut into three pyramids, and by the lemma they have pairwise equal volumes.

Let $ABCA'B'C'$ be a triangular prism, with vertices $A'$, $B'$, $C'$ above $A$, $B$, $C$; the base area is $S$ and the height $h$. The first cut is by the plane $A'BC$: it cuts off the pyramid $A'ABC$ with apex $A'$ and base $ABC$. What’s left is a four-sided pyramid with apex $A'$ and the parallelogram $BCC'B'$ as its base. The second cut is by the plane $A'BC'$ through the diagonal $BC'$: it divides it into the pyramids $A'BB'C'$ and $A'BCC'$. Compare the first and the second. Regard the second, $A'BB'C'$, as a pyramid with base $A'B'C'$ and apex $B$. Its base equals the base of the prism, and the apex $B$ lies in the plane of the lower base, so its height is $h$. The first has base $ABC$ and height $h$ too. By the lemma the volumes are equal. Compare the second and the third, taking $A'$ as the apex of both. Their bases, the triangles $BB'C'$ and $BCC'$ into which the diagonal $BC'$ divides the parallelogram $BCC'B'$, are congruent. The height of both is the distance from $A'$ to the plane of this parallelogram. By the lemma the volumes are equal again. Three pyramids of equal volume make up the prism of volume $Sh$, so each is $\frac13 Sh$. Any other pyramid or cone with base area $S$ and height $h$ has the same volume, by the lemma. The area of the base. For a cone, $S = \pi r^2$. The height: the distance from the apex to the plane of the base, not the length of a lateral edge. Example: when it was built, the Great Pyramid of Khufu had a square base with a side of about $230.3$ m and a height of about $146.6$ m. Its volume is $\frac13 \cdot 230.3^2 \cdot 146.6 \approx 2.59$ million m³, the same as a cube with an edge of about $137$ m.

An ice-cream cone is a cone with base radius $3$ cm and height $10$ cm. How many cubic centimetres of ice cream does it hold, level with the rim? You can write the answer using $\pi$ or as a decimal.

$V = \frac13 \pi r^2 h = \frac13 \cdot \pi \cdot 9 \cdot 10 = 30\pi \approx 94.2$ cm³. A cylindrical tub of the same size would hold three times as much, $90\pi \approx 282.7$ cm³.

A ball in a cylinder

In 75 BC the quaestor of Sicily, Marcus Tullius Cicero, tracked down Archimedes’ grave near Syracuse, overgrown with brambles. He recognised it by a column on which a sphere and a cylinder around it were carved. According to Plutarch, Archimedes himself asked for this figure to be put on his grave: the volume of a ball is two thirds of the volume of the cylinder around it, and he valued this result above all others. Let’s prove it with the tools we already have: cross-sections.

We need a solid that has the same cross-section as a hemisphere at every height, but whose volume we already know how to compute.

Stand a hemisphere of radius $R$ on a plane. Next to it put a cylinder of radius $R$ and height $R$, and remove from it a cone whose apex is the centre of the cylinder’s lower base and whose base is the cylinder’s upper base. Cut the hemisphere with a plane at height $z$. The cross-section is a disc. Its radius $\rho$ is a leg of a right triangle with hypotenuse $R$ (from the centre of the ball to the edge of the cross-section) and other leg $z$. By Pythagoras’ theorem $\rho^2 = R^2 - z^2$, and the area of the cross-section is $\pi\left(R^2 - z^2\right)$. The same plane cuts the cylinder-minus-cone in a ring. The outer radius of the ring is the cylinder’s radius $R$. The inner one is the radius of the cone at height $z$: the cone’s height and base radius are both $R$, so the radius of its cross-section grows with the height and at height $z$ equals $z$. The area of the ring is $\pi R^2 - \pi z^2$. The areas match: $\pi\left(R^2 - z^2\right) = \pi R^2 - \pi z^2$ at every height from $0$ to $R$. By Cavalieri’s principle the volume of the hemisphere equals the volume of the cylinder minus the cone: $\pi R^2 \cdot R - \frac13 \pi R^2 \cdot R = \frac23 \pi R^3$. A ball is two hemispheres, so its volume is $\frac43 \pi R^3$. The cylinder around it (radius $R$, height $2R$) has volume $2\pi R^3$, and the ball takes up exactly two thirds: what was on Archimedes’ grave. Two thirds of two: the ball takes up $\frac23$ of the cylinder around it, and the cylinder’s volume is $2\pi R^3$. The cube of the radius times $\pi$. Twice the radius means $2^3 = 8$ times the volume. Example: a ball of radius $3$ has volume $\frac43 \pi \cdot 27 = 36\pi \approx 113.1$. The Earth, with a mean radius of about $6371$ km, is about $1.08 \cdot 10^{12}$ km³.

One watermelon is $30$ cm across, another $20$ cm. How many times more flesh does the first have, treating watermelons as balls?

$1.5$ times $2.25$ times About $3.4$ times $10$ times

The ratio of the radii is $\frac32$, and volume is proportional to the cube of the radius: $\left(\frac32\right)^3 = \frac{27}{8} \approx 3.4$. The first watermelon is only one and a half times “wider”, but it has more flesh than three of the second.

What’s left is the area of the sphere. It too can be got from the volume, by the same trick that gets the area of a disc from the circumference.

Four great circles. A kitchen experiment: peel an orange and lay the peel in pieces into circles the size of the orange; it fills about four. The area of a great circle, a cross-section through the centre. Example: for the Earth $R \approx 6371$ km, and $S \approx 4\pi \cdot 6371^2 \approx 5.1 \cdot 10^8$ km², about $510$ million square kilometres. The curved surface of the cylinder around the sphere, $2\pi R \cdot 2R = 4\pi R^2$, is exactly the same. Archimedes knew this too.

The radius of a ball is doubled. By what factor does its volume grow?

$\frac43 \pi (2R)^3 = 8 \cdot \frac43 \pi R^3$: by a factor of $8$. The area of the sphere grows only by $4$, and each square centimetre of surface now has twice as much volume behind it. That’s why large animals cool down more slowly than small ones.

Take-home orders

The workshop is closing, but there are orders left. The trainer has three levels: areas of polygons by the formulas of this chapter; squared paper, where Pick’s theorem comes to the rescue; volumes of prisms, pyramids, cylinders, cones and balls.

Where next

Everything we measured in this chapter could be touched: cut up, rearranged, put on the scales. But some sizes are out of reach of both scissors and tape measure. The height of Everest was calculated by the surveyors of British India in 1856, almost a hundred years before Edmund Hillary and Tenzing Norgay first reached the summit in 1953. The surveyors themselves never went near the mountain: their instruments stood on the plains of India, more than a hundred kilometres away, and measured only angles. How to find distances and heights from angles, and the size of the whole Earth along with them, is the business of trigonometry.