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Bonaventura Cavalieri

Bonaventura Francesco Cavalieri · Bonaventura Cavalerius

A course in two strands: the life of a Jesuate friar, crippled by gout, who held the chair of mathematics at Bologna for eighteen years, and the method of indivisibles, the bridge between Archimedes and the integral calculus.

Born 1598 (probably), Milan Died 30 November 1647, Bologna Geometria written by 1627, printed 1635 Level advanced undergraduate
“All the lines” of two figures between the same parallels. Each chord on the right equals the chord at the same height in the ellipse, so the two areas are equal.
Lesson 00

How to use this course

Most students meet Bonaventura Cavalieri once, in a school geometry lesson: two solids with equal cross-sections at every height have equal volume. The principle carries his name, but his work was much wider. He built a whole geometry on comparing “all the lines” and “all the planes” of figures, found the area under every curve \(y=x^n\) with \(n\) a positive integer, introduced logarithms into Italy, and wrote on conic mirrors, telescopes and projectiles. He did this as a member of a small religious order, in constant pain from gout, while serving as the first-ranked mathematician of the University of Bologna.

Prerequisites for Part II
  • Single-variable calculus: definite integrals, substitution, volumes of revolution.
  • Euclidean geometry of similar figures, and the conic sections in their focal form.
  • For Lesson 07, the statement of Fubini’s theorem helps; it is recalled there.
  • For Lesson 09, a little spherical geometry.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern language, sketches of Cavalieri’s arguments and of their modern counterparts, and exercises with folded solutions. Where Cavalieri’s own reasoning is reconstructed rather than quoted, the text says so. Historians still debate exactly what his “collections of lines” were; Lesson 07 discusses the problem.

Reference

Timeline

1598Born in Milan, then a Spanish Habsburg possession, probably in this year; christened Francesco.
1615Receives minor orders with the Jesuati in Milan and takes the name Bonaventura.
1616Moves to the Jesuati house of San Girolamo in Pisa. Benedetto Castelli teaches him geometry and introduces him to Galileo.
1619Applies without success for the chair at Bologna left vacant by Giovanni Antonio Magini; judged too young. Correspondence with Galileo begins about this time.
1620–23Back in Milan. Ordained deacon to Cardinal Federico Borromeo in 1621; teaches theology at San Girolamo. First ideas on indivisibles.
1623–26Prior of San Pietro at Lodi.
1626–29Prior of the Jesuati house in Parma. Attacks of gout begin. Applications for chairs at Rome and Pisa fail.
162716 December: tells Galileo and Borromeo that the Geometria is complete.
1629Appointed to the chair of mathematics at Bologna, on a three-year trial, with Galileo’s backing. Also prior of Santa Maria della Mascarella.
1632Directorium generale uranometricum (logarithmic tables) and Lo specchio ustorio (conics and mirrors).
1635Geometria indivisibilibus continuorum nova quadam ratione promota published in Bologna.
1636Summer near Florence, in conversation with Galileo at Arcetri. Declines a chair at Pisa and a post at the Biblioteca Ambrosiana.
1638–39Compendio delle regole dei triangoli; Centuria di varii problemi, with the quadrature of \(y=x^n\) for all positive integers \(n\); Nuova pratica astrologica.
1641Paul Guldin’s attack on indivisibles appears in the final volume of his Centrobaryca.
1643Trigonometria plana et sphaerica, linearis et logarithmica.
1646Trattato della ruota planetaria perpetua, under the pseudonym Silvio Filomantio. Illness stops his teaching.
1647Exercitationes geometricae sex. Dies in Bologna on 30 November.
1653Second edition of the Geometria.
1668Pope Clement IX suppresses the order of the Jesuati.
Part I · Formation
Lesson 01

Milan and the Jesuati, 1598–1616

Cavalieri was born in Milan, probably in 1598. The year is not certain, and neither is much else about his family. He was christened Francesco. Milan was then the capital of a duchy ruled by the Spanish Habsburgs, and its religious life was dominated by the Borromeo family: Carlo Borromeo, the reforming archbishop who had died in 1584 and been canonized in 1610, and his cousin Cardinal Federico Borromeo, archbishop from 1595 and founder of the Biblioteca Ambrosiana.

As a boy he joined the Jesuati, a small religious congregation founded in Siena in the fourteenth century by Giovanni Colombini. The name came from the habit of opening and closing sermons with the cry of “Jesus”. Its members cared for the sick, lived austerely and followed the rule of St Augustine. The order has nothing to do with the far larger and better known Jesuits, though the similarity of names confused contemporaries and still confuses readers. In 1615 Cavalieri received minor orders in Milan and took the religious name Bonaventura, which he kept for the rest of his life.

Membership of the Jesuati shaped his career in two ways. It gave him a home, a vocation and a sequence of administrative posts in the order’s houses. It also, he believed, counted against him when universities chose professors. Later sources say his first taste of mathematics came from reading Euclid, and that it was immediate.

Lesson 02

Pisa, Castelli and Galileo, 1616–1620

In 1616 his superiors sent him to the Jesuati house of San Girolamo in Pisa. There he met the Benedictine monk Benedetto Castelli, who had been Galileo’s student at Padua and now taught mathematics at the University of Pisa. Castelli taught him Euclid, Archimedes and Apollonius, and opened the circle of Galileo to him. Cavalieri spent about a year in Florence around 1617. The introduction to Galileo was smoothed by Cardinal Borromeo, who corresponded with Castelli.

Castelli was the great teacher of the Galilean school. His other pupils included Evangelista Torricelli, who later became Cavalieri’s ally and correspondent. From this time Cavalieri regarded himself as a disciple of Galileo. Over the next two decades, until 1641, he wrote Galileo more than a hundred letters; 112 survive in the national edition of Galileo’s works, against two of Galileo’s replies. The letters are a record of his mathematical development, his search for a post and his health.

In 1619 he applied for the chair of mathematics at Bologna, vacant since the death of the astronomer Giovanni Antonio Magini in 1617. The Bologna authorities judged him too young. The episode set the pattern for the next decade: a talented geometer without a university position.

Lesson 03

Prior and geometer, 1620–1629

In 1620 he was called back to Milan. In 1621 he was ordained deacon to Cardinal Federico Borromeo, and he taught theology at the house of San Girolamo in Milan until 1623. During these years he formed his first ideas about indivisibles. He then served as prior of San Pietro at Lodi (1623–1626) and of the Jesuati house in Parma (1626–1629). Applications for chairs at Rome and at Pisa, where Castelli’s departure for Rome in 1626 left a vacancy, came to nothing. Cavalieri blamed his membership of the Jesuati.

In the autumn of 1626 he suffered a severe attack of gout, the disease that troubled him throughout his life. It kept him in Milan for several months. From then on illness, especially of the legs, was a constant theme of his letters.

The Geometria and the Bologna chair

On 16 December 1627 he announced to Galileo and to Borromeo that he had finished a work on the geometry of indivisibles. The manuscript contained a version of what became Books I–VI of the Geometria. He hoped it would win him a chair. In 1629 it did: Galileo wrote in his support to Cesare Marsili, a Bolognese nobleman and member of the Accademia dei Lincei, and Cavalieri was appointed to the chair of mathematics at Bologna. The appointment was for three years on trial and was renewed thereafter; he held it until his death. He was made prior of the Jesuati house attached to the church of Santa Maria della Mascarella in Bologna. Part III returns to his Bologna years.

Publication of the Geometria was delayed until 1635. Part of the reason was deference to Galileo, who had long promised a work on the continuum and whose approval Cavalieri wanted. Galileo’s own discussion of indivisibles appeared only in the Two New Sciences of 1638.

Part II · The mathematics
Lesson 04

All the lines: the method of indivisibles

For the general reader. A deck of cards has the same volume whether it is stacked neatly or pushed sideways into a slanted pile. Cavalieri turned this observation into a method. Think of a flat figure as made of all its parallel lines, and a solid as made of all its parallel planes. If two figures have equal lines at every height, they have equal areas. Comparing “all the lines” of different figures, he found areas and volumes that had taken Archimedes ingenious and separate proofs.

Goals
  • Understand Cavalieri’s “all the lines” of a figure and the role of the regula.
  • State Cavalieri’s principle in its classical and modern forms.
  • Use it to compute areas and volumes by comparison with known figures.

Omnes lineae

In Book II of the Geometria, Cavalieri takes a plane figure \(F\) and a fixed direction, the regula. Two lines parallel to the regula touch \(F\) on either side; between them, a moving line parallel to the regula sweeps across \(F\) and cuts it in a chord at each position. The collection of all these chords he calls the omnes lineae of \(F\), “all the lines”. For a solid he uses a plane moving parallel to itself and takes “all the planes”. He was careful never to say that a figure is the sum of its lines. Instead he treated the collection as a new kind of magnitude and argued that two such collections have a ratio, like any two magnitudes of the same kind.

Theorem (Geometria II.3)

For two plane figures \(F_1,F_2\) and the same regula, \(F_1:F_2\) equals the ratio of all the lines of \(F_1\) to all the lines of \(F_2\).

From this he derived the principle that bears his name. In the form of Book VII, where a second, “distributive” proof avoids collections of lines altogether, it reads as follows.

Cavalieri’s principle

Let two plane figures lie between the same two parallel lines. If every line parallel to these cuts the two figures in segments of equal length, the figures have equal area. Let two solids lie between the same two parallel planes. If every plane parallel to these cuts them in sections of equal area, the solids have equal volume. More generally, if corresponding sections are always in the ratio \(p:q\), so are the figures.

In modern notation, if \(\ell_1(t)\) and \(\ell_2(t)\) are the lengths of the chords at height \(t\), the principle says \(\int\ell_1\,dt=\int\ell_2\,dt\) whenever \(\ell_1=\ell_2\). It is a special case of the formula \(\mathrm{Area}(F)=\int\ell(t)\,dt\), which Cavalieri never wrote but which his method presupposes.

The distributive proof

Book VII proves the principle by superposition, in the spirit of Euclid. Place one figure on the other so that their chords lie along the same lines. The parts that coincide are equal. The parts that do not coincide still have equal chords at each height, so the argument can be repeated on the remainders, and Cavalieri argued that the process exhausts the figures. The weak point, which critics noticed, is that the process need not end after finitely many steps; Cavalieri replied that the residues vanish in the limit.

A first example: the ellipse

The ellipse \(x^2/a^2+y^2/b^2\le1\) has horizontal chord \(2a\sqrt{1-y^2/b^2}\) at height \(y\). The circle of radius \(b\) has chord \(2b\sqrt{1-y^2/b^2}\) at the same height. The ratio is always \(a:b\), so the ellipse has area \(\tfrac ab\cdot\pi b^2=\pi ab\).

Exercise 4.1A sphere of radius \(R\) is drilled through its centre by a cylindrical hole, leaving a ring of height \(h\). Use Cavalieri’s principle to show that the volume of the ring depends only on \(h\), and compute it.

Show solution
Let the hole have radius \(r\), with \(r^2=R^2-h^2/4\). At height \(z\) (with \(|z|\le h/2\)) the section of the ring is an annulus between radii \(r\) and \(\sqrt{R^2-z^2}\), of area \(\pi(R^2-z^2-r^2)=\pi(h^2/4-z^2)\). That is exactly the section at height \(z\) of a ball of radius \(h/2\). By the principle the ring has the volume of that ball, \(\tfrac43\pi(h/2)^3=\tfrac{\pi h^3}{6}\), whatever \(R\) is.

Exercise 4.2An oblique cylinder has a circular base of radius \(r\), height \(h\), and axis inclined at angle \(\alpha\) to the vertical. Find its volume, and explain why it does not depend on \(\alpha\).

Show solution
Every horizontal plane between the two bases cuts the oblique cylinder in a disk of radius \(r\), translated sideways by \(z\tan\alpha\). A right cylinder of the same base and height has the same sections. By the principle the volumes are equal: \(\pi r^2h\). This is the stack-of-coins picture in the hero image.
Lesson 05

All the squares: cones, pyramids, parabolas

For the general reader. Cavalieri did not stop at lines. He also compared the squares built on the lines of figures. If a square is erected on every line of a triangle, the squares stack into a pyramid. Comparing the squares of a triangle with those of the surrounding parallelogram, he found the ratio 1 to 3, which gives at once the volumes of the pyramid and the cone and the area under a parabola.

Goals
  • Understand Cavalieri’s “all the squares” and its meaning as a volume.
  • State and prove the 3:1 theorem of Book II.
  • Derive the volume of cone and pyramid and the area of the parabolic segment.

The 3:1 theorem

Take a parallelogram \(ABCD\) with diagonal \(AC\), and take the regula parallel to \(AB\). Every line of the parallelogram has length \(a=|AB|\); the line of the triangle \(ACD\) at the same position has length \(x\) running from 0 to \(a\). Cavalieri’s “all the squares” of a figure is the collection of squares on its lines. In modern terms, with the lines spaced uniformly, “all the squares” of the triangle is proportional to \(\int_0^a x^2\,dx\), and of the parallelogram to \(\int_0^a a^2\,dx\).

Theorem (Geometria II.24)

All the squares of a parallelogram are three times all the squares of either triangle into which a diagonal divides it, the regula being one of its sides. In modern form: \(\int_0^a x^2\,dx=\tfrac13a^3\).

Proof sketch

The argument is a symmetry trick that Lesson 06 generalizes. Each line of the parallelogram is split by the diagonal into two parts \(x\) and \(a-x\), one in each triangle. The two triangles are congruent, so all the squares of one equal all the squares of the other. Write \(x=\tfrac a2+z\), \(a-x=\tfrac a2-z\). Then

\[ x^2+(a-x)^2 = 2\Big(\frac a2\Big)^2+2z^2 . \]

Collect over all lines. On the left are all the squares of both triangles, \(2T\). On the right, the constant \(2(a/2)^2\) collected over all the lines gives half of all the squares of the parallelogram, \(\tfrac12P\). The segments of length \(|z|\) are the lines of two small triangles on either side of the centre line, each similar to the original triangle in ratio \(\tfrac12\). All the squares of a triangle scale like a volume, by \(\tfrac18\), so the term \(2z^2\) contributes \(2\cdot2\cdot\tfrac18T=\tfrac12T\). Hence \(2T=\tfrac12P+\tfrac12T\), so \(P=3T\). In modern form: \(2T=\tfrac{a^3}2+2\int_{-a/2}^{a/2}z^2\,dz\).

Consequences

  • Pyramid. Erect on each line of the parallelogram a square in a perpendicular plane. The squares fill a prism; those on the triangle fill a pyramid with the same base and height. So a pyramid is one third of its prism, which is Euclid XII.7 without exhaustion.
  • Cone. Replace squares by circles: circles are to squares on the same line in a fixed ratio \(\pi:4\), so the cone is one third of the cylinder.
  • Parabola. In a parabola \(y=cx^2\) the abscissas vary as the square roots of the ordinates. The later books on conic sections apply the 3:1 theorem to show that the region between the parabola \(y=x^2\), the axis and the ordinate \(x=a\) is one third of the rectangle \(a\times a^2\).

Exercise 5.1Show that the parabolic segment cut from \(y=x^2\) by the line \(y=1\) has area \(\tfrac43\), which is \(\tfrac43\) of the inscribed triangle with vertices \((\pm1,1)\) and \((0,0)\). This is Archimedes’ result.

Show solution
The region under \(y=x^2\) on \([0,1]\) is one third of the unit square, so the region above it and below \(y=1\) is \(\tfrac23\). By symmetry the full segment has area \(\tfrac43\). The triangle has base 2 and height 1, area 1. The ratio is \(\tfrac43\).

Exercise 5.2Use the cone result and Cavalieri’s principle to find the volume of a ball of radius \(r\). (Compare the hemisphere with a cylinder of radius and height \(r\) from which a cone is removed.)

Show solution
At height \(z\in[0,r]\) the hemisphere has section \(\pi(r^2-z^2)\). The cylinder with a cone removed, vertex at the bottom, has section \(\pi r^2-\pi z^2\): the same. So the hemisphere has volume \(\pi r^3-\tfrac13\pi r^3=\tfrac23\pi r^3\), and the ball \(\tfrac43\pi r^3\).
Lesson 06

The quadrature of the higher parabolas

For the general reader. The curves \(y=x^2\), \(y=x^3\), \(y=x^4\) and so on were called the higher parabolas. In the 1630s several mathematicians found, independently, that the area under \(y=x^n\) from 0 to \(a\) is \(a^{n+1}/(n+1)\). Cavalieri announced it for every positive whole number \(n\) in his Centuria di varii problemi of 1639, and gave his fullest treatment in the fourth of his Exercitationes in 1647. It was the first general formula of the integral calculus.

Goals
  • State the quadrature formula and its history.
  • Follow a proof by symmetry and the binomial theorem in the style of Cavalieri.
  • Relate it to the limit of sums of powers.
Theorem (Cavalieri’s quadrature formula)

For every positive integer \(n\) and every \(a>0\),

\[ \int_0^a x^n\,dx=\frac{a^{n+1}}{n+1}. \]

In Cavalieri’s language: all the \(n\)-th powers of a parallelogram are \(n+1\) times all the \(n\)-th powers of the triangle cut off by its diagonal.

Cavalieri had the cases \(n=1,2\) in the Geometria; later accounts report that he worked the cases one by one up to \(n=9\). Pierre de Fermat and Gilles Personne de Roberval had found it by 1636 with arithmetic methods based on sums of powers, and Torricelli extended the method in the 1640s. John Wallis, in the Arithmetica infinitorum of 1656, extended the formula to fractional and negative exponents other than \(-1\).

A proof by symmetry

The following argument uses only the ingredients available to Cavalieri: the symmetry of the parallelogram, similarity, and the binomial expansion. It is a modern rendering in his spirit; historians differ on how closely each step follows his text. Write \(T_n(a)=\int_0^a x^n\,dx\), and suppose we know only that \(T_n(a)=c_n a^{n+1}\) for a constant \(c_n\) (this is the scaling of similar figures). Split each line of the square of side \(a\) as \(x=\tfrac a2+z\), \(a-x=\tfrac a2-z\). By the symmetry of the two triangles,

\[ 2T_n(a)=\int_0^a\big[x^n+(a-x)^n\big]dx=\int_{-a/2}^{a/2}\Big[\Big(\tfrac a2+z\Big)^n+\Big(\tfrac a2-z\Big)^n\Big]dz . \]

Odd powers of \(z\) cancel, and \(\int_{-a/2}^{a/2}z^k\,dz=2T_k(a/2)=2c_k(a/2)^{k+1}\) for even \(k\). So

\[ \begin{aligned} 2c_na^{n+1}&=2\sum_{k\ \mathrm{even}}\binom nk\Big(\frac a2\Big)^{n-k}\,2c_k\Big(\frac a2\Big)^{k+1}\\ &=\frac{2a^{n+1}}{2^{n}}\sum_{k\ \mathrm{even}}\binom nk c_k , \end{aligned} \]

giving the recursion

\[ c_n=\frac1{2^n}\sum_{\substack{0\le k\le n\\ k\ \mathrm{even}}}\binom nk c_k,\qquad c_0=1 . \]

For odd \(n\) the right side involves only earlier \(c_k\). For even \(n\) the term \(k=n\) contains \(c_n\) itself and is moved to the left. Either way the recursion determines every \(c_n\). The values \(c_k=1/(k+1)\) satisfy it, because

\[ \sum_{k\ \mathrm{even}}\binom nk\frac1{k+1}=\frac12\int_{-1}^{1}(1+t)^n\,dt=\frac{2^n}{n+1}. \]

Sums of powers

The arithmetic route of Fermat and Roberval, and later Wallis, replaces the lines by finitely many equally spaced lines and uses

\[ \lim_{N\to\infty}\frac{1^n+2^n+\cdots+N^n}{N^{n+1}}=\frac1{n+1}, \]

which follows from the inequalities \(\sum_{i=1}^{N-1}i^n<\frac{N^{n+1}}{n+1}<\sum_{i=1}^{N}i^n\). The two routes are the same theorem seen geometrically and arithmetically.

Exercise 6.1Use the recursion to compute \(c_3\) and \(c_4\) from \(c_0=1\) and \(c_2=\tfrac13\).

Show solution
\(n=3\): \(c_3=\tfrac18\big(\binom30c_0+\binom32c_2\big)=\tfrac18(1+1)=\tfrac14\). \(n=4\): \(c_4=\tfrac1{16}\big(1+6\cdot\tfrac13+c_4\big)\), so \(\tfrac{15}{16}c_4=\tfrac3{16}\) and \(c_4=\tfrac15\).

Exercise 6.2The curve \(y=x^n\) on \([0,1]\) divides the unit square into two parts. Find the ratio of the part above the curve to the part below it, and compute the volume of the solid obtained by rotating the part below the curve about the \(x\)-axis.

Show solution
Below: \(\tfrac1{n+1}\); above: \(\tfrac n{n+1}\). The ratio is \(n:1\). The solid has volume \(\pi\int_0^1x^{2n}\,dx=\tfrac{\pi}{2n+1}\), one \((2n+1)\)-th of the circumscribed cylinder. Cavalieri used such cubatures of solids of revolution in the fourth Exercitatio.

Exercise 6.3Prove \(\sum_{i=1}^{N-1}i^n<\frac{N^{n+1}}{n+1}<\sum_{i=1}^{N}i^n\) by comparing with \(\int_0^N x^n\,dx\), and deduce the limit above.

Show solution
On \([i-1,i]\) we have \((i-1)^n\le x^n\le i^n\), with strict inequality on most of the interval. Summing over \(i=1,\dots,N\): \(\sum_{i=0}^{N-1}i^n<\int_0^Nx^n\,dx<\sum_{i=1}^Ni^n\). Divide by \(N^{n+1}\). The two sums differ by \(N^n\), so after division they differ by \(1/N\to0\), and both tend to \(1/(n+1)\).
Lesson 07

Paradoxes, Guldin, and modern rigour

For the general reader. Can a surface really be made of lines, which have no width? Critics said no, and they produced paradoxes in which careless use of indivisibles proves false results. The Jesuit mathematician Paul Guldin also accused Cavalieri of borrowing from Kepler. Cavalieri answered at length. Three centuries later, measure theory showed exactly when his principle is true.

Goals
  • See how paradoxes arise from indivisibles.
  • Understand Guldin’s objections and Cavalieri’s replies.
  • Prove Cavalieri’s principle from Fubini’s theorem.

The objections

Paul Guldin, a Swiss Jesuit who taught in Vienna and Graz, attacked the Geometria in the fourth and last volume of his Centrobaryca (1641). He made three charges. First, that Cavalieri’s method came from Johannes Kepler’s Nova stereometria doliorum (1615), a work on the volumes of wine barrels, and from Bartolomeo Sovero’s Curvi ac recti proportio (1630). Second, that a figure cannot be composed of lines. Third, that “all the lines” of a figure are infinite in number, and that between one infinite and another there is no ratio.

Cavalieri replied briefly in his Trigonometria of 1643. He drafted a dialogue against Guldin, whose name he disguised by the anagram “Ginulous”, and then abandoned it. His full answer is the third of the Exercitationes (1647). On plagiarism he was on firm ground: his manuscript predated Sovero’s book. On composition he repeated that his method never assumed a figure to be made of its lines; it compared collections of lines as magnitudes in their own right, and the distributive method of Book VII did not need even that. On infinite ratios he argued that what matters is the relative infinity of two collections taken under the same regula. Most contemporaries found this last answer unconvincing.

The paradoxes

In 1644 Torricelli forwarded to Cavalieri an objection claiming that the method proves all triangles of equal height to be equal. Cavalieri’s answer was that Theorem II.3 applies only when both collections of lines are taken with the same regula and the same “passage”, so that corresponding lines are equally spaced. Galileo, in the Two New Sciences (1638), had produced a different puzzle: a bowl and a cone whose sections are always equal, but whose last sections are a circle and a point (Exercise 7.1). Galileo used it to argue that the infinite cannot be handled by the rules of finite quantities.

The modern statement

Theorem (Cavalieri’s principle, measure-theoretic form)

Let \(A,B\subset\mathbb R^{n}\) be Lebesgue measurable, and for \(t\in\mathbb R\) let \(A_t=\{y\in\mathbb R^{n-1}:(y,t)\in A\}\) be the section at height \(t\). If \(\lambda^{n-1}(A_t)=\lambda^{n-1}(B_t)\) for almost every \(t\), then \(\lambda^n(A)=\lambda^n(B)\).

The proof is one line from Tonelli’s theorem: \(\lambda^n(A)=\int_{\mathbb R}\lambda^{n-1}(A_t)\,dt\). The word “measurable” and the almost-everywhere clause are the precise answer to Galileo’s bowl: a single section, a circle or a point, has measure zero in the integral and cannot affect it. The answer to the triangle paradox is the factor \(dt\): the principle compares sections at the same heights, which is Cavalieri’s “same passage”.

Exercise 7.1 (Galileo’s bowl)From a cylinder of radius \(r\) and height \(r\), remove a hemisphere of radius \(r\) whose flat face is the top of the cylinder. Compare the remaining bowl with a cone of base radius \(r\) and height \(r\), standing on the same base plane with its vertex at the top. Show that the sections at each height have equal area, and describe the sections at the top.

Show solution
Put the base at \(z=0\) and the top at \(z=r\). The hemisphere’s centre is at \((0,0,r)\), so at height \(z\) the removed disk has radius\(^2\) \(r^2-(r-z)^2\), and the bowl’s section is an annulus of area \(\pi r^2-\pi\big(r^2-(r-z)^2\big)=\pi(r-z)^2\). The cone’s section at height \(z\) is a disk of radius \(r-z\), with the same area. At \(z=r\) the bowl’s section is the rim, a circle, and the cone’s is the vertex, a point. Both have area zero, so “the circle equals the point” in Galileo’s phrase is harmless: equal volumes follow, \(\tfrac13\pi r^3\) each.

Exercise 7.2Here is a false argument of the kind Cavalieri’s critics raised. The diagonal from \((0,0)\) to \((1,2)\) cuts the rectangle \([0,1]\times[0,2]\) into two congruent triangles. Each point \(P=(x,2x)\) of the diagonal determines a horizontal segment from \(P\) to the left side, lying in the upper triangle, and a vertical segment from \(P\) down to the base, lying in the lower triangle. The segments pair off one to one and the vertical one is always twice the horizontal one. So “all the lines” of the lower triangle are twice those of the upper one, and the lower triangle has twice the area. Find the error.

Show solution
The horizontal segment at \(P\) has length \(x\) and the vertical one \(2x\), so the ratio really is \(1:2\). But the horizontal segments are indexed by height \(y=2x\) and the vertical ones by \(x\). As \(P\) moves, the horizontal segments are spaced \(dy=2\,dx\) apart, twice as far as the vertical ones. The areas are \(\int x\,dy=\int_0^1 x\cdot2\,dx=1\) and \(\int_0^1 2x\,dx=1\), equal. Cavalieri’s rule compares lines taken with the same regula and the same spacing; the argument compares two different regulae, which is what his reply to such objections pointed out.
Lesson 08

Centres of gravity and Guldin’s theorem

For the general reader. The volume of a doughnut equals the area of its cross-section times the distance travelled by the centre of that cross-section. This rule, stated in late antiquity by Pappus of Alexandria and restated by Guldin, is now called the Pappus–Guldin theorem. Cavalieri proved it with indivisibles, turning his critic’s best-known result into an advertisement for his own method. His fifth Exercitatio extended centres of gravity to bodies of variable density.

Goals
  • State Pappus–Guldin for volumes and surfaces.
  • Prove it by slicing.
  • Use it to locate centroids and compute volumes.
Theorem (Pappus–Guldin)

Let a plane region \(F\) of area \(A\) lie on one side of a line \(L\) in its plane, and let \(\bar x\) be the distance from its centroid to \(L\). The solid swept out by rotating \(F\) once about \(L\) has volume \(V=2\pi\bar x\,A\). Likewise a plane curve of length \(s\), with centroid at distance \(\bar x\) from \(L\), sweeps a surface of area \(S=2\pi\bar x\,s\).

Proof by indivisibles

Take the regula parallel to \(L\). The line of \(F\) at distance \(x\) from \(L\), of length \(\ell(x)\), sweeps a cylindrical surface of area \(2\pi x\,\ell(x)\). The solid is composed of these surfaces, so

\[ V=\int 2\pi x\,\ell(x)\,dx=2\pi\int x\,\ell(x)\,dx=2\pi\,\bar x\,A, \]

since \(\bar x=\frac1A\int x\,\ell(x)\,dx\) is the definition of the centroid’s distance from \(L\). Cavalieri proved the theorem in the Exercitationes with the help of a lemma supplied by his pupil Giannantonio Rocca. Guldin had stated the rule in the Centrobaryca without a satisfactory proof.

Exercise 8.1A circle of radius \(r\) centred at distance \(R>r\) from an axis in its plane is rotated about the axis. Find the volume and surface area of the torus.

Show solution
The centroid of the disk and of the circle is the centre, at distance \(R\). \(V=2\pi R\cdot\pi r^2=2\pi^2Rr^2\) and \(S=2\pi R\cdot2\pi r=4\pi^2Rr\).

Exercise 8.2Use Pappus–Guldin in reverse to find the centroid of a half-disk of radius \(r\), and of a semicircular arc of radius \(r\), measured from the diameter.

Show solution
Rotating the half-disk about its diameter gives a ball: \(\tfrac43\pi r^3=2\pi\bar x\cdot\tfrac12\pi r^2\), so \(\bar x=\tfrac{4r}{3\pi}\). Rotating the arc gives a sphere: \(4\pi r^2=2\pi\bar x\cdot\pi r\), so \(\bar x=\tfrac{2r}{\pi}\).
Lesson 09

Logarithms and spherical trigonometry

For the general reader. Logarithms turn multiplication into addition. John Napier published them in 1614 and Henry Briggs produced base-ten tables in the 1620s. Cavalieri brought them to Italy. His Directorium generale uranometricum of 1632 contained tables of logarithms and of logarithms of trigonometric functions, and rules for solving triangles on the sphere, the daily work of astronomers. Galileo praised the work.

Goals
  • Use base-ten logarithms as a seventeenth-century computer did.
  • Solve right spherical triangles with Napier’s rules.
  • State and prove the area formula for spherical triangles.

The books

The Directorium had three parts: logarithms, plane trigonometry and spherical trigonometry. It included demonstrations of Napier’s rules for right spherical triangles and a theorem on the area of spherical triangles. Its renewal of his Bologna appointment depended partly on it. Cavalieri followed it with the Compendio delle regole dei triangoli (1638), the Centuria di varii problemi (1639), a hundred problems showing logarithms at work in astronomy, geography and the design of sundials, and the Trigonometria plana et sphaerica, linearis et logarithmica (1643).

Napier’s rules for a right spherical triangle

On a sphere of radius 1, let a triangle have a right angle at \(C\), legs \(a,b\), hypotenuse \(c\), and angles \(A,B\) opposite \(a,b\). Then, among other relations,

\[ \cos c=\cos a\cos b,\qquad \sin a=\sin c\,\sin A,\qquad \tan b=\tan c\,\cos A . \]
Theorem (area of a spherical triangle)

A spherical triangle with angles \(A,B,C\) on a sphere of radius \(R\) has area \(R^2(A+B+C-\pi)\). Albert Girard had published the theorem in 1629; the Directorium contains Cavalieri’s treatment of the area of spherical triangles.

Proof sketch. A lune between two great semicircles meeting at angle \(\theta\) has area \(2\theta R^2\), a fraction \(\theta/2\pi\) of the sphere. The three great circles through the sides of the triangle form, for each angle, a lune and its antipodal lune. These six lunes cover the sphere, covering the triangle and its antipode three times each and everything else once. So \(4\cdot2R^2(A+B+C)=4\pi R^2+4\,\mathrm{Area}\), which gives the formula.

Exercise 9.1Using \(\log_{10}2\approx0.30103\), find the number of decimal digits of \(2^{64}\) and its first three digits. (Also given: \(\log_{10}1.84\approx0.26482\), \(\log_{10}1.85\approx0.26717\).)

Show solution
\(\log_{10}2^{64}=64\times0.30103=19.26592\). The integer part 19 means 20 digits. The fractional part 0.26592 lies between the logarithms of 1.84 and 1.85, so \(2^{64}\approx1.84\ldots\times10^{19}\). (In fact \(2^{64}=18\,446\,744\,073\,709\,551\,616\).)

Exercise 9.2On the unit sphere, a right triangle has both legs equal to \(60^\circ\). Find the hypotenuse and the area.

Show solution
\(\cos c=\cos^2 60^\circ=\tfrac14\), so \(c=\arccos\tfrac14\approx75.52^\circ\). From \(\sin a=\sin c\sin A\): \(\sin A=\frac{\sqrt3/2}{\sqrt{15}/4}=\frac{2}{\sqrt5}\), so \(A=B\approx63.43^\circ\). The area is \(A+B+C-\pi\approx(63.43+63.43+90-180)^\circ=36.87^\circ\approx0.6435\).

Exercise 9.3Check the area formula on the triangle bounded by the equator and two meridians \(90^\circ\) apart.

Show solution
All three angles are right angles, so the area is \(R^2(\tfrac{3\pi}2-\pi)=\tfrac{\pi R^2}2\). Eight such triangles tile the sphere, and \(8\cdot\tfrac{\pi R^2}{2}=4\pi R^2\).
Lesson 10

The burning mirror: conics, telescopes, projectiles

For the general reader. Legend says Archimedes set Roman ships on fire with mirrors. Cavalieri’s first book, Lo specchio ustorio, overo trattato delle settioni coniche (Bologna, 1632), used the focal properties of the parabola, ellipse and hyperbola to examine the legend. Along the way it sketched designs for a telescope that uses mirrors rather than lenses, and stated that a projectile moves along a parabola, a result Galileo regarded as his own.

Goals
  • Prove the reflective properties of the conics.
  • Understand two-mirror telescope designs.
  • Derive the parabolic trajectory and the angle of maximum range.
Reflective properties

A ray parallel to the axis of a parabola reflects through its focus. A ray from one focus of an ellipse reflects through the other focus. A ray aimed at one focus of a hyperbola, striking the convex outer side of the branch around that focus, reflects toward the other focus.

Mirrors and telescopes

The book discussed parabolic, hyperbolic and elliptical mirrors, spherical mirrors as optical and acoustic instruments, and the equivalence of lens and mirror systems. It treated the focal length of a lens whose two faces have different curvature, a problem whose modern answer, for a thin biconvex lens with radii \(r_1,r_2\) and refractive index \(n\), is the lensmaker’s formula \(1/f=(n-1)(1/r_1+1/r_2)\). Cavalieri described three arrangements of a large concave mirror with a smaller secondary mirror, convex in two of them. Several historians therefore credit him with the idea of a reflecting telescope, decades before James Gregory (1663) and Isaac Newton (1668). Cesare Marsili had written to Galileo about mirror telescopes as early as 1626.

Projectiles and a quarrel

The book also stated that a body thrown horizontally or obliquely moves, neglecting air resistance, along a parabola. Galileo had found this result long before but had not published it, and he felt that Cavalieri had used his ideas and demonstrations. Cesare Marsili mediated, and the friendship was restored. Galileo published his own proof in the Two New Sciences in 1638.

Exercise 10.1Show that the tangent to the parabola \(y=x^2/(4f)\) at \(P=(x_0,y_0)\) makes equal angles with the vertical line through \(P\) and with the line from \(P\) to the focus \(F=(0,f)\). Deduce the reflective property.

Show solution
The tangent meets the \(y\)-axis at \(T=(0,-y_0)\), since its slope is \(x_0/(2f)\) and \(y_0-\frac{x_0}{2f}x_0=-y_0\). Then \(|FT|=f+y_0\), and \(|FP|=\sqrt{x_0^2+(y_0-f)^2}=\sqrt{4fy_0+(y_0-f)^2}=y_0+f\). So triangle \(FPT\) is isosceles, and the angle at \(P\) between \(PT\) and \(PF\) equals the angle at \(T\), which equals the angle between the tangent and the vertical through \(P\) (alternate angles). A vertical ray arriving at \(P\) therefore reflects along \(PF\).

Exercise 10.2A projectile leaves the ground with speed \(v\) at angle \(\theta\). Show that its path is \(y=x\tan\theta-\dfrac{g\,x^2}{2v^2\cos^2\theta}\), find its range, and show that the range is greatest at \(45^\circ\).

Show solution
\(x=vt\cos\theta\), \(y=vt\sin\theta-\tfrac12gt^2\). Eliminate \(t=x/(v\cos\theta)\) to get the parabola. Setting \(y=0\) gives the range \(x=\frac{2v^2\sin\theta\cos\theta}{g}=\frac{v^2\sin2\theta}{g}\), maximal when \(2\theta=90^\circ\).

Exercise 10.3A concave paraboloidal mirror has focus \(F\). Before the light reaches \(F\), it meets a small convex mirror shaped like one branch of a hyperboloid with foci \(F\) and \(F'\), where \(F'\) lies behind the main mirror. Where does the light come to a focus? Why would a designer put \(F'\) there?

Show solution
The light reflected by the paraboloid converges toward \(F\). By the reflective property of the hyperbola, rays aimed at one focus reflect toward the other, so they converge at \(F'\). Putting \(F'\) behind the main mirror (seen through a hole in it) folds a long focal length into a short tube. This is the arrangement later called the Cassegrain telescope (1672).
Part III · Life and legacy
Lesson 11

Bologna, 1629–1647

The Bologna chair was a heavy teaching post with an astronomical tradition. The curriculum required Euclid’s Elements, the medieval Theorica planetarum and Ptolemy’s Almagest, and the university authorities pressed Cavalieri to produce astronomy rather than pure geometry. He did both. Of his eleven books published in Bologna, several served students and astronomers: tables, trigonometry and two works with astrological titles. In the Nuova pratica astrologica (1639) he stated that astrology cannot predict the future; the book and the Trattato della ruota planetaria perpetua (1646) are works of computational astronomy and chronology. He was one of the first Bologna professors to teach the Copernican system, presented, as censorship required, as a hypothesis.

He was a well-placed and respected figure. The Bologna Senate, Cardinal Borromeo, Pope Urban VIII and Grand Duke Ferdinando II de’ Medici of Tuscany recognized his work, and Galileo called him a second Archimedes. In 1636 he spent the summer near Florence talking mathematics with Galileo, then under house arrest at Arcetri. He declined Galileo’s offer of a chair at Pisa and Borromeo’s invitation to the Ambrosiana in Milan.

Pupils

His pupils included Stefano degli Angeli, a fellow Jesuate who wrote many of his letters when his hands could no longer hold a pen and who later continued work on indivisibles; Giannantonio Rocca, whose lemma he used for Guldin’s theorem; and Pietro Mengoli, his successor at Bologna, who in 1650 proved that the harmonic series diverges and posed the problem of summing \(\sum1/n^2\), later solved by Euler. Through his letters he was in contact with Torricelli in Florence and with Marin Mersenne in Paris.

Illness

Gout, which had afflicted him since youth, worsened. Problems with his legs began about 1629 and never left him. By the 1640s he could barely move, and in 1646 he had to stop teaching. He worked in these years on the Exercitationes geometricae sex, 543 pages in six parts: a revised account of the collective method, the distributive method, the reply to Guldin, the extension to higher powers and higher curves, centres of gravity, and miscellaneous problems.

DiscussionCavalieri held a chair that required astronomy, wrote tables and astrological almanacs, and still produced a new geometry. Compare his position with that of a modern academic expected to teach service courses and publish research. What did each kind of work give him?

Lesson 12

Death and legacy

The Exercitationes appeared in 1647. Cavalieri died in Bologna on 30 November 1647, unable to walk, a month after Torricelli, the other great practitioner of indivisibles. He was buried in the church of Santa Maria della Mascarella, where a memorial tablet records him. Some older sources give other days in November or December; 30 November is the date of the standard biographies. Twenty-one years later, in 1668, Pope Clement IX suppressed the Jesuati.

Legacy

  • Method. The Geometria (second edition 1653) and the Exercitationes were the main sources for indivisibles in the seventeenth century. Torricelli extended the method to curved indivisibles and to infinite solids. Wallis read Cavalieri and Torricelli and turned their geometry into the arithmetic of the Arithmetica infinitorum. Through Wallis and Isaac Barrow, the tradition reached Newton; Leibniz knew it from Pascal and others.
  • The quadrature formula. \(\int_0^ax^n\,dx=a^{n+1}/(n+1)\), found independently in the 1630s by Cavalieri, Fermat and Roberval, is the first general theorem of the integral calculus. More than a century later, Joseph-Louis Lagrange wrote notes reconstructing a complete proof along Cavalieri’s lines; they were studied in 2025.
  • Controversy. Indivisibles were controversial in Catholic Italy. The historian Amir Alexander has argued that opposition to them among the Jesuits had theological as well as mathematical motives. The debate is part of the history of how mathematics came to accept the infinitely small.
  • Rigour. Cavalieri’s principle survives intact as a corollary of Fubini’s theorem and is taught in school geometry worldwide.
  • Logarithms. His tables and textbooks brought logarithmic computation into Italian astronomy; Giovanni Domenico Cassini later consulted his trigonometry.
  • Honours. The lunar crater Cavalerius is named after him.

Final projectTake one result from Part II (the 3:1 theorem, the quadrature of \(y=x^n\), or Pappus–Guldin). Write a two-page account that gives Cavalieri’s argument with indivisibles, identifies the step a critic like Guldin would attack, and supplies a modern proof that repairs it.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Bonaventura Francesco Cavalieri, MacTutor History of Mathematics.
  2. E. Carruccio, Cavalieri, Bonaventura, Dictionary of Scientific Biography, vol. 3.
  3. Cavalieri, Bonaventura, Biographical Encyclopedia of Astronomers.
  4. Bonaventura Cavalieri, Wikipedia.
  5. Bonaventura Cavalieri, Encyclopaedia Britannica.
  6. W. W. Rouse Ball, Bonaventura Cavalieri, from A Short Account of the History of Mathematics (1908).
  7. Cesare Marsili, Wikipedia (on the Bologna appointment and the 1632 dispute).

Mathematics

  1. B. Cavalieri, Geometria indivisibilibus continuorum nova quadam ratione promota, Bologna, 1635; 2nd ed. 1653.
  2. B. Cavalieri, Exercitationes geometricae sex, Bologna, 1647.
  3. B. Cavalieri, Lo specchio ustorio, overo trattato delle settioni coniche, Bologna, 1632; Directorium generale uranometricum, 1632; Centuria di varii problemi, 1639.
  4. K. Andersen, Cavalieri’s method of indivisibles, Archive for History of Exact Sciences 31 (1985), 291–367.
  5. Ph. Henry, on Lagrange’s manuscript notes about Cavalieri’s geometry of indivisibles, Comptes Rendus. Mathématique 363 (2025), 1407–1427.
  6. Cavalieri’s quadrature formula and Cavalieri’s principle, Wikipedia.
  7. A. Alexander, Infinitesimal: How a Dangerous Mathematical Theory Shaped the Modern World, Scientific American / Farrar, Straus and Giroux, 2014.
  8. C. B. Boyer, The History of the Calculus and its Conceptual Development, Dover, 1959.
  9. E. Giusti, Bonaventura Cavalieri and the Theory of Indivisibles, Cremonese, Rome, 1980.