How to use this course
Gilles Personne de Roberval was the one professional mathematician in the Paris circle of Marin Mersenne, the colleague of Fermat, the Pascals and Descartes, and the enemy of the last. He found the area under the cycloid, a general way to draw tangents by composing motions, and the quadrature of \(y=x^n\) and of the sine, all in the 1630s. He published almost none of it. His name survives today mostly on a kitchen scale.
- Single-variable calculus: integrals, parametric curves, arc length, volumes of revolution.
- Vectors in the plane: addition, the parallelogram rule.
- Elementary statics: forces, moments, equilibrium.
- For Lesson 09, the idea of air pressure and of a pendulum.
Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern notation, sketches of Roberval’s methods, and exercises with folded solutions. Many of his results survive only in the posthumous Divers ouvrages of 1693 and in letters, so dates are often given as “before” a letter in which a result is mentioned. The course follows the dates established by Kokiti Hara in the Dictionary of Scientific Biography.
Timeline
| 1602 | Born 10 August at Noël-Saint-Martin, near Villeneuve-sur-Verberie and the village of Roberval (Oise), to Pierre Personne and Jeanne Le Dru, farmers. |
| c. 1616 | At about 14, begins Latin and mathematics with the parish priest of Rhuis. |
| 1620s | Travels through France, earning his living by teaching mathematics; meets Fermat in Bordeaux. |
| 1627 | At the siege of La Rochelle, studying fortification and ballistics. |
| 1628 | Arrives in Paris; joins the circle of Mersenne. Takes the name “de Roberval”. Mersenne draws his attention to the cycloid. |
| 1632 | Professor of philosophy at the Collège de Maître Gervais. |
| 1634 | 24 June: wins the competition for the Ramus chair of mathematics at the Collège Royal. |
| by 1636 | Area of the cycloid and volume of its solid of revolution about the base; tangents by composition of motions; quadrature of \(y=x^n\). Traité de mécanique published. |
| 1637–38 | With Fermat, in dispute with Descartes over tangents and the Géométrie. |
| 1642–43 | Compares the arc lengths of the spiral and the parabola. |
| 1643 | October: Torricelli writes to him on the cycloid. A priority dispute follows the publication of Torricelli’s Opera geometrica in 1644. |
| 1644 | Aristarchi Samii de mundi systemate, a heliocentric system with universal attraction. |
| 1646 | Dispute with Descartes over the centre of oscillation of a compound pendulum. |
| 1647–48 | Two Narrationes on the vacuum experiments; the carp-bladder experiment; apparatus for the “vacuum within a vacuum”. |
| 1655 | At the end of the year succeeds Gassendi in a second mathematics chair at the Collège Royal. |
| 1658 | Pascal’s Histoire de la roulette credits him with the first quadrature of the cycloid. |
| 1666 | Founding member of the Académie royale des sciences. |
| 1669 | 21 August: presents the balance that bears his name to the Académie. |
| 1675 | Dies in Paris on 27 October, aged 73. |
| 1693 | Divers ouvrages de M. de Roberval published by the Académie. |
A harvest field in the Oise, 1602–1627
Gilles Personne was born on 10 August 1602 at Noël-Saint-Martin, a hamlet of Villeneuve-sur-Verberie near Senlis, north-east of Paris. His parents, Pierre Personne and Jeanne Le Dru, worked the land. According to family tradition his mother gave birth in the fields during the harvest. He had brothers and sisters; only his sister Marie outlived him.
At about 14 he began to study Latin and mathematics with the parish priest of Rhuis, a chaplain of the queen mother Marie de’ Medici. He probably also learned some Greek. There was no question of a university. Instead, by his early twenties he left home and travelled through France, supporting himself by giving lessons in mathematics and seeking out university teachers to discuss harder questions. In Bordeaux he met Pierre de Fermat; the two would correspond for years.
In September 1627 he was at the siege of La Rochelle, the Huguenot stronghold besieged by the army of Louis XIII. There he studied fortification and the motion of projectiles. Descartes was present at the same siege.
Paris and the Mersenne circle, 1628–1634
In 1628 he arrived in Paris and found his way into the circle of the Minim friar Marin Mersenne. Its members met to read papers and propose problems: Claude Mydorge, Claude Hardy, Étienne Pascal and later his son Blaise. Mersenne valued him highly. Around this time he added “de Roberval” to his name, after the village near his birthplace, with local permission.
Mersenne collected open problems and passed them on. One of them was the cycloid, the curve traced by a point on the rim of a rolling wheel. Galileo had named it and had tried to find its area by weighing cut-out shapes. Mersenne put the problem to Roberval in 1628. Roberval later called the curve the trochoid, from the Greek for wheel; Pascal’s French name roulette also stuck. Lesson 05 tells what he found.
Before 1632 he had also studied the symbolic algebra of François Viète. He wrote a short treatise on the theory of equations, De recognitione aequationum, probably before the publication of Descartes’s Géométrie in 1637, though it goes only as far as the basics. In 1632 he was appointed professor of philosophy at the Collège de Maître Gervais in Paris, where he lived for the rest of his life.
The Ramus chair and the habit of secrecy
The philosopher Petrus Ramus, killed in the massacre of St Bartholomew’s Day in 1572, had endowed a chair of mathematics at the Collège Royal, the royal foundation that is now the Collège de France. Its terms were unusual. The chair was filled by public competition, and every three years the holder had to defend it against challengers. The holder set problems, and anyone who solved them better could claim the chair.
On 24 June 1634 Roberval won the competition. He then held the chair for 41 years, until his death, defending it every three years. The likely consequence, argued by historians since the eighteenth century, is his secrecy. A method that only he knew was a weapon for the next competition. He let results be known through Mersenne and in letters, kept his methods hidden, and published very little. In his lifetime he printed only two works: a short Traité de mécanique (1636), which appeared in Mersenne’s Harmonie universelle, and the astronomical treatise attributed to Aristarchus (1644).
The price was credit. Bonaventura Cavalieri published his indivisibles in 1635 and his name became attached to them; Evangelista Torricelli published the cycloid in 1644. Roberval spent much of the rest of his career claiming priority, often angrily. Contemporaries described his temper as difficult, and he had a reputation for meanness, though he helped to provide dowries for his relatives. He lived austerely and never married.
At the end of 1655, on the death of Pierre Gassendi, he was also appointed to Gassendi’s mathematics chair at the Collège Royal. This one did not require competition, and he held both chairs until his death.
Indivisibles and sums of powers
For the general reader. Roberval, like Cavalieri, found areas by cutting figures into very many thin strips. He said he took the idea from Archimedes, not from Cavalieri, whose book he had not seen. His strips were thin but not quite zero: he thought of a line as made of many small lines, and a surface as made of many small surfaces. This made his method arithmetical. Areas became sums of numbers, and the key was a formula for sums of powers.
- Understand Roberval’s view of indivisibles as small elements.
- Prove the power-sum inequalities and deduce \(\int_0^a x^n\,dx=a^{n+1}/(n+1)\).
- See how the same method gives areas by comparing curves.
The Traité des indivisibles
The Traité des indivisibles, printed in 1693 from his papers, explains that one may treat a line as composed of infinitely many lines, a surface of infinitely many surfaces, a solid of infinitely many solids, as long as the elements are homogeneous with the whole. In practice he divided a segment into \(N\) equal parts, built figures on the parts, computed, and let \(N\) grow. Hara dates his quadrature of the higher parabolas to about 1636. In a letter to Fermat he stated the inequalities that make it work.
For all positive integers \(n\) and \(N\),
\[ 1^n+2^n+\cdots+(N-1)^n \;<\; \frac{N^{n+1}}{n+1} \;<\; 1^n+2^n+\cdots+N^n . \]For every positive integer \(n\), the region under \(y=x^n\) between \(x=0\) and \(x=a\) has area \(\dfrac{a^{n+1}}{n+1}\), that is, \(\dfrac1{n+1}\) of the rectangle \(a\times a^n\).
Proof
Divide \([0,a]\) into \(N\) equal parts of width \(a/N\). The region is squeezed between the inscribed and circumscribed stacks of rectangles:
\[ \frac aN\sum_{i=0}^{N-1}\Big(\frac{ia}N\Big)^n \;<\; \text{Area} \;<\; \frac aN\sum_{i=1}^{N}\Big(\frac{ia}N\Big)^n . \]Both bounds equal \(\frac{a^{n+1}}{N^{n+1}}\) times a power sum. By the lemma, the lower bound is less than \(\frac{a^{n+1}}{n+1}\) and the upper bound greater, and the bounds differ by \(\frac{a^{n+1}}{N}\). Since this is true for every \(N\), the area is exactly \(\frac{a^{n+1}}{n+1}\). This is an exhaustion argument in the Greek style with an arithmetical engine, which is why Roberval could honestly say he followed Archimedes.
To prove the lemma, expand \((k+1)^{n+1}-k^{n+1}=(n+1)k^n+\binom{n+1}{2}k^{n-1}+\cdots+1\). This is greater than \((n+1)k^n\). By the mean value theorem, or by the identity \(b^{n+1}-c^{n+1}=(b-c)\sum b^jc^{n-j}\), it is less than \((n+1)(k+1)^n\). Summing over \(k=0,\dots,N-1\) telescopes the left side to \(N^{n+1}\).
Exercise 4.1Verify the lemma for \(n=2\), \(N=4\), and find the exact formula for \(\sum_{i=1}^N i^2\) from the identity \((k+1)^3-k^3=3k^2+3k+1\).
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Exercise 4.2Using the theorem, find the area enclosed between the curves \(y=x^2\) and \(y=x^3\) on \([0,1]\), and the ratio in which the curve \(y=x^3\) divides the region under \(y=x^2\).
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The roulette: area and solids
For the general reader. Roll a coin along a ruler and follow a point on its rim. The point traces a row of arches, the cycloid. Galileo guessed, by weighing paper models, that one arch encloses about three times the area of the coin. Roberval proved that it is exactly three times, by inventing a second curve that splits the arch into pieces he could measure. The second curve, his “companion of the cycloid”, is the first sine curve drawn in the history of mathematics.
- Parametrize the cycloid and its companion curve.
- Prove that one arch has area three times the generating circle.
- Compute the volume of the solid generated by rotating the arch about its base.
The curves
Let a circle of radius \(a\) roll along the \(x\)-axis, and let \(P\) be the point of the rim that starts at the origin. After the circle has turned through angle \(t\), its centre is at \((at,a)\), and
\[ P(t)=\big(a(t-\sin t),\ a(1-\cos t)\big),\qquad 0\le t\le2\pi . \]Roberval’s companion point \(Q(t)\) has the same height as \(P\) but sits directly below the centre of the circle’s horizontal position:
\[ Q(t)=\big(at,\ a(1-\cos t)\big). \]So \(Q\) traces \(y=a(1-\cos(x/a))\), a sine curve. The horizontal segment \(QP\) has length \(a\sin t\), which is exactly the half-chord of the rolling circle at height \(y\).
One arch of the cycloid generated by a circle of radius \(a\) encloses, with its base, the area \(3\pi a^2\), three times the area of the generating circle. Rotated about its base, the arch generates a solid of volume \(5\pi^2a^3\).
Roberval’s argument for the area
Take half an arch, \(0\le t\le\pi\), inside the rectangle of width \(\pi a\) and height \(2a\), area \(2\pi a^2\).
- Between the cycloid and the companion curve, the horizontal segment at height \(y\) equals the half-chord of the generating circle at that height. By the principle of indivisibles, the region between them has the area of a half-disk, \(\tfrac12\pi a^2\).
- The companion curve passes through the centre of the rectangle and is symmetric about it, so it cuts the rectangle into two equal parts of area \(\pi a^2\).
- The region under the half-arch is therefore \(\pi a^2+\tfrac12\pi a^2=\tfrac32\pi a^2\), and the whole arch \(3\pi a^2\).
In modern terms the calculation is \(\int_0^{2\pi}a(1-\cos t)\cdot a(1-\cos t)\,dt=a^2\int_0^{2\pi}(1-2\cos t+\cos^2t)\,dt=a^2(2\pi+\pi)=3\pi a^2\).
Roberval told Descartes the result. Descartes answered that it was a pretty result but would give no trouble to any moderately skilled geometer, and set Roberval to find the tangent instead. Roberval also extended his results to the general trochoid, traced by a point inside or outside the rim, and between 1644 and 1645 found the volume of the solid generated by rotating the arch about its axis of symmetry (Lesson 10).
Exercise 5.1Verify the volume \(5\pi^2a^3\) of the solid generated by rotating one arch about its base.
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Exercise 5.2Check step 1 of Roberval’s argument by computing \(\int_0^{2a}|QP|\,dy\) over the half-arch.
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Exercise 5.3Find the centroid height \(\bar y\) of the region under one arch, and use the Pappus–Guldin theorem to check the volume in Exercise 5.1.
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Tangents by composition of motions
For the general reader. Swing a stone on a string and let go: it flies off along the tangent to the circle. Roberval made this into a method. Regard a curve as the path of a moving point. If the motion is a combination of simpler motions, add their velocities like forces, by the parallelogram rule. The result points along the tangent. He developed the method before 1636 and taught it from 1639 to 1644; his pupil François du Verdus wrote up the lessons, published in 1693 as Observations sur la composition des mouvemens.
- State the kinematic principle of tangents.
- Find the tangents to the cycloid, the parabola and the ellipse.
- Compare the method with the derivative.
Let a curve be traced by a point whose motion is composed of motions with velocities \(\mathbf v_1,\dots,\mathbf v_k\) at a given instant. Then the tangent at that point has the direction of \(\mathbf v_1+\cdots+\mathbf v_k\).
In modern terms, if \(\gamma(t)=\gamma_1(t)+\gamma_2(t)\) then \(\gamma'(t)=\gamma_1'(t)+\gamma_2'(t)\), and \(\gamma'(t)\) is tangent to \(\gamma\) when it is non-zero. The principle is the linearity of the derivative, used before the derivative existed. The difficulty, which critics noted, is to choose the component motions correctly and to know their relative speeds.
The cycloid
The tracing point moves with the centre of the wheel, velocity \(a\omega\) horizontally, and turns about the centre, velocity \(a\omega\) along the circle. The two components have equal length, so their sum bisects the angle between them. A short calculation shows that the bisector passes through the highest point of the rolling circle; equivalently, the normal passes through the point of contact with the ground.
Conics
A point on an ellipse moves so that the sum of its distances to the foci is constant. As it moves, it recedes from one focus exactly as fast as it approaches the other. So its velocity has equal components along the two focal radii, one away from \(F_1\), one toward \(F_2\), and the tangent bisects the angle between \(\overrightarrow{F_1P}\) and \(\overrightarrow{PF_2}\), the external angle at \(P\) of the triangle \(F_1PF_2\). For the parabola the same reasoning uses the focus and the directrix.
Exercise 6.1With \(P(t)=(a(t-\sin t),a(1-\cos t))\), show that \(P'(t)\) is parallel to the vector from \(P\) to the top of the rolling circle, \(T=(at,2a)\).
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Exercise 6.2For the parabola with focus \(F\) and directrix \(d\), use Roberval’s method to show that the tangent at \(P\) bisects the angle between \(PF\) and the perpendicular from \(P\) to \(d\).
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Exercise 6.3Show that the method needs care: the Archimedean spiral \(r=c\theta\) is traced by a point moving outward along a rotating ray at speed \(c\omega\) while the ray turns at rate \(\omega\). Find the angle \(\psi\) between the tangent and the radius at distance \(r\).
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The sine, and the lengths of curves
For the general reader. The companion curve of the cycloid led Roberval to the area under a sine curve, one of the first integrals of a trigonometric function. He also compared lengths of curves, a question the Greeks had abandoned as hopeless. He showed that a spiral and a parabola can have equal length, and he claimed, before 1640, to know the length of the cycloid.
- Compute \(\int\sin\) as Roberval did, by indivisibles on the circle.
- Prove the equality of arc lengths of the Archimedean spiral and a parabola.
- Find the length of the cycloid arch.
For \(0\le\theta\le\pi\), \(\displaystyle\int_0^\theta\sin\varphi\,d\varphi=1-\cos\theta\). In particular one arch of the sine curve \(y=\sin x\) has area 2.
Roberval’s idea, reconstructed. Divide the arc of the unit circle from angle 0 to \(\theta\) into equal small arcs \(d\varphi\). Each small arc, almost straight, has a horizontal projection of length \(\sin\varphi\,d\varphi\) onto the horizontal diameter, by similar triangles. The projections tile the segment of the diameter from \(\cos\theta\) to 1. So the sum of the sines times the arcs equals \(1-\cos\theta\). He used this to solve a problem he liked to set: to trace on a right circular cylinder, with a single motion of a compass, a region equal in area to a given square.
The arc of the Archimedean spiral \(r=c\theta\) from \(\theta=0\) to \(\theta=\Theta\) has the same length as the arc of the parabola \(y=x^2/(2c)\) from \(x=0\) to \(x=c\Theta\).
Proof. In polar coordinates, \(ds^2=dr^2+r^2d\theta^2\), so along the spiral \(ds=c\sqrt{1+\theta^2}\,d\theta\). Along the parabola, \(ds=\sqrt{1+x^2/c^2}\,dx\). Substituting \(x=c\theta\) gives the same element \(c\sqrt{1+\theta^2}\,d\theta\). Roberval obtained this by comparing the motions that generate the two curves.
One arch of the cycloid generated by a circle of radius \(a\) has length \(8a\), four times the diameter of the circle.
Roberval claimed this before 1640, by reducing it to an integral of the sine; Hara notes there is no independent evidence of the claim. Christopher Wren found a proof in 1658 in response to Pascal’s challenge, and Torricelli had a version around 1645. The proof is Exercise 7.2.
Exercise 7.1Compute the area under \(y=a(1-\cos(x/a))\) on \([0,\pi a]\) and confirm step 2 of Lesson 05.
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Exercise 7.2Prove that one arch of the cycloid has length \(8a\).
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Exercise 7.3Compute the length of the first turn of the spiral \(r=\theta\) (\(0\le\theta\le2\pi\)) in closed form.
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Statics: inclined planes and the balance
For the general reader. Roberval’s first publication, the Traité de mécanique of 1636, asked how much effort holds a weight on a slope, and answered by showing how forces pulling in different directions combine. Thirty-three years later he showed the Académie a balance with an odd property: the weighing comes out the same wherever you put the load on the pan. Versions of it are still in shops and kitchens.
- Derive the law of the inclined plane and the composition of forces.
- Explain the Roberval balance by moments and by virtual work.
The inclined plane and two cords
The treatise studies a weight held on an inclined plane by a force, then the same weight hanging from two cords. Following a route close to that of Simon Stevin, and reducing the inclined plane to a balance in the manner of Galileo, Roberval obtained what is now the parallelogram law for forces. He also gave a clear account of the pressure that a body exerts on the plane supporting it.
A weight \(W\) rests on a smooth plane inclined at angle \(\alpha\). A force \(F\) along the plane holds it in equilibrium if and only if \(F=W\sin\alpha\). The plane then exerts a normal reaction \(N=W\cos\alpha\).
The balance
On 21 August 1669 Roberval presented to the Académie the plan of a balance whose pans sit above the beam. Two parallel horizontal beams, pivoted at their middles on a central post, are joined at each end by a vertical rod that carries a pan. The four bars form a parallelogram on each side, so the rods stay vertical and the pans stay level as the beams tip.
In an ideal frictionless Roberval balance, the condition for equilibrium is equality of the two loads, independent of where each load sits on its pan.
Virtual-work proof. Tip the beams through a small angle \(\delta\theta\). Each vertical rod is translated without rotation, so every point of the left pan, wherever it is, rises by the same \(\ell\,\delta\theta\), and every point of the right pan falls by \(\ell\,\delta\theta\), where \(\ell\) is the half-length of the beams. The work done by gravity is \((W_2-W_1)\ell\,\delta\theta\), which vanishes for all \(\delta\theta\) exactly when \(W_1=W_2\). Hara notes that Roberval’s own analysis contains this idea of “moment” as weight times displacement, which is an early form of the principle of virtual work.
Exercise 8.1A weight \(W\) on a smooth plane inclined at \(\alpha\) is held by a force \(F\) making angle \(\beta\) with the plane (pulling up and away). Find \(F\) and the normal reaction.
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Exercise 8.2A load \(W\) sits on the left pan of a Roberval balance at horizontal distance \(d\) outside the vertical rod. The upper and lower beams are a vertical distance \(h\) apart. Find the horizontal forces that the rod exerts on the two beams, and explain why they do not affect the balance.
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The vacuum, the pendulum and the system of the world
For the general reader. In the 1640s Roberval took part in three debates in physics. He helped interpret the new mercury tube of Torricelli, with an experiment using a fish’s bladder. He argued with Descartes about where the “centre of oscillation” of a swinging body lies. And under the name of an ancient Greek astronomer he published a sun-centred universe held together by mutual attraction.
The vacuum
Torricelli’s experiment of 1644 left an apparently empty space above a column of mercury. In Rouen and Paris, Blaise Pascal repeated it with tubes of many shapes. Roberval wrote two Latin reports, or Narrationes. The first, dated 20 September 1647, described Pascal’s experiments and his own, and concluded that the space, if not empty, did not contain the substances the philosophers proposed. The second, from 1648, explained the suspension of the mercury by the pressure of the outside air. He placed a carp’s bladder, almost emptied of air and tied off, in the space at the top of the tube, and it swelled: the little air shut inside it expanded, so the space around it pressed on it hardly at all. His apparatus was the prototype of Pascal’s experiment of the “vacuum within a vacuum”. He avoided declaring whether a true vacuum exists, but he rejected Descartes’s identification of space with matter.
A column of liquid of density \(\rho\) and height \(h\), closed at the top by a near-vacuum, is held by atmospheric pressure \(p_0\) when \(p_0=\rho g h\). For mercury (\(\rho\approx13\,600\ \mathrm{kg/m^3}\)) at sea level (\(p_0\approx101\ \mathrm{kPa}\)), \(h\approx0.76\ \mathrm m\).
The compound pendulum
Mersenne asked for the length of a simple pendulum that swings in step with a given rigid body, the body’s centre of oscillation. In 1646 Descartes and Roberval argued over it. Descartes had the better idea of what the centre of oscillation is but neglected the directions of the forces on the parts of the body; Roberval corrected this but doubted that the point he found was exactly right. Christiaan Huygens settled the problem in 1673.
A rigid body of mass \(m\), swinging about a horizontal axis at distance \(d\) from its centre of mass, with moment of inertia \(I\) about the axis, oscillates for small amplitudes like a simple pendulum of length \(L=I/(md)\).
Aristarchus
In 1644 Roberval published Aristarchi Samii de mundi systemate, presented as a work of the Greek astronomer Aristarchus of Samos, recovered from an Arabic manuscript, with notes. The text was his own. It described a heliocentric system and grounded it on a universal attraction between all parts of matter, an idea he drew from Kepler. In the dedication he wrote that perhaps all three systems, those of Ptolemy, Tycho and Aristarchus, were false and the true one unknown, though that of Aristarchus seemed the simplest. The disguise gave cover in the years after Galileo’s condemnation.
Exercise 9.1Water has density \(1000\ \mathrm{kg/m^3}\). How tall would a water barometer be? A bladder holding \(0.1\ \mathrm{cm^3}\) of air at atmospheric pressure (760 mmHg) is carried into a space where the pressure is 10 mmHg. Assuming Boyle’s law (stated two decades later), what volume would the air occupy?
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Exercise 9.2Find the centre of oscillation of a uniform rod of length \(\ell\) hung from one end, and of a uniform disk of radius \(R\) hung from a point of its rim, swinging in its own plane.
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Trochoids and the solid about the axis
For the general reader. A point on a wheel’s spoke, inside the rim, traces a wavy curve that never touches the ground; a point on a train wheel’s flange, outside the rim, traces loops. Roberval called all these curves trochoids and extended his cycloid results to them. Between May 1644 and October 1645 he also solved a harder problem: the volume of the solid made by spinning the cycloid arch about its axis of symmetry, the vertical line through its top. He set his results down in a piece titled Ad trochoidem, ejusque solida.
- Parametrize the general trochoid and find its area by the companion-curve method.
- Compute the volume of the cycloidal solid about the axis by shells.
The general trochoid
Let a circle of radius \(a\) roll on the \(x\)-axis and follow a point at distance \(b\) from its centre. The point traces
\[ P(t)=\big(at-b\sin t,\ a-b\cos t\big). \]For \(b<a\) the curve is a curtate trochoid, for \(b=a\) the cycloid, for \(b>a\) a prolate trochoid with loops. The companion point \(Q(t)=(at,\ a-b\cos t)\) lies at the same height, and the horizontal gap \(|QP|=b\sin t\) is the half-chord at that height of the circle of radius \(b\) described by the point about the centre.
For \(b\le a\), the region between one arch of the trochoid (\(0\le t\le2\pi\)) and the line \(y=a-b\) through its lowest points has area \(2\pi ab+\pi b^2\).
Proof by the companion curve. Between the trochoid and its companion, the horizontal segments are the half-chords of a circle of radius \(b\), so over the two halves of the arch the regions between them add up to a full disk, \(\pi b^2\). Measured from the line \(y=a-b\), the companion \(y=b(1-\cos(x/a))\) has area \(2\pi ab\) under it on \([0,2\pi a]\), half of the rectangle \(2\pi a\times2b\) by central symmetry. With \(b=a\) the total is \(3\pi a^2\), as in Lesson 05.
The solid about the axis
The region under one arch of the cycloid of radius \(a\), rotated about the vertical line through the top of the arch, generates a solid of volume
\[ V=\pi a^3\Big(\frac{3\pi^2}{2}-\frac83\Big). \]Derivation by shells. The half-arch \(0\le t\le\pi\) sweeps the whole solid. A vertical strip at abscissa \(x\), of height \(y\), lies at distance \(\pi a-x\) from the axis and sweeps a cylindrical shell of volume \(2\pi(\pi a-x)\,y\,dx\). With \(x=a(t-\sin t)\), \(y=a(1-\cos t)\), \(dx=a(1-\cos t)\,dt\):
\[ V=2\pi a^3\int_0^{\pi}\big(\pi-t+\sin t\big)(1-\cos t)^2\,dt . \]Expand \((1-\cos t)^2=\tfrac32-2\cos t+\tfrac12\cos2t\) and use \(\int_0^\pi(\pi-t)\,dt=\tfrac{\pi^2}{2}\), \(\int_0^\pi(\pi-t)\cos t\,dt=2\), \(\int_0^\pi(\pi-t)\cos2t\,dt=0\), and \(\int_0^\pi\sin t(1-\cos t)^2dt=\tfrac83\). The first part gives \(\tfrac32\cdot\tfrac{\pi^2}2-2\cdot2=\tfrac{3\pi^2}4-4\). So \(V=2\pi a^3\big(\tfrac{3\pi^2}{4}-4+\tfrac83\big)=\pi a^3\big(\tfrac{3\pi^2}{2}-\tfrac83\big)\approx38.13\,a^3\).
Exercise 10.1Verify the integral \(\int_0^\pi(\pi-t)\cos t\,dt=2\) and \(\int_0^\pi\sin t\,(1-\cos t)^2\,dt=\tfrac83\).
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Exercise 10.2Check the trochoid area formula directly from \(\int(y-(a-b))\,dx\) for \(b\le a\).
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Exercise 10.3For which \(b\) does the trochoid have self-intersections, and why does the area formula then need care?
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Quarrels: Descartes, Torricelli, Pascal
Descartes
Descartes was Roberval’s “irreconcilable adversary”, in Hara’s phrase. In 1637–38, after the publication of the Discours de la méthode with the Géométrie and the Dioptrique, Roberval sided with Fermat and Jean de Beaugrand against Descartes in a long dispute, conducted through Mersenne, over methods for tangents and maxima. Descartes challenged Roberval to find the tangent to the cycloid; Roberval did not produce it at once, while Fermat and Descartes did. The curve now called the folium of Descartes entered the same exchanges; Roberval thought it formed four petals and called it the fleur de jasmin, but only one loop is real. They quarrelled again in 1646 over the compound pendulum, and in 1647 over the vacuum. Jacqueline Pascal described in a letter of 25 September 1647 a meeting at which Descartes visited her brother and Roberval defended Blaise’s experiments against him with some heat.
Torricelli
In October 1643 Torricelli wrote to Roberval about the cycloid. In 1644 his Opera geometrica published the area of the cycloid and its tangent. Roberval, who had the area by 1636 but had not published it, accused Torricelli of taking the results from papers sent from Paris to Galileo’s circle. Modern historians generally find that Torricelli worked independently. The dispute was unresolved at Torricelli’s death in 1647. Their correspondence also touched mechanics: in a 1647 letter Roberval outlined a treatise in eight books on forces, the balance, cords, machines, compound motions and the centre of percussion, which was never published.
Pascal
Roberval had known Blaise Pascal since Blaise was a boy in Mersenne’s circle and supported him in the vacuum controversy. In 1658 Pascal set his challenge problems on the cycloid under the name Amos Dettonville, and his Histoire de la roulette (10 October 1658) took Roberval’s side against Torricelli on the priority for the area. Roberval was among the judges who awarded no prize. Pascal also showed in 1659 that arcs of a generalized cycloid equal arcs of an ellipse, a result Roberval may have known before August 1648.
Exercise 11.1The folium of Descartes is \(x^3+y^3=3axy\). Parametrize it by \(y=tx\), find the tangent at the point where \(t=1\), and explain why Roberval’s four petals are impossible.
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DiscussionRoberval protected his chair by keeping methods secret and lost priority for most of his discoveries. Torricelli and Cavalieri published and gained it. Was Roberval’s choice rational under the rules of the Ramus chair? What does the episode suggest about how institutions shape credit in science?
The Académie, death and legacy
On 22 December 1666 the Académie royale des sciences, founded under Jean-Baptiste Colbert, held its first meeting in the king’s library. Roberval was one of the founding mathematicians, with Christiaan Huygens, Pierre de Carcavi, Adrien Auzout, Bernard Frénicle de Bessy and Jacques Buot. He received a royal pension and was active in the Académie’s early years, presenting his balance there in 1669 and drafting around the same time a plan for a book of mechanics founded on composed motions. He also contributed on optics to Mersenne’s posthumous L’optique et la catoptrique (1651), with a study of the oval curves of Descartes’s Dioptrique.
He died in Paris on 27 October 1675, aged 73, still holding both chairs at the Collège Royal. A considerable sum of money was found in his rooms. He left his papers to the Académie, which published a selection in 1693 as Divers ouvrages de M. de Roberval: the treatises on indivisibles, on the composition of motions and tangents, on the trochoid and on equations, and letters.
Legacy
- Calculus before calculus. The quadrature of \(y=x^n\) and of the sine, the cycloid’s area and solids, and the comparison of arc lengths place him with Cavalieri, Fermat and Torricelli among the immediate precursors of Newton and Leibniz.
- Kinematic geometry. Tangents by composition of motions anticipate Newton’s fluxions in spirit and are standard today as the rule that the velocity of a sum is the sum of the velocities.
- The sine curve. His companion curve of the cycloid is the first appearance of the graph of the sine.
- Mechanics. The parallelogram of forces in 1636 and the balance of 1669, with its early form of the principle of virtual work.
- Physics. The interpretation of the barometer by air pressure, and a universal attraction stated four decades before the Principia.
- Names. The Roberval balance, the lunar crater Roberval, and the town of Roberval in Quebec, named after a different man (Jean-François de La Rocque de Roberval).
Final projectWrite a two-page history of the cycloid from Galileo to Johann Bernoulli’s brachistochrone. Include Roberval’s area proof in full, a tangent construction, the arc length, and a paragraph on the priority disputes.
Sources and further reading
Biography
- J. J. O’Connor and E. F. Robertson, Gilles Personne de Roberval, MacTutor History of Mathematics.
- K. Hara, Roberval, Gilles Personne (or Personier) de, Dictionary of Scientific Biography.
- Gilles de Roberval and Gilles Personne de Roberval, Wikipedia (English and French).
- Linda Hall Library, Scientist of the Day: Gilles Personne de Roberval.
- L. Auger, Un savant méconnu : Gilles Personne de Roberval (1602–1675), Blanchard, Paris, 1962.
- D. J. Sturdy, Science and Social Status: The Members of the Académie des Sciences 1666–1750, Boydell, 1995.
- Museo Galileo, Roberval and the vacuum experiments.
Mathematics and mechanics
- Divers ouvrages de M. de Roberval, Académie royale des sciences, Paris, 1693.
- G. P. de Roberval, Aristarchi Samii de mundi systemate, Paris, 1644.
- MacTutor, Cycloid and Folium of Descartes.
- B. Hughes, The companion curves of Gilles Personne de Roberval (2002).
- P. R. Wolfson, The crooked made straight: Roberval and Newton on tangents, American Mathematical Monthly 108 (2001).
- J. Itard, La lettre de Torricelli à Roberval d’octobre 1643, Revue d’histoire des sciences (1975).
- V. Jullien, Descartes–Roberval, une relation tumultueuse (1998).
- Roberval balance, Wikipedia.
- E. A. Whitman, Some historical notes on the cycloid, American Mathematical Monthly 50 (1943).