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A Translation of Evangelista Torricelli’s Quadratura Parabolae per
novam indivisibilium Geometriam pluribus modis absoluta.

Introduction

One of the most important threads in seventeenth century mathematics
was the introduction of infinitesimal methods for solving area and
volume problems. Foremost among these newfound techniques was
Cavalieri’s Principle, a creation
of Bonaventura
Cavalieri (1598-1647) that used Eudoxus’ theory of proportions (as
found
in Book
V
and Book
VI of Euclid’s Elements) to study infinitesimal slices of
objects and derive a proportion between the objects
themselves. Ironically, when mathematicians in the seventeenth century
wanted to understand Cavalieri’s method, often they didn’t turn to the
work of Cavalieri himself. Instead, they turned to the work of his
contemporary
Evangelista
Torricelli (1608-1647). Torricelli’s sole published work,
his Opera
Geometrica (1644), was a lengthy and wide-ranging tome, but
one of the most important sections for understanding Cavalieri’s
method was de Dimensione Parabolae.

De Dimensione Parabolae consisted of two parts–both of which
were devoted to alternate proofs of the classical result from
Quadrature
of the Parabola by
Archimedes
of Syracuse (287-212 BCE). The first part of de Dimensione
Parabolae was Quadratura Parabolae Pluribus modis per
duplicem positionem, more antiquorum, absolutae (“The Quadrature
of the Parabola solved with many methods through a two-fold placing in
the manner of the ancients”). As the name implies, this section was
devoted to finding the area of a segment of the parabola using
Archimedean techniques–i.e., the method of exhaustion. However, in
the second section, Quadratura Parabolae per novam indivisibilium
Geometriam pluribus modis absoluta (“The Quadrature of the
Parabola solved with many methods through the new geometry of
indivisibles”), Torricelli presented 11 Propositions which gave unique
and wide-ranging approaches to solving the quadrature of the parabola,
all of which involve Cavalieri’s method or similar infinitesimal
methods. As Torricelli himself noted, in the process he demonstrated
that

it is certain that this wonderful geometry is a shortcut for
invention, and that it confirms countless almost inscrutable theorems
with brief, direct, and affirming demonstrations, which is certainly
not able to be done easily through the ancient teaching. To be sure,
this is truly the Royal Road in the mathematical thorn hedges, that
Cavalieri, creator of these wonderful inventions, first among everyone
opened up and made public for the common good [Torricelli, p. 56].

The purpose of this paper is to give a translation of
Torricelli’s Quadratura Parabolae per novam indivisibilium
Geometriam pluribus modis absoluta. To put the work in context,
we will first look briefly at the 17th century mathematical figures
who provided a milieu for its development. Then, following an
overview of the results in Quadratura Parabolae per novam
indivisibilium Geometriam pluribus modis absoluta, we will
present a slightly modernized introduction to
Archimedes’ Quadrature of the Parabola–the work that was the
focus of Torricelli’s de Dimensione. With an eye toward
understanding how Torricelli’s work might be used in a classroom
today, we also look closely at an example from the translation which,
while not strictly an application of Cavalieri’s method, demonstrates
just how conceptually different these new infinitesimal methods were
from the traditional approach presented in Quadrature of the
Parabola that had been the gold standard for mathematical rigor
for centuries.

Some historical background

It’s impossible to talk about Torricelli without introducing some
other figures who were undoubtedly his most important mathematical
influences: Galileo
Galilei, Bonaventura
Cavalieri, and the relatively unknown figure
of Benedetto
Castelli. Galileo (1563-1642) was, of course, the famous Italian
scientist whose trouble with the Roman Inquisition is discussed even
to this day. Cavalieri (1598-1647) was the 17th century mathematician
whose work formalizing indivisibles would be the basis for
Torricelli’s work in de Dimensione Parabolae. However, it
was Castelli (1578-1643) who would undoubtedly have the most important
social and mathematical impact on Torricelli’s life.

Benedetto Castelli

Born Antonio Castelli to relatively wealthy parents in Brescia,
Castelli took the name Benedetto when he entered the Benedictine
monastic order in 1595. Eventually, he was transferred to a monastery
in Padua, where Galileo had been teaching since 1592. Castelli became
Galileo’s student, and eventually the two became close friends, as
evidenced by their frequent correspondence after Castelli was
transferred to another monastery near Naples in 1607. In 1610, when
the publication of Sidereus Nuncius enabled Galileo to obtain
a position as Chief Mathematician to the Medicis with a nominal
position in mathematics at the University of Pisa, Castelli quickly
obtained a transfer to a Benedictine Abbey in Florence. In 1613, with
Galileo’s help, Castelli too obtained a professorship in mathematics
at the University of Pisa, where he remained until 1626, when he took
a position teaching mathematics and solving problems in hydraulics for
the Pope in Rome. Significantly, there was no Benedictine monastery
in Pisa, so Castelli resided in a Jesuati monastery which from 1616 to
1620 was also the home of a young Jesuati named Bonaventura Cavalieri.
Castelli was widely regarded as an outstanding teacher, and it was
from Castelli that Cavalieri learned much of his mathematics.
Castelli also introduced Cavalieri to Galileo, with whom Cavalieri
exchanged more than 100 letters.

One of the interests these men shared was a mathematical technique
called the method of indivisibles, which determined ratios between
plane or solid figures by dividing them up into parallel, indivisible
slices and studying the ratios of these slices. They weren’t the
first to stumble on this approach. (Unbeknownst to seventeenth century
mathematicians, Archimedes had discovered a similar technique in his
lost work on The
Method of Mechanical
Theorems. Johannes
Kepler (1571-1630), who discovered the laws of planetary motion
and was a Galileo correspondent, had also utilized similar ideas in
his Nova stereometria doliorium vinariorum.) But they were
probably the ones most responsible for promoting it. In
his Dialogues Concerning Two New Sciences (1638) Galileo
demonstrated the equality between the volume of a “soup
bowl” AFBED of revolution to a cone DCE of the same
height by demonstrating that on each planar slice perpendicular to the
shared axis CF of the figures, the annulus on ON had
the same area as the circle on radius PL.

(See [Baron, p. 119] and [Katz, p. 515].)
Cavalieri’s Geometria
indivisibilibus continuorum nova quadam ratione promota was
intended to be the most thorough exposition of the theory of
indivisibles that was available at the time. However, as Andersen
notes, it was “so difficult that [nineteenth century mathematical
historian] Maximilien Marie suggested that if a prize existed for the
most unreadable book, it should be awarded to Cavalieri
for Geometria” [Andersen, p. 294].

Evangelista Torricelli

Torricelli was a latecomer to this group of mathematicians. Born in
1608 in Faenza to a family with limited means, his education was
overseen by his uncle Jacopo, a Benedictine monk. He entered a Jesuit
college in 1624 and, when Castelli moved to Rome in 1626, Torricelli
spent the next six years as both Castelli’s student and his secretary.
From 1632 to 1641, Torricelli was secretary to Giovanni Ciampoli,
another close friend of Galileo, and in 1641, three month’s before
Galileo’s death, Torricelli joined Galileo as an assistant in Arcetri
(across the Arno from Florence), where Galileo had been living under
house arrest since his run-in with the Inquisition in 1633. After
Galileo’s death, Torricelli would go on to hold his position as court
mathematician to the Medicis. Though he is now most famous for his
discovery of barometric pressure, his only published work,
the Opera Geometrica of 1644, which contained de
Dimensione Parabolae and several other works, was very well
regarded at the time. In particular, because Cavalieri’s work on
indivisibles was so difficult to understand, Torricelli’s Opera
Geometrica, and in particular Quadratura Parabolae per novam
indivisibilium Geometriam pluribus modis absoluta, was often the
source from which mathematicians learned about the method of
indivisibles–hence its importance in the history of mathematics.

Overview of the Translation

As noted above, the Quadratura Parabolae per novam indivisibilium
Geometriam pluribus modis absoluta is the second half of
de Dimensione Parabolae. The original Latin version (see
here
or here)
runs for 29 pages and consists of an introduction and Propositions 11
through 21 of de Dimensione Parabolae, interspersed with
an additional 14 lemmas and several scholia and “Aliter” sections.
Torricelli notes his focus at the outset of this section:

Until now, the matter about the measurement of the parabola has
been related in the manner of the ancients. It remains that we should
approach the same measurement of the parabola with a certain new but
marvelous system–namely, by the aid of the Geometry of Indivisibles
and with diverse methods in this manner.

There is no formal subdivision of the section, but the “diverse
methods” that Torricelli cites can be roughly subdivided as follows:

Propositions 11-13: Solving the quadrature of the parabola
through proportions from classical geometry. For Proposion 11,
Torricelli uses Euclid’s proportion
(Euclid XII.10)
between the cone and the cylinder. For Proposition 12, he uses
Euclid’s result
(Euclid XII.2)
that two circles are in the same ratio as the squares of the diameters
(see Lemma 20 of Torricelli). For Proposition 13, he uses Archimedes’
proportion (see [Heath, p. 43]) between the sphere and the cylinder.

Proposition 14: Solving the quadrature of the parabola
using Archimedes’ Equilibrium of Planes I, which determines the
center of gravity of a triangle (See [Heath, p. 201]). He also makes
use of a more recent result on the location of the center of
equilibrium of a cone, which had been established
by Federico
Commandino (1506-1575) in his Liber de Centro Gravitatis
Solidorum (1565).

Propositions 15-16: Solving the quadrature of the parabola
by several approaches which generalize Archimedes’ geometric summation
method found in Quadrature of the Parabola to an infinite
geometric series. This will be discussed in the section below.

Proposition 17: Solving the quadrature of the parabola using
Archimedes’ proportion between the first turn of a spiral and the
circumscribing cylinder. Logically, this belongs with
Propositions 11-13, but it requires a more general version of
Cavalieri’s Principle that Torricelli develops in Lemma 29.

Proposition 18-21: Solving the quadrature of the parabola
using centers of gravity. Determining centers of gravity was a
major area of research in the sixteenth and seventeenth centuries, and
there were several major results extending the results of Archimedes
(in Plane Equilibrium I and II), who determined the center of
gravity of the triangle and the parabola. Proposition 18 finds the
quadrature of the parabola with the center of gravity of a triangle
and Proposition 19 finds the quadrature with the center of gravity of
a parabola. It makes use of a very brief proof (Lemma 30)
which shows how infinitesimals can demonstrate quite easily
Archimedes’ result on the center of gravity of a parabola found
in On the Equilibrium of Planes II. Lemma 31 is the first
published proof of the Pappus-Guldin Theorem, and it is used in
Proposition 20 to establish a proof of the quadrature of the parabola
using the Pappus-Guldin Theorem. Proposition 21 presents a quadrature
of the parabola very similar in spirit to Proposition 1 of Archimedes’
On The Method (which was unknown at the time), and
Proposition 22 presents a quadrature proof using the center of gravity
of a hemisphere,
which Luca
Valerio (1552-1618) had established in his work de Centro
Gravitatis Solidorum libri tres (1603).

Some Prerequisites for Reading the Translation

De Dimensione Parabolae builds on the geometrical tradition
that mathematicians in the seventeenth century inherited from Euclid
and Archimedes. The primary goal of the work is to establish
equivalences between many classical results in geometry and
Archimedes’ well-known result from the Quadrature of the
Parabola, so it’s necessary to have some familiarity with these
results. These, in turn, rely on results
in Book
V
and
Book VI of Euclid’s Elements. However, our translation
begins in the middle of a larger work, and Torricelli sometimes refers
to earlier results in this work. For the sake of completeness, we
summarize these here. More details on the mathematical background of
this work can be found in [Leahy, pp. 175-177].

Lemma 3: If a parabola were to have three tangents, two at the
base and a third through the vertex, the triangle comprised by the
tangents will be eight times the triangle which arises from a fourth
tangent having been drawn through the vertex of either semiparabola.

Lemma 3 is referenced in Proposition 15, where Torricelli generalizes
Archimedes’ result expressing the quadrature of the parabola in terms
of a triangle inscribed inside of a parabola to the triangle
comprised of the three given tangent lines to the parabola.

Lemma 7: If a triangle having the same base and the same
altitude as the parabola is inscribed in a parabola, two other
triangles are also inscribed in like manner in the remaining portions.
The triangle inscribed first will be eight times either of the
triangles inscribed afterwards.

As Torricelli notes, “This Lemma is demonstrated by Archimedes in
Proposition 21 of Quadrature of the Parabola”. It will also
be demonstrated by us below.

Lemma 11: An entire semiparabola hangs at equilibrium from a
point on the base in which it is thus divided so that the part
terminated at the curve is to the remaining part as five is to three.

Lemma 11 generalizes Archimedes’ result on the center of gravity of a
segment of the parabola found in Book II of On the Equilibrium of
Planes.

Lemma 18: If a first magnitude were to a second thus as a third
is to a fourth, and thus however many times it is pleasing, and if all
of the firsts and also all of the thirds are equal among themselves,
then all of the first magnitudes together are to all of seconds
together, as all of the thirds together are to all of the fourths.

Lemma 18 is Torricelli’s version of Cavalieri’s Principle. It is
generalized in Lemma 29, which is stated and proved in the translation
below.

Torricelli also cites, e.g., “the explanation of Proposition 9” at
several points, referring to the proof of this Proposition. This proof
establishes a result (also established in Propositions 16 and 17 of
Archimedes’ Quadrature of the Parabola [Heath, pp. 244-246])
which is equivalent to Archimedes’ result on the area of a parabola:
Construct the “tangent triangle” to the segment of the parabola,
consisting of the tangent line \(DQ\) at one end of the base \(Qq\) of
the segment of the parabola, the base itself, and a line \(Dq\)
parallel to the diameter \(PV\) through the other endpoint \(q\) of
the base.

Then \( {1 \over 4} \Delta DqQ = \Delta QPq\). Why? Proposition I.35
of the Conics of Apollonius of Perga (262-190 BCE)
[Apollonius, p. 62] shows that \(TP = PV \). Thus, triangles \(
\Delta TQV\) and \( \Delta QPq\) have the same area. A similarity
argument shows that \( \Delta DqQ\) has four times the area of \(
\Delta TVQ\). Thus, \( {1 \over 4} \Delta DqQ = \Delta QPq\), and so
to show that \( area(par(QPq)) = {4 \over 3} \Delta QPq\) it is
necessary and sufficient to show that \( area(par(QPq)) = {1 \over 3}
\Delta QDq\).

Archimedes’ Quadrature of the Parabola

Torricelli’s de Dimensione establishes the amazing result
that Archimedes’ computation of the area of a segment of a parabola is
logically equivalent to most of the major geometrical results known at
the time. Given its centrality to Torricelli’s work, it’s useful to
review in modern terms the key result in
Archimedes’ Quadrature
of the Parabola, which determines the area of a segment of a
parabola in terms of an inscribed triangle:

Proposition 24 of QP: Every segment bounded by a
parabola and a chord \(Qq\) is equal to four-thirds of the triangle
which has the same base as the segment and equal height [Heath, p. 251].

Archimedes’ proof of this result (see also, e.g., [Katz, pp. 108-109]
or [Boyer, pp. 52-53]) is an application of the method of exhaustion,
which was the traditional approach among Greek mathematicians for
solving area and volume problems. The proof also hinges on what is
essentially a summation formula for a finite geometric series and an
ingenious observation about how triangles can be inscribed inside a
segment of a parabola. This result about inscribed triangles in turn
depends on two more fundamental properties about the nature of the
parabola which can be found in [Heath, p. 248]:

Given any chord \(Qq\) in a parabola and the point \(P\) on the
parabola where the tangent at \(P\) is parallel to \(Qq\), the line \(PV\)
parallel to the axis of the parabola bisects \(Qq\).

Given \(Qq\) and \(PV\) as above, if the points \(W\) on \(PV\) and \(R\) on
the parabola are such that \(RW\) is parallel to \(Qq\), then
\[
{PV \over PW} = {QV^2 \over RW^2}.
\]

With these results in hand, Archimedes constructs a sequence of
successively smaller and non-overlapping triangles inscribed inside of
the segment of the parabola. Starting with the original triangle
\(\Delta PQq\) from the proposition, another triangle \(\Delta PRQ\)
with one-eighth the size of \(\Delta PQq\) and inscribed in the
parabola along the side of \(PQ\) of \(\Delta PQq\) is constructed as
follows: With \(Qq\) bisected at \(V,\) consider the point \(M\)
bisecting \(QV\) and the point \(R\) on the parabola such that \(RM\)
is parallel to \(PV\) (and hence parallel to the axis of the parabola
itself by definition of \(PV\)). Label the intersection of \(RM\) and
\(PQ\) by \(Y\) and let \(RW\) be parallel to \(Qq\).

It follows that \(RM\) is 3/4 of \(PV\). Why? From property (2) above,
\[
{PV \over PW} = {QV^2 \over RW^2} = {(2 MV)^2 \over RW^2} = 4 {RW^2
\over RW^2} = 4,
\]

because \(RMVW\) is a parallelogram (and thus \(MW = RV\)). This
implies that \( {1 \over 4} PV = PW\) and so \( RM = WV = PV – PW = {3
\over 4} PV\). Now note that the triangles \(\Delta QYM\) and \(\Delta
QPV\) are similar, because \( RM\) and \( PV\) are parallel. Since \(
QM = {1 \over 2} QV\), this similarity implies that \( YM = {1 \over
2} PV\) as well. Hence \(YR = RM – YM = {1 \over 4} PV\) and so \(YM
= 2 YR\). Thus, \(\Delta PRQ\) and \(\Delta PMQ\) have the same base
(\( PQ\)), but the height of \(\Delta PMQ\) is twice the height of
\(\Delta PRQ\), and it follows that in terms of area \(\Delta PMQ = 2
\Delta PRQ\).

Likewise, \(\Delta PQM\) and \(\Delta PQV\) have the same height but
\(\Delta PQV\) has twice the base of \(\Delta PQM\), so we also have
\(\Delta PQV = 2 \Delta PQM\) in terms of area. A similar argument
shows that \(\Delta PQq = 2 \Delta PQV\). It follows that \( \Delta
PRQ = {1 \over 8} \Delta QPq\), because

\[ 8 \Delta PRQ = 4 \Delta QPM = 2 \Delta PQV = \Delta QPq. \]

A triangle \(\Delta Pqr\) with the same ratio to \(\Delta PQq\) can
likewise be constructed on the side \(Pq\), so altogether the original
triangle has four times the total area of the two newly-constructed
triangles.

Also, note the geometrically obvious fact that the area of the three
triangles together make an even better approximation to the area of
the segment of the parabola (which we will denote by \( par(PQq) \) in
what follows) than the first triangle alone.

But why stop there? Each line \(PQ\) and \(Pq\) is itself the base of
a segment of the parabola and \(rm\) and \(RM\) are parallel to the
axis of the parabola (just like \(PV\)), so the same construction can
be applied to the sides of \(\Delta PRQ\) and \(\Delta Prq\),
resulting in four new triangles with combined area \((1/4)^2 \Delta
PQq\). Obviously, the sum of these seven triangular areas is an even
better approximation to the area of the segment of the parabola. And
so on. As Archimedes will show eventually (see the discussion of the
method of exhaustion below), the difference between the sums of these
triangular areas and the area of the sector of the parabola is going
to zero as \(n\) gets larger. At the \(n\)th step of the
approximation another \( (1/4)^n \Delta PQq\) is added to the
approximation, so the total area of the approximation after \(n\)
stages is the finite geometric series
\[
\Delta PQq + (1/4) \Delta PQq + (1/4)^2 \Delta PQq + \cdots
+ (1/4)^n \Delta PQq. \]

The next observation in Archimedes’ proof is essentially (from our
perspective today) an algebraic identity that was designed to
understand how this finite geometric series determined by the
triangular areas grows as more triangles are added. Note first that
for any quantity \(A\), simple arithmetic shows that

\[ {1 \over
4} A + {1 \over 3}\cdot {1 \over 4} A= {1 \over 4} A + {1 \over 12} A
= {1 \over 3} A. \]

Next, consider a finite sequence \(A_1, A_2, \ldots, A_n\) such that
\(A_i = {1 \over 4} A_{i-1}\). Since \({1 \over 4} A_i +{1 \over 12}
A_i = {1 \over 3} A_i\) for each \(i\),

\[
{1 \over
4} A_1 + \cdots + {1 \over 4} A_n + {1 \over 12} A_1 + \cdots + {
1\over 12} A_n = {1 \over 3} A_1 + \cdots + {1 \over 3} A_n.
\]

But since \(4 A_i = A_{i-1}\) for \(i = 2, 3, \ldots n\), we have \({1
\over 12} A_{i-1} = {1 \over 3} A_i\) and this equation can be
rewritten as

\[
{1 \over 4} A_1 + \cdots + {1 \over 4} A_n + {1
\over 3} A_2 + \cdots + { 1\over 3} A_{n+1} = {1 \over 3} A_1 + \cdots
+ {1 \over 3} A_n.
\]

By canceling, this becomes
\[
{1 \over 4} A_1 +\cdots + {1 \over 4} A_n + { 1\over 3} A_{n+1} = {1
\over 3} A_1.
\]

Adding \(A_1\) to both sides, we have Archimedes’ geometric summation
formula:
\[
A_1 + {1 \over 4} A_1 +
\cdots + {1 \over 4} A_n + { 1\over 3} A_{n+1} = {4 \over 3} A_1.
\]

If \(A_1 = area(\Delta PQq)\), then \((1/4) A_1\) is the area of the
triangles added at the second step, \((1/4)A_2\) is the area of the
triangles added at the third step, and so on. So from a modern
perspective, the identity
\[
A_1 + {1 \over 4} A_1 + \cdots + {1 \over 4} A_n + { 1\over 3} A_{n+1} =
{4 \over 3} A_1
\]

makes Archimedes’ area formula geometrically obvious. The right-hand
side never changes as \(n \rightarrow \infty\). On the left-hand
side, since \(A_{n+1} = (1/4)^n \Delta PQq\), \({1 \over 3} A_{n+1}
\rightarrow 0\) as \(n \rightarrow \infty\). The remaining terms are
the sum of the inscribed, non-overlapping triangular areas, which
converge to the area of the parabola. Thus, Archimedes’ area formula
\( par(PQq) = {4 \over 3} \Delta PQq\) follows.

However, Archimedes would never take that step. Greek mathematicians,
perhaps stung by the criticisms of Zeno of Elea (490-430BCE) in his
paradoxes (or perhaps not–see [Baron, p. 22-25]), did not use
infinite processes in formal proofs and never developed the concept of
the limit. Instead, Archimedes employed the method of exhaustion, a
technique found first in Antiphon, later perfected by Eudoxus, and
adopted by Archimedes and others [Boyer, p. 32]. This was the
standard method of solving area and volume problems for Greek
mathematicians and, unlike our modern approach, doesn’t require
passage to the limit at any step. But it pays for this conceptual
clarity with a lengthy double reductio ad absurdum proof, and
that’s exactly the sort of proof that Archimedes submits in
the Quadrature of the Parabola.

In particular, to show that the area equation \(par(QPq) = {4 \over 3}
\Delta QPq\) is true, Archimedes supposes that the equation isn’t
true. Then there are two possible relations between the areas:

\(par(QPq) > {4 \over 3} \Delta QPq\)
\(par(QPq) < {4 \over 3} \Delta QPq\).

Both of these will be shown to lead to a contradiction.

To show that \(par(QPq) > {4 \over 3} \Delta QPq\) yields a
contradiction, Archimedes notes that since each inscribed triangle is
greater than half of the parabolic segment in which it lies (see
[Heath, p. 248]), the sum of the triangular areas \(A_1 + {1 \over 4}
A_1 + \cdots + {1 \over 4} A_n\) can be made as close to \(par(QPq)\)
as we want for large enough \(n\). In particular, since \({4 \over 3}
\Delta QPq < par(PQq)\) by assumption, we must also have

\[
{4 \over 3} \Delta QPq < A_1 + {1 \over 4} A_1 +
\cdots + {1 \over 4} A_n
\]

for some finite number \(n\). However,
\[
A_1 + {1 \over 4} A_1 + \cdots + {1 \over 4} A_n < A_1 + {1
\over 4} A_1 + \cdots + {1 \over 4} A_n + { 1\over 3} A_{n+1}
\]
and his algebraic identity for the geometrical
series shows:

\[
A_1 + {1 \over 4} A_1 + \cdots + {1 \over 4} A_n + { 1\over 3} A_{n+1}
= {4 \over 3} A_1 = {4 \over 3} \Delta PQq
\]

and so \({4 \over 3} \Delta QPq < {4 \over 3} \Delta QPq\). Hence a
contradiction.

To show that \(par(QPq) < {4 \over 3} \Delta QPq\) yields a
contradiction, note that since \(A_{n+1} = \left({1 \over 4}\right)^{n}
A_1\), it follows that \(A_{n+1} \) can be made arbitrarily small for
some finite \(n\). Since (by assumption) \(par(QPq)
< {4 \over 3} A_1\), we have
\[
A_1 + {1 \over 4} A_1 + \cdots + {1 \over 4} A_n + { 1\over 3} A_{n+1}
= {4 \over 3} A_1 > par(QPq).
\]

Because \(A_{n+1}\) can be made arbitrarily small, this shows we must
eventually have \( A_1 + {1 \over 4} A_1 + \cdots + {1 \over 4}
A_n > par(QPq)\) for some finite \(n\). Hence another
contradiction–this time of the fact that each of the
non-overlapping regions \(A_1, \ldots, A_n\) is inscribed in the
segment of the parabola \(par(QPq)\), so their sum must be smaller
than the area of \(par(QPq)\).

By trichotomy, \(par(QPq) = {4 \over 3} \Delta QPq\) is the only
possibility.

Torricelli’s Quadrature of the Parabola (One Version)

The method of exhaustion was a logically sound approach to solving
area and volume problems. However, as Boyer notes, “the
cumbersomeness of its application led later mathematicians to seek a
more direct approach to problems in which the application of some such
procedure would have been indicated” [Boyer, p. 35]. In the
seventeenth century, Cavalieri’s method was the premier example of
this “more direct approach”, and in the section of de Dimensione
Parabolae, given in translation below, Torricelli presented
eleven different examples of how to find the quadrature of the
parabola using Cavalieri’s method and other infinitesimal methods.
Modern interpretations of some of these Cavalierian approaches have
already been given in [Anderson, p. 356] and [Leahy, pp.178-183]. As a
specific example we will discuss another approach found in
Proposition 15 of the work. (See also [Baron, p. 183-185] for another
discussion of this result.)

What was noteworthy about the approach in Proposition 15 was that it
directly extended Archimedes’ own technique for finding the quadrature
of the parabola. More precisely, whereas Archimedes stopped after
finitely many terms of his geometric series and let the method of
exhaustion take care of the rest, Torricelli used a geometrical
argument to demonstrate how to add the infinitely many terms of the
same geometric series and arrive at the same area formula.

To justify this infinite summation Torricelli introduced a geometrical
figure he called a flexilineum, which was constructed as
follows: Consider two lines \(B_1D\) and \(C_1D\) that intersect at
\(D\), and a sequence of points \(B_1, B_2, \ldots\) on \(B_1D\), and
\(C_1, C_2, \ldots\) on \(C_1D\) such that for \(i = 1, 2\ldots \) the
segments \(B_iC_i\) are mutually parallel, as are the segments
\(B_{i+1}C_i\). The flexilineum consisted of the infinitely many
conjoined segments \(B_1 C_1 B_2 C_2 \ldots\)

Torricelli made several observations about the flexilineum that were
key to his proof. First, he noted that the terms of the sequence of
lengths \(B_iC_i\) are in a continuous geometric proportion. That is,
\[
{B_1 C_1 \over B_2 C_2} = {B_2 C_2 \over B_3 C_3} = {B_3 C_3 \over B_4
C_4} = \cdots
\]
To see why, note that since both families of line segments
\(B_i C_i\) and \(B_{i+1} C_i\) are parallel among themselves,
Euclid
VI.2 implies that for each \(i\) \(\Delta B_i C_i D\) is similar
to \( \Delta B_{i+1} C_{i+1} D\), and \(\Delta B_{i+1} C_i D \) is
similar to \(\Delta B_{i+2} C_{i + 1} D\). So by similar triangles
we have

\[
\eqalign{ {B_i C_i \over B_{i +1} C_{i +1}} &= { C_i D \over C_{i+1}
D} \cr
&= {B_{i+1} D \over B_{i+2} D} \cr
&= {B_{i+1} C_{i+1} \over B_{i+2} C_{i+2}} \cr}
\]

and the result follows. A similar argument shows that the line
segments \(B_{i+1} C_i\) are also in a continuous geometric proportion.

Torricelli’s second and most important result about the flexilineum was
the derivation of a geometrical representation for the sum of the
lengths of the infinitely many line segments \(B_i C_i\). To find
this sum geometrically, he extended each of the segments \(B_{i+1}
C_i\) so that they intersected the extension of \(B_1 C_1\) in a point
we will call \(D_{i+1}\). (Note: \(D_1 = B_1\) and \(D_2 = C_1\) by
definition as well.) Similarly, he let the line parallel to all the
\(B_{i+1} C_i\) and passing through \(D\) intersect the extension in
a point \(L\).

All of the lines \(B_i D_i\) are parallel to each other. By
assumption, the lines \(B_i C_i\) are parallel to each other as well.
Consequently the quadrilateral \(D_i B_i C_i D_{i+1}\) is a
parallelogram, so \(D_i D_{i+1} = B_i C_i\) and the length of the
segment \(B_1 L\) (that is, the infinite sum of the lengths of all of
the \(D_i D_{i+1}\) segments) is clearly the same as the infinite sum
of the lengths of all of the segments \(B_i C_i\).

Torricelli was so concerned to get this proof right that he spent
several pages filling in details. He noted, for instance, that even
though the proof deals with ratios of lengths, it is still valid for
ratios of any other type of magnitude [Torricelli, p. 67]. It’s also
an easy exercise to show that given any ratio \(a/b\) it’s possible to
construct a flexilineum such that \( B_i C_i / B_{i+1} C_{i+1} =
a/b\).

Returning to the original problem of the quadrature of the parabola,
let \( B_i C_i\) represent \(A_i\), the area of the triangles added at
the \(i^{th}\) stage of Archimedes’ construction. Then by Archimedes’
construction the segments \(B_i C_i\), \(i = 1,2,\ldots\), are in a
continuous geometrical proportion in a ratio of 1 to 4. That is,

\[
{1 \over 4 } = {B_1 C_1 \over B_2 C_2} = {B_2 C_2 \over B_3 C_3} = \cdots
\]

With this interpretation, Torricelli’s flexilineum showed that the sum
of the areas \(A_1, A_2, \ldots\) (i.e., the area of the segment of
the parabola) is the same as the length \(B_1 L\).

The final step in Torricelli’s proof was to compute \( B_1 L\). To
find the length of \(B_1 L\), he established a proportion between:

the first length \(B_1 C_1\) (which is the area of the first triangle
\(\Delta QPq\) in Archimedes’ construction)
the extended length \(B_1 L\) (which is the area of the segment
\(par(QPq)\) of the parabola)
the difference \(B_1 C_1 – B_2 C_2\) (which is the
difference between the first triangle and the triangles added at the
second step–that is, \( {3 \over 4} \Delta PQq\)).

To see this proportional relation, note that if \(V\) is placed on
\(B_1 L\) so that \(V B_2\) is parallel to \(C_1 D\), then \(V C_1 =
B_2 C_2\), since \(V C_1 C_2 B_2\) is a parallelogram.

Likewise, we have \(\Delta B_1 V B_2\) similar to \( \Delta B_1 C_1
D\) and \(\Delta B_1 C_1 B_2\) similar to \(\Delta B_1 L D\).
With some work, the similarity of these triangles can be used to show:
\[
{B_1 C_1 – B_2 C_2 \over B_1 C_1} = {B_1 V \over B_1 C_1}
= {B_1 B_2 \over B_1 D} = {B_1 C_1 \over B_1 L},
\]

which we can solve for
\[
B_1 L = { (B_1 C_1)^2 \over (B_1 C_1 – B_2 C_2)}.
\]

In terms of the quadrature of the parabola, this means

\[
par(PQq) = {(\Delta PQq)^2 \over {3 \over 4} \Delta PQq} =
{4 \over 3} \Delta PQq,
\]

and thus Torricelli derived Archimedes’ quadrature formula for a
segment of a parabola.

In the Classroom and Conclusion

In the Classroom

Solutions to area problems are some of the most fundamental and
easy-to-grasp results in calculus. The Fundamental Theorem of
Calculus is an amazing tool for solving these problems, but it isn’t
the only tool. Historically, the quest to solve general area problems
can be traced to Archimedes’ particular solution to the area problem
for a parabola, and there is a rich history of diverse methods for
solving this specific area problem which stretches from ancient Greece
to the modern calculus. If a detailed discussion of Archimedes’
solution to the area problem for the parabola is included in a history
class, the Quadratura Parabolae per novam indivisibilium
Geometriam pluribus modis absoluta can also provide examples of
the many different approaches to solving this particular problem, and
so is an important original source in the history of calculus. It is
also accessible to any student who is comfortable with Books V and VI
of Euclid’s Elements and has experience reading original
sources up to the level of Archimedes. Here are some specific
examples of how it might be used as an original source in an upper
level history of mathematics class:

For those students who have read Archimedes’ Quadrature of
the Parabola, in particular the proof of the area formula given
in Propositions 21-14, Torricelli’s Lemmas 24-27 and Proposition 15
present a proof identical in spirit but without resort to the method
of exhaustion. What is the purpose of Torricelli’s painstaking
attempt to justify his infinite summation in geometrical terms? How
does this contrast with the method of exhaustion that Archimedes
utilized? Does it lack anything in rigor? (See the section
“Torricelli’s Quadrature of the Parabola” above for a modern
explanation of how Torricelli was attempting to wrestle with
infinite summations.)
As an extension of the previous exercise, Torricelli’s Lemma 28
(with the accompanying “In Another Way” discussion) and Proposition
16 give a nearly identical proof of Archimedes’ result using
triangles that are tangent to the parabola instead of inscribed in
it. What would an Archimedean proof using the method of exhaustion
for this result look like?
For those students who have read the first Proposition of
Archimedes’ The Method, Torricelli’s Lemmas 32-33 and
Proposition 20 should be familiar, because (with some minor
differences) Torricelli’s proof is the same as the proof Archimedes’
gives in the method. Has Torricelli indepedently rediscovered
Archimedes’ Method more than 250 years before it finally
came to light again in 1906?
Those students who have read the proof of Proposition 8 of Book
II of Archimedes’ On the Equilibrium of Planes know that it
depends on the formula for the area of a segment of a parabola given
in Quadrature of the Parabola. Torricelli’s Proposition 30
demonstrates (using the Law of the Lever as found in Book I
of On the Equilibrium of Planes) that conversely one can
find the area formula for a segment of a parabola once one knows the
center of gravity of a parabola (a result which Torricelli proves
quite easily using infinitesimals in Lemma 30 of his work). Do you
think Archimedes was aware that the two results were equivalent?
Perhaps the most important part of Torricelli’s work is that it
articulates Cavalieri’s method of indivisibles better than Cavalieri
himself did. Moreover, whereas Cavalieri restricted himself to
comparing figures having equal altitudes, Torricelli used
indivisibles much more freely. Those students who have read about
Cavalieri’s Principle at the level of, say, [Katz, p. 514-517],
would benefit from reading Torricelli’s most general expression of
Cavalieri’s Principle as it is found in Lemma 29. How does
Torricelli’s statement of Cavalieri’s Principle compare to
Cavalieri’s own statement?
Immediately following Lemma 29, Torricelli uses this version of
Cavalieri’s Principle to show (in Proposition 17) that Proposition
24 of Archimedes’ On Spirals can be used to find
Archimedes’ area formula for a parabolic segment. Was Archimedes
aware that these two results were also equivalent?

One issue that students may have with our translation is that we have
made no attempt to modernize the language of proportionality used in
the text. Students may find it easier to read if they understand how
to make the translation to modern fractional notation themselves.
Thus, the proportion “DF will be to FB as CE is to DF” may be easier
to understand when written as “DF/FB = CE/DF”. For a model of how to
construct modern interpretations such as this of some of Torricelli’s
arguments, see [Leahy, pp.178-183].

Conclusion

Infinitesimal methods for solving area and volume problems were some
of the most important developments in mathematics during the
seventeenth century. Cavalieri’s method of indivisibles was the most
well known of these techniques, but his works were not widely
understood. As Andersen notes, Torricelli is “an important link
between Cavalieri’s method and the general understanding of it”, and
his sole work, Opera Geometrica, in particular was
“influential in spreading knowledge of the method of indivisibles”
[Andersen, p. 356]. Among the various parts of this
work, Quadratura Parabolae per novam indivisibilium Geometriam
pluribus modis absoluta stands out as one of the most clear,
concise, and compelling presentations of Cavalieri’s method that was
available at the time. We hope that the translation presented here
wil aid in understanding the importance of this work.

Acknowledgments and References

Acknowledgments

The authors of the translation would like to thank Brenda Fineberg for
her assistance with this translation. We would also like to
acknowledge the indirect support of Google, which made the original
Latin version of Torricelli’s Opera Geometrica available
through
books.google.com.
The original Latin is also available
here.
The authors would also like to thank the reviewers, whose meticulous
reading of the translation and its preamble found numerous errors and
yielded many ideas for improvement.

References

Kirsti Andersen, Cavalieri’s method of indivisibles, Archive for
History of Exact Science 31 (1985) 291–367.
Apollonius, Densmore, Dana, ed. Taliaferro, Catesby,
trans. Conics Books I-III. Santa Fe, NM: Green Lion Press,
1998.
Archimedes, Thomas Little Heath, ed. The Works of
Archimedes. Dover, New York, 2002.
Margaret E. Baron, The Origins of the Infinitesimal Calculus,
Pergamon Press, Oxford, 1969.
Boyer, Carl B. The History Of The Calculus And Its Conceptual
Development. New York: Dover 1959.
Katz, Victor, A History of Mathematics: An Introduction (3rd
edition). Boston: Addison-Wesley 2009.
Leahy, Andrew, Evangelista Torricelli and the ‘Common Bond of
Truth’ in Greek Mathematics, Mathematics Magazine 87 (2014)
174-184.
Evangelista Torricelli, de Dimensione Parabolae, Opera
Geometrica, 1644. Translated by Kasandara Sullivan and the
author.

Appendix: The Quadrature of the Parabola
solved by many methods through the new geometry of indivisibles

Translated by Andrew Leahy and Kasandara Sullivan

Until now, the matter about the measurement of the parabola has been
related in the manner of the ancients. It remains that we should
approach the same measurement of the parabola with a certain new but
marvelous system–namely, by the aid of the Geometry of Indivisibles
and with diverse methods in this manner. In fact, with the principal
theorems of the ancients assumed, as much of Euclid as of Archimedes
(they may concern very diverse matters themselves), it is amazing that
from every single one of these the quadrature of the parabola is able
to be elicited with so little trouble and vice versa, as if there were
a certain common bond of truth. In fact, from the premise that a
cylinder is three times its own inscribed cone, hence it follows that
the parabola is 4/3 of its own inscribed triangle. Indeed, if you
prefer to assume that the cylinder is 3/2 of its own inscribed sphere,
the quadrature of the parabola is immediately inferred. The same
result is concluded by supposing the demonstration which shows that
the center of gravity of a cone is placed on its axis, so that the
part which is near the vertex is three times the remainder. The
parabola is no less squared by also supposing that the space bound by
a spiral line in the first revolution and by the line which is the
beginning of the revolution is 1/3 of the first circle. On the other
hand, with the quadrature of the parabola supposed, all the
aforementioned theorems are able to be demonstrated easily. Moreover,
for my part I would not dare to assert that this Geometry of
Indivisibles is a thoroughly new invention. Rather, I would have
believed that the old geometers used this one method in the discovery
of the most difficult theorems, although they would have produced
another way more acceptable in their demonstrations, either for
concealing the secret of the art or lest any opportunity for
contradiction be proffered to envious detractors. Whatever it is, it
is certain that this wonderful geometry is a shortcut for invention,
and that it confirms countless almost inscrutable theorems with brief,
direct, and affirming demonstrations, which is certainly not able to
be done easily through the ancient teaching. To be sure, this is
truly the Royal Road in the mathematical thorn hedges, that Cavalieri,
creator of these wonderful inventions, first among everyone opened up
and made public for the common good.

Proposition 11.

The parabola is 4/3 of a triangle that has the same base and
height.1

Let ABC be a parabola with tangent CD and let AD be parallel to the
diameter. Let the parallelogram AE be drawn, and let a circle with
diameter AD be conceived which is the base of a cone having vertex at
point C and likewise is the base of some cylinder ACED of the same
height with the aforementioned cone.

Now let some line FG be drawn parallel to AD, and let a plane parallel
to the circle on AD be conceived to pass through the line itself. FG
will then be to IB as the line DA is to IB–that is, as the square on
DC is to the square on CI (because of the parabola). Or as the square
on DA is to the square on IG (because of similar triangles)–that is,
as the circle on DA is to the circle on IG–namely, as the circle on
FG is to the same circle on IG. And it is this always. All the first
magnitudes are equal to the line DA and therefore equal among
themselves. Also, all the thirds are equal to the circle on DA, and
on account of this equal among themselves. Therefore, by Lemma 18,
all the firsts together–namely the parallelogram AE–will be to all
the seconds together–namely, to the trilineum ABCD–as all the thirds
together–namely, the cylinder AE–are to all the fourths together
together–that is, to the cone ACD.2
Therefore, the parallelogram AE is three times the trilineum
ABCD.3
With the parallelogram AE cut in half, the triangle ACD will be 3/2 of
the trilineum ABCD. By conversion of the ratio, the triangle ACD will
be three times the parabola itself. On account of this, from the
explanation of Proposition 9, the parabola will be 4/3 of its own
inscribed triangle. Quod erat &c.

We will also square the parabola by a different computation with the
principles of indivisibles demonstrated previously, by which it will
be able to be done with brevity. Moreover, we, scraping the earth
with less daring, will resolve it by means of the immense ocean of
Cavalieri’s Geometry. Whoever wishes will be able to see all these
things (shall I say in a fountain, or in a sea?) around the middle of
the second book of Cavalieri’s Geometry of Indivisibles.

Lemma 20.

The squares of all the parts of a straight line together are in a
ratio of 1/3 to the same number of squares of the whole line together.

Let AB be a straight line. I say that all the squares of all the parts
of the line AB together are 1/3 of the same number of the squares of
that same line AB.

Indeed, let ACDB be a square with the diameter AD drawn. Let the
figure be revolved around the axis AB until it returns to its starting
point. It is clear that a cylinder CH is described by the square and
also the cone DAH which has vertex at A is described by the triangle
ABD. Now let EF be drawn parallel to CA and let AF–or FG (indeed
they are equal)–be one of the infinite parts of the whole line AB.

Now the square on the whole line AB is to the square on the part AF as
the square on EF is to square on FG (because of equality)–namely, as
the circle on diameter EL is to the circle on diameter GI (Euclid
XII.2).
And it will always be so. Also, the first magnitudes
individually are equal to the square on AB, and the thirds are always
equal to the circle on DH. Therefore, by Lemma 18 all the firsts
together (that is, as many lines AB squared as the line itself has
parts) will be to all the squares of the parts as all the thirds
together (that is, as the cylinder CH) will be to all fourths together
(namely, to the cone DAH). Therefore, just as many squares of some
line as the line itself has parts are to all squares of its very parts
as the cylinder CH is to the cone DAH–namely, 3 to 1. By transposing,
the proposition stands. Quod demonstrandum fuerat &c.

Lemma 21.

All rectangles which are bounded both by a straight line together with
its own individual pieces and by the remainders are in a ratio of 2/3
to the same number of squares of the same straight lines.

With the figure in the preceding lemma assumed, some point F should be
taken on the line AB. The rectangle contained by BAF (as one straight
line) and by FB will be one of all the aforementioned rectangles. (In
other words, one side is composed from the whole line AB together with
the part AF; the other side is clearly FB, without a doubt the
remaining part.)

Moreover, the aforementioned rectangle contained by BAF as one line
and by FB is the same as the rectangle EIL (because of the equality of
sides). And this is always true in this manner wherever the point F
is. But all the rectangles contained both by the lines intersecting
in the trapezoid CAHD (one of which is EI) and by the remainders (one
of which is IL) together with all the squares of the intermediate
section (one of which is FI) are equal (because of
Euclid II.5)
to all the squares on the halves (one of which
is FL). In fact, all squares of the intermediate sections (one of
which is FI) are to all the squares of the halves (one of which is FL)
as 1 is to 3 (by the preceding lemma). Therefore, if all squares of
the intermediate sections are taken away, all the rectangles will
remain (one of which is EIL). That is, all the rectangles contained
both by AB together with its parts and by the remainders will remain
as 2/3 of all the squares which are made from the halves–that is, the
same number of all the squares of the whole AB. Quod fuerat
ostendendum &c.

Proposition 12.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose diameter is BE, and around the parabola
let there be a parallelogram DC. Let a line FG be drawn parallel to
the diameter. FG will be to GI as BE is to GI, or as the rectangle CEA
is to CGA–that is, as the square on CE is to the rectangle CGA. And
it will always be so. Also, the first magnitudes are always equal to
the line BE. Moreover, the thirds are always equal to the square on
CE. Therefore (by Lemma 18) all the firsts together (that is, the
parallelogram AB) will be in relation to all the seconds together
(namely, the semiparabola AIBE) as all the thirds together (clearly
just as many squares on the line CE as the line itself has parts) in
relation to all the fourths together (namely, all the rectangles
contained by CE with its parts and by the remaining parts). Therefore
(from the preceding lemma) the parallelogram AB will be 3/2 of its own
semiparabola. Also, the whole parallelogram DC will be 3/2 of the
whole parabola–indeed, as 6 is to 4. Therefore the parabola will be
to its own inscribed triangle (which is in fact 1/2 of the
parallelogram DC) as 4 is to 3–that is, 4/3. Quod erat &c.

We are able to square the parabola by the same argument without the
trouble of the former lemmas, with yet a different
supposition–namely, by supposing the proportion which the cylinder
has in relation to its inscribed sphere–a proportion which is indeed
3/2, as shown from Archimedes’ On the Sphere and the Cylinder,
Book I

Proposition 13.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola, around which is the parallelogram AD. Let
there be a semicircle on the diameter AC, around which is the
rectangle AE. Then with the axis AC fixed, let it be conceived that
the semicircle itself is revolved around the axis in such a way that
from its own revolution a sphere is circumscribed. A cylinder likewise
arises from the revolution of the rectangle AE.

With some point G now assumed, let a line GF be drawn parallel to the
diameter HB, and through the same point G let the plane GL be drawn
perpendicular to the axis AC.

The line FG will be to GI as BH is to GI (because of equality). That
is, as the rectangle CHA is to the rectangle CGA, or as the square on
HN is to the square on GM (because of the circle), or as the square on
GL is to the square on GM–namely, as the circle on the semidiameter
GL in the cylinder is to the circle on the semidiameter GM in the
sphere. And it will always be thus wherever the point G is
assumed. Moreover, all the firsts are equal among themselves, as are
all the third magnitudes among themselves. Therefore, all the firsts
(namely, the parallelogram AD) will be to all the seconds (namely, to
the parabola ABC) as all the thirds (that is, the cylinder) are to all
the fourths together (clearly to the sphere). But the cylinder to the
sphere is 3/2. Therefore, the parallelogram AD will certainly be 3/2
of the parabola, and the parabola itself will be 4/3 of its own
inscribed triangle, as was concluded in the preceding lemma. Quod
&c.

Lemma 22.

Suppose some number of magnitudes will have been hung from given
points on a balance, and the same number of magnitudes of another
type, equally proportional with the aforementioned magnitudes, will
hang from the same points. The center of equilibrium of each type of
the magnitudes will be one and the same.

Let some number of magnitudes C, D, E and F of the first type be hung
from whatever points on a balance AB. Let just as many magnitudes G,
H, I, and L of a second type hang from the same points and let them be
proportional–namely, as C is to D so is G to H, and as C is to E so
is G to I, etc. I say that the same point on the balance is the common
center of equilibrium of each type of suspended magnitudes.

Indeed, since as C is to D so is G to H, they will counterbalance
from the same point, so the two magnitudes C and D will be as the
two magnitudes G and H.

Further, since as C is to D so is G to H, by convertendo and
componendo DC will be to C as HG is to G. Moreover, C is to E as G is
to I. Therefore, CD together will be to E as GH together is to I by
equality. Wherefore the magnitudes CD and E will counterbalance from
the same point from which the two GH and I counterbalance.

Additionally, since as CD is to E so is GH to I by the things just
mentioned, by componendo CDE will be to E as GHI is to I. But E is to
C as I is to G, and C is to F as G to L. Wherefore CDE together will
be to F as GHI together is to L by equality. Therefore, the two
magnitudes CDE and F will have the same point of equilibrium which the
two magnitudes GHI and L have. And thus also if there are many
magnitudes, all the way to infinity, quod erat propositum &c.

Lemma 23.

If a parabola should have a tangent at the base and indeed a line
parallel to the diameter for the other side, the trilineum bound by
the parabolic curve, the tangent, and the aforementioned parallel will
balance at equilibrium from the point on the tangent where it is
divided so that the part nearest to the point of tangency is three
times the remaining part.

Let ABC be a parabola whose tangent at the base is CD and let AD be
parallel to the diameter. I say that the mixed trilineum ABCD
balances at equilibrium from the point on the tangent CD where it is
divided so that the part towards the tangent C is three times the
remaining part.

Let the figure be imagined in such a way that DA is perpendicular to
the horizontal, and let a circle on the diameter DA be conceived which
is the base of a cone having vertex at point C.

With some point E already assumed, let EF be drawn parallel to DA
itself, and let a plane parallel to the base of the cone pass through
the EF.

Therefore the line DA will be to EB as the square on DC is to the
square on CE (because it is a parabola), or as the square on DA is to
the square on EF–that is, as the circle on DA is to the circle on
EF. And it is always so wherever the point E. Therefore, since
proportional magnitudes of two types are suspended from the same
points, as was required in the preceding lemma, all the magnitudes of
the first type together (that is, all the lines of the trilineum ABCD,
or the trilineum itself) will have the same point of equilibrium as
all the magnitudes of the second type together (that is, all the
circles of the cone ACD, or the cone itself) have. Moreover, the cone
balances at equilibrium from the point which divides CD in such a way
that the part toward C is three times the remaining part, since the
line DA is perpendicular to the
horizontal.4 Therefore, the trilineum
ABCD also balances at equilibrium from the same point. Quod erat
propositum, &c.

Proposition 14.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose diameter DE is understood to be
perpendicular to the horizontal. Let CF and AD be tangents, and let
AF in fact be parallel to the diameter.

Let FH then be assumed as 1/4 of the whole FC. The mixed trilineum
ABCF will balance at equilibrium from the point H (by the preceding
Lemma). Also, let FI be taken as 1/3 of the whole FC. The whole
triangle AFC will balance at equilibrium from I.5
Clearly the parabola
will balance at equilibrium from D, since it has a center on the
diameter. Therefore, the trilineum ABCF will be reciprocally to the
parabola itself as DI is to IH–namely, 2 to 1. (In particular, if FC
is 12 of some parts, FD is 6, FI is clearly 4, and FH is 3.
Therefore, DI is 2 and IH is 1.) Consequently, by componendo, the
whole triangle AFC will be three times the parabola. The remainder of
the quadrature is completed as was done in Proposition 9. Quod
erat &c.

In another way.

With the same things assumed as above, let FH be supposed 1/4 part of
the whole FC. The mixed trilineum ABCF will balance at equilibrium at
the point H. Also, let FI be supposed 1/3 part of FD itself. Then
indeed the triangle FDA will balance at equilibrium from the point
I. Clearly, the mixed trilineum ABCD will balance at equilibrium from
the point D. (For the entire triangle ADC balances at equilibrium from
the point D, and the removed parabola balances at equilibrium from the
same point D. Therefore, it is also necessary that the remaining
trilineum ABCD balance at equilibrium from the point D.) And so the
triangle FDA will be to the trilineum ABCD reciprocally as DH is to
HI–namely, as 3 is to 1. By conversion of the ratio, the triangle
ADC will be to the parabola as 3 is to 2, or as 6 is to 4. Wherefore,
the parabola will be to the triangle ABC as 4 is to 3–namely,
4/3. Quod erat propositum demonstrare &c.

With the following observation on geometric progressions said first,
we may approach the quadrature of the parabola from yet another
supposition.

Lemma 24.

If two straight lines should intersect each other and between these
should be described a certain constant flexilineum from alternately
parallel lines, all lines which are parallel among themselves will be
in continuous proportion.

Let two straight lines AB and CB intersect each other at point B, and
between these let a flexilineum CADEFG etc. be described in such a
way that CA, DE, FG, etc. are parallel among themselves, and likewise
AD, EF, and the remaining in turn are assumed parallel among
themselves. I say that AC, ED, and GF are in continuous proportion.

Indeed, since they are parallel, as AC is to ED, so is AB to BE, or DB
to BF–that is, ED to GF.
(Euclid VI.2 and
VI.4.)
Constat ergo quod propositum fuerat.

Lemma 25.

With two straight lines intersecting each other as shown above, if the
two lines AC and DE are parallel between themselves and with CD
joined, let it be conceived that the flexilineum ACDE is continued
in infinitum all the way to the point B of intersection. I
say that in a flexilineum of this kind are to a hair each and all the
terms which are in the progression of the proportion of AC to DE
continued in infinitum.6

Let F be assumed equal to AC itself and G equal to DE. Also, let the
proportion of F to G be conceived as continued in its own infinitely
many terms from F to H.

Now if it is possible, let there be some term or terms in the
progression from F to H which are not found in the flexilineum and let
I be greatest term of those terms which, though they are in the
progression FH, are not in the flexilineum. Therefore, the term L
preceding such an I will be in the flexilineum. Let it be MN. Since L
is to I as F is to G, or as AC is to DE, or as NM is to PO, following
next, and L and NM are equal, I and PO will also be equal. Therefore,
the term I, which was assumed to not be in the flexilineum, was found
in the same.

We would demonstrate in completely the same method that there is no
term in the flexilineum which is not also in the progression FH etc.
We would then conclude that all the terms in the flexilineum are
precisely the terms of the proportion of AC to DE continued in
infinitum, since it was demonstrated that no term which is in the
progression FH is missed in the flexilineum nor is any term which
would not also be found in the progression FH present in the
flexilineum.

Lemma 26.

With infinitely many straight lines of a greater than inequality
assumed in continuous proportion, to find a straight line which is
equal to all the aforementioned lines together.

Let A and B be the first two lines of the given progression, in which
CD is supposed equal to the greater A and EF to the lesser B. Let CD
and EF be parallel, and let DF and CE, which intersect by necessity,
be joined. Thus let them intersect at point G and, with CF drawn, let
GL be parallel to the CF.

I say that the line DL is equal to all the infinite terms of the
progression ABM taken together.

Indeed, let it be conceived that the flexilineum DCFE etc. is
continued in infinitum all the way to the point G. All the
lines will be in the flexilineum itself, or the terms of the given
progression ABM.

Now let HE, NI, and the remaining lines parallel to them be extended
all the way to DL. EF will be equal to CP itself, HI equal to PQ, and
NO equal to QR, and so on one at a time (by
Euclid I.34).
Indeed, each line which is in the flexilineum
will have its own corresponding small part on the line DL equal to
itself until the flexilineum should arrive at the ultimate point G.
At that time, moreover, there will not be anything from the
flexilineum or from the line DL that will be left over, but the
flexilineum itself as much as the line DL will also have been
completely used up. Indeed, GL itself, which is drawn from the final
point G of the flexilineum, is the last of all the parallels which are
drawn all the way to DL. Therefore, all the lines of the flexilineum
together, of which the first is CD, taken alternately (that is, all
the lines of the progression ABM) are equal to all the small parts of
the line DL taken together (that is, DL itself).
Quod erat ostendendum &c.

Lemma 27.

With infinite magnitudes supposed in a continuous geometric proportion
of a greater than inequality, the first magnitude will be the mean
proportional between the first difference and the aggregate of all the
magnitudes.

Indeed, with the preceding construction assumed, let FV be drawn
parallel to the GC. Then DV will be the first difference. But DV is
to the first magnitude DC as FD is to DG (Euclid
VI.4)–that
is, as DC is to the aggregate DL of all of them. Quod erat
demonstrandum &c.

Comment.

We will not hesitate to assert that this is also true for numbers and
any sort of magnitudes. We will convey a demonstration even more
universal, especially since it is quite brief. The conclusion of this
truth, when it had been contrived in passing by us for the most
celebrated Cavalieri, also itself established the same Theorem with
the following demonstration, which has been already supplied by us in
the first proof.

This is said in advance. But if there will be however many
magnitudes, either finite or infinite in number, the antecedents of
which are always larger than the consequents, the first magnitude
of all will be equal to all the differences taken together with the
smallest magnitude itself.

This was known among geometers and is demonstrated as was done by us
in Lemma 15, where we show that the parallelogram AE is equal to all
the differences between the following parallelograms and the smallest
parallelogram OC.

Now let magnitudes be supposed infinite in number in a continuous
geometric proportion of a greater than inequality. It is evident that
the smallest magnitude of all either will or will not be a point.
Therefore, in this case the first magnitude will be equal to all such
differences.

Moreover, since the magnitudes are assumed in a continuous geometric
proportion, the differences will also be proportional in the same
ratio and therefore (with a conversion having been done) as the first
difference is to the first magnitude, so will the second difference be
to the second magnitude, and thus always. Consequently, as one is to
one, so summarily will all be to all–namely as the first difference
is to the first magnitude, so are all the differences together (that
is, the first magnitude itself) to all the magnitudes
together. Therefore, it stands that the first magnitude is the mean
proportion between the first difference and the aggregate of all of
them.

Proposition 15.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola in which is inscribed the triangle ABC. I say
that the parabola is 4/3 of the triangle ABC.

Indeed, let the two triangles ADB and BEC also be inscribed in the
remaining portions ADB and BEC of the parabola. The triangle ABC will
be four times the two triangles ADB and BEC together (Lemma 7). Let
four inscribed triangles also be conceived in the remaining small
parts AD, DB, BE, and EC. Both triangles ADB and BEC together will be
four times the aforementioned subsequent triangles together (Lemma 7),
and always in this manner. Therefore, the parabola is nothing other
than a certain aggregate of magnitudes infinite in number in a 4 to 1
proportion, of which the first is the triangle ABC, and indeed the
second consists of the two triangles ADB and BEC. Therefore, the first
magnitude ABC will be the mean proportional between the first
difference and the aggregate of all of them–namely, the parabola.

Consequently, suppose the triangle ABC is 4. Therefore, the two
triangles ADB and BEC together are 1, and the first difference (of
course between 4 and 1) is 3. Therefore, the aggregate of all the
infinite magnitudes (namely the parabola itself) will be (by lemma 27)
to the first magnitude (that is, to the inscribed triangle ABC) as the
first magnitude itself is to the first difference–clearly, as 4 is to 3.
Namely, 4/3. Quod erat propositum demonstrare &c.

In Another Way.

Let ABC be a parabola whose diameter is DB, let AD and CD be tangents
to the base, and also let EF be tangent to the. Moreover, in the
remaining trilinei ABE and BCF let the two triangles GEH and IFL be
inscribed (as was required for the construction in the third and
fourth lemmas). Likewise, let four triangles be conceived in the four
remaining mixed trilinei, and always in this way. So the entire
trilineum ABCD will be nothing other than a certain aggregate of
magnitudes infinite in multitude in a 4 to 1 proportion (Corollary 1
of Lemma 3), of which the first is the triangle EDF, the second in
truth consists of the two triangles GEH and IFL, and the third in
truth from the following four, etc. Therefore, the aggregate of all
(namely, the mixed trilineum ABCD) will be to the first magnitude
(namely, to the triangle EDF) as the first magnitude itself is to the
first difference (Lemma 27)–clearly, as 4 is to 3.

Thus, since the trilineum ABCD is to the triangle EDF as 4 is to 3,
the same trilineum will be to the triangle ADC as 4 is to 12.
Therefore, the parabola will be to the triangle ADC as 8 is to 12 and
to its own inscribed triangle as 8 to 6–namely, 4/3. Quod erat
demonstrandum &c.

Lemma 28.

If straight lines AB, CD, EF, etc. infinite in number are in a
continuous geometric proportion of a greater than inequality and
moreover another progression BG, DH, FI, etc. is supposed so that just
as the first AB is to the first BG, so the second CD is to the second
DH, and so the third EF is to the third FI, and thus always, I say
that the entire aggregate of the progression AB, CD, EF, etc. is to
the aggregate of the progression BG,DH, FI as AB is to BG.

Let all the terms of the two progressions be conceived to be in the
flexilinei (in like manner with Lemma 25). With AD and GD joined, let
OL be drawn parallel to AD itself and OM parallel to DG itself. BL
will be equal to all the infinite terms AB, CD, EF, etc. (Lemma 26),
and indeed OM will be equal to all the infinite terms of the remaining
progression BG, DH, and FI (Lemma 26).

Now as LB is to BA, so is OB to BD–that is, MB to BG
(Euclid
VI.4).
Therefore, by permutando, the aggregate LB is
to the aggregate BM as AB is to BG–namely, as one magnitude is to the
other. Quod erat &c.

This theorem had been able to be substituted for the demonstration in
Proposition 12 of Book 5 of Euclid. Indeed, it is one and the same
with the Theorem in the aforementioned Proposition. But in fact,
since almost all are of the opinion that Euclid there supposed a
multitude finite in number, we have determined with the help of
flexilinei that,

Proposition 16.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose diameter is DE, whose tangents at the base
are AD and CD, and whose tangent through the vertex is in fact FBG.
Let ABC be the inscribed triangle. I say that the parabola is 4/3 of
the triangle ABC.

For since the EB is equal to the BD (because it is a parabola),the
line AC is in fact twice the line FG and the inscribed triangle ABC
will be twice the triangle FDG contained by the tangents. And this is
also always true around the remaining parabolic parts AIB and
BOC. (Indeed, AIB is a parabola whose tangents at the base are AF and
BF. Therefore, the inscribed triangle AIB will be twice the tangent
triangle LFM. The same is also true for the other part. Therefore,
the two triangles AIB and BOC together are twice the two triangles LFM
and NGP together.) Therefore, since the two progressions are both in
a continuous proportion of magnitudes infinite in multitude (namely,
the one inside the parabola whose first term is the triangle ABC and
second term is the two triangles AIB and BOC, together, etc, and also
the other progression outside the parabola whose first term is namely
the triangle FDG and second moreover is the triangles LFM and NGP
together, etc.) and since the individual terms of the progression
inside the parabola are twice the individual terms of the progression
outside the parabola, therefore the entire aggregate of the first
progression will be twice the entire aggregate of the second
progression (Lemma 28). In particular, the parabola itself will be
twice the mixed trilineum ABCD. Therefore, by componendo and
conversion of the ratio, the triangle ADC will be 3/2 of its own
parabola–namely, as 6 is to 4. Therefore, the parabola will be to
the triangle ABC as 4 is to 3–clearly, 4/3. Quod erat ostendendum
&c.

With the aid of infinitesimals the quadrature of the parabola is able
to be gained with yet other results assumed. We suppose Propositions
14 and 25 which Archimedes demonstrated in the book on Spiral
lines with a lemma of this kind set out beforehand.

Lemma 29.

If a first magnitude is to a second as a third is to a fourth, and
thus however often it will have been pleasing, and if all the firsts
and also all the thirds are proportional in the same order, then all
the firsts together will be to all the seconds together as all the
thirds together are to all the fourths together.

Let the first A be to the second B as the third C is to the fourth D,
and E to F as G to H, and thus however often it will have been
pleasing. And let all the firsts A, E, I, etc. and all the thirds C,
G, M etc. be proportional in order–namely, as A is to E so should C
be to G. Moreover, as A is to I, so should C be to M, etc., and thus
always. I say that all the firsts A, E, I etc. together are to all the
seconds B, F, L etc. together as all the thirds C, G, M etc. together
are to all the fourths D, H, N etc. together.

Let O, P, Q etc. be taken as individuals equal to the first of the
firsts–that is, to A itself–and let there be just as many as there
are all the firsts A, E, I, etc. Likewise, let R, S, T etc. be assumed
to be just as many as there are all the thirds, and let the
individuals R, S, T be equal to the first of the thirds–namely, C
itself.

Now on account of equality, as O is to A so will R be to C. Further,
since P is equal to A itself, and S to C itself (because of the
supposition) as P is to E so will S be to G. And this always. All O,
P, Q are equal, and likewise all R, S, T are equal. Therefore, all O,
P, Q, etc. together are to all A, E, I, etc. as all R, S, T together
are to all C, G, M together (Lemma 18). Finally, by convertendo, all
R, S, T are to all O, P, Q as all C, G, M are to all R, S, T.
Remember this.

Since indeed as O is to A so is R to C and as A is to B so is C to D,
O will be to B as R is to D (by equality). For completely the same
reason, we conclude that as P is to F so is S to H (by equality), and
thus concerning the rest. Therefore, all O, P, Q etc. together will
be to all B, F, L etc. as all R, S, T etc. together are to all D, H, N
etc. (Lemma 18). Wherefore, by equality, all A, E, I etc. will be to
all B, F, L etc. as all C, G, M etc. are to all D, H, N etc. Quod
erat ostendendum.

Proposition 17.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose tangent is AE. Indeed, let the diameter
be parallel to CE, and let some FD be drawn parallel to CE itself. EC
will be to FB in length as EA is to AF2 (because
it is a parabola), or as EC2 is to FD. Therefore
EC, FD, and FB will be in a continuous proportion.

Then let the intervals AC and AD become two circles with center A.
Let the beginning of a spiral be placed on the semidiameter AC, and
let the spiral itself be AGC.

So DF will be to FB as CE is to DF, or as CA is to AD–that is, as CA
is to AG, or as the entire circumference CLHC is to the arc CLH
(Proposition 14 of On Spiral Lines). That is, as the entire
circumference DPGD is to the arc DPG. Indeed, it will always be thus
wherever the point D is taken. All the first magnitudes and likewise
all the thirds, are proportional in the way in which they ought to be
(as we will show below). Moreover, all the firsts together (namely,
the triangle AEC) will be to all the seconds together (namely, the
mixed trilineum ABCD) as all the thirds together (namely, the circle
CLH) are to all the fourths together (that is, to the remainder of the
circle itself with the region inside the spiral CAGC subtracted) by
the preceding lemma. Moreover, the circle CLH is 3/2 of the
aforementioned space with the region inside the spiral subtracted
(Proposition 25 of On Spiral Lines). Therefore, the triangle
ACE will also be 3/2 of the mixed trilineum ABCE. By conversion of the
ratio, the triangle ACE will be three times the parabola ABC. The
remainder of the quadrature can be completed as in Proposition
9. Propositione factum est.

Moreover, we will now show what was assumed–namely, that all the
firsts and all the third magnitudes are proportional in the way that
is required in the preceding lemma.

Let some MO be drawn parallel to FD itself as in the previous figure,
and let us suppose that the FD is the first of the firsts and that the
circumference DPG is itself first of the thirds. Therefore, DF will be
to OM as DA is to AO, or as the circumference DPG is to the
circumference whose semidiameter is AO etc. And thus always. Quod
oportebat &c.

We will also square the parabola by a way as of yet untried–of
course, with its center of gravity found with the prior help of
indivisibles. Moreover, we suppose a lemma which Archimedes showed in
the Book 2 of on the Equilibrium of Planes–that is, that
the centers of gravity of parabolas divide their own own diameters in
the same proportion.

Lemma 30.

The center of gravity of a parabola divides the diameter so that
the part ending at the vertex is 3/2 of the
remainder.7

Let ABC be a cone whose base is AMC, axis is BC, and the axial
triangle is in fact ABC. Let the cone be cut by the plane EFG as was
demanded in Proposition 11 of The Conics, Book I. The
section will be that which is called a parabola, and its diameter will
be FH. Now let the center of gravity of the parabola EFG be some
point, call it I. It must be shown that the line FI is 3/2 of the IH.

Let the line AIL be drawn through the point I, and let the cone be cut
by another plane MNO parallel to the EFG. The section on MNO will be a
parabola, and its center of gravity will be P. (For since they are
parabolas, as FI is to IH so is NP to PR. But I is assumed the center
of gravity of the parabola EFG. Therefore, by Proposition 7 of
of On the Equilibrium of Planes, Book II, P will be the
center of gravity of the parabola MNO.) And thus always, wherever the
plane MNO. Therefore, the centers of gravity of all the parabolas
which are in the cone ABC are found individually on the line AL.
Wherefore, the common center of gravity of all the same aforementioned
parabolas together will also be on the line AL. Moreover, all the
parabolas are the same as the cone itself. Therefore the center of
the cone is on the line AL, but since it is also on the axis BD, the
center of the cone will be on the common point of intersection S.
Therefore, BS will be three times SD itself.

Let the line DQ be drawn from the center of the base parallel to
AL. CQ and QL will be equal (Euclid
VI.2).
Moreover, since
BS is three times SD (because S is the center of the cone), BL
will also be three times LQ (Euclid
VI.2).
Therefore, BL is
3/2 of LC. Wherefore, FI will also be 3/2 of IH. Quod
erat propositum &c.

Proposition 18.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose diameter is BD, and indeed let the
inscribed triangle be ABC. I say that the parabola is 4/3 of the
triangle ABC.

Let AD and DC be divided in half at the points E and F, and let EG and
FH be drawn parallel to the diameter. These will be the diameters of
the segments AGB and BHC. Let the centers of gravity of the
aforementioned segments be O and N. GO and HN will each be 3/2 of the
remainders OI and NL (by the preceding lemma). Let ON be joined. The
common center of gravity of the two segments will be on ON itself (by
Proposition 8 of On the Equilibrium of Planes, Book I). But
it is also on BD. (For as the center of the whole parabola is on BD,
so also is the center of the triangle ABC.) Wherefore, the point P
will be the center of the segments AGB and BHC. Assume BD is 60 parts.
GE will be 45 parts (since it is 3/4 of BD), the IE will be 30 parts,
and the EO–that is, DP–36 parts. Let Q be the center of gravity of
the triangle ABC; DQ will be 20 parts. Let R be the center of the
parabola; RD itself will be 24 parts (by the preceding lemma).
Therefore, PR will be 12 parts and RQ 4 parts. But as PR is to RQ so
reciprocally is the triangle ABC to the two segments AGB and BHC.
Wherefore, the triangle ABC will be to the two portions AGB and BHC as
12 is to 4–namely, as 3 is to 1. By componendo and conversion of the
ratio, the parabola ABC will be to the triangle inscribed in it as 4
is to 3–namely, 4/3. Quod erat propositum &c.

We will undertake the quadrature of the parabola by a hitherto new
calculation, with the following lemma assumed, which indeed was said
to have come forth from the Cavalerian School. Indeed, it was devoted
to the measurement of a certain solid arising from the parabola itself
having been revolved around its axis. Moreover, the lemma is of a
type from the author Ionas Antonio Roccha, a distinguished geometer.8

Lemma 31.

If a plane figure is balanced above some straight line of its own
which divides the figure itself, the moments of the segments of the
figure are as the solids of revolution described by the segments
having been revolved around the dividing line.

Let ACDBFE be a plane figure which the straight line AB divides, and
let the figure be imagined to be balanced above the line AB. I say
that the moment of the segment ACDB is to the moment of the segment
AEFB as the solid of revolution arising from the revolution of the
segment ACDB around the axis AB is to the solid of revolution arising
from the revolution of the remaining segment around the same axis of
revolution.

For with two given points H and I assumed on the line AB, let the
lines CE and DF be drawn through H and I perpendicular to the AB, and
let the segments DH and HF be cut in half at the points L and M.

Therefore the moment of the line DH will have a ratio to the moment of
the line HF compounded from the ratio of the magnitudes DH to HF, and
from the ratio of the distances LH to HM–or DH to
HF.9 Therefore, the
moment of the line DH will be to the moment of the line HF as the
square on DH is to the square on HF.

It will be shown in the same manner that the moment of the line CI is
to the moment of the line IE as the square on CI is to the square on
IE, and thus always.

Further, the moment on DH is to the moment on CI (on account of the
same reason as above) as the square on DH is to the square on CI, and
in this way always. Therefore, (by Lemma 29) all the first magnitudes
together, (namely, all the moments of the figure ACDB) will be to all
the seconds together (namely, to all the moments of the remaining
figure AEFB) as all the thirds together (namely, all the squares of
the figure ACDB) are to all the squares of the remaining figure, or as
all the circles of the figure ACDB (namely, the circular solid
described by its rotation around the axis AB) are to all the circles
of the remaining figure AEFB (namely, to the circular solid arising
from its revolution around the same axis AB). Quod erat
ostendendum &c.

With this having been set forth (which, as we indeed declared
thoroughly above, was undertaken by others, was introduced here as
another’s, and which I believe was not well known until now), we will
square the parabola, with the demonstration, which is proved in many
ways, assuming that the Cylinder is twice its own inscribed parabolic
conoid.10

Proposition 19.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABCD be a semiparabola, around which is a rectangle DE. Let a
point F be assumed such that AF is to FD as 5 is to 3, and let FG be
drawn parallel to the diameter. The center of gravity of the
semiparabola will be on the FG (by Lemma 11). Let it be some point,
call it I, let LIM be drawn parallel to AD through I, and let IN be
taken equal to IM itself. Also, let PQ (wherever it may fall) be
understood as extended parallel to the diameter CD in such a way that
the rectangular parallelogram DP is equal to the semiparabola
itself. Then let a rectangle DR be conceived applied to the line CD in
such a way that it balances at equilibrium with the semiparabola with
the balance above the line CD. Let the center of the aforementioned
rectangle be the point S and, with TSX drawn parallel to the AD, let
the line IS be joined.

Now it is clear from the lemma above that the cylinder made from the
rectangle DR revolved around the axis DC will be equal to the
parabolic conoid made from the rotation of the semiparabola ACD around
the same axis of revolution CD, since the moments of the plane figures
are assumed equal. Therefore, the cylinder made from the rectangle DR
will be 1/2 of the cylinder made from the rectangle DE, and
consequently the square on TX will be 1/2 of the square on ML. (For
cylinders of equal height are among themselves as the squares of the
bases.) Remember this.

In fact, MN is to TX as IM is to TS (Euclid
VI.4)–for they
are halves of the same things), or as IV is to VS–in particular,
(because the plane figures balance at equilibrium above the line CD,
or from the point V) reciprocally as the rectangle DR is to the
semiparabola (or to the rectangle DP equal to the semiparabola
itself), or as their bases TX to MO. Therefore, TX is the mean
proportional between MN and MO. Wherefore, the rectangle NMO will be
1/2 of the square on LM, since it is equal to the square on TX.

Clearly the ratio of the square on LM to the rectangle NMO is composed
from the ratio of LM to MN (which is 4/3 by construction, for we
assumed the point F is such that AF is to FD as 5 is to 3) and from
the ratio of LM to MO, which indeed was unknown, but now shows itself
as 3/2 by necessity. For the ratio of 2 to 1 is composed from 4/3 and
3/2, as also was made known by the Cantorians themselves, as was seen
in the three numbers 4,3,2.

Therefore, the rectangle DE to DP itself, or to the semiparabola, will
be 3/2, and the semiparabola to the triangle ACD will be
4/3. Quod erat ostendendum &c.

Lemma 32.

Let ABC be a parabola with base AC and tangent CD. Let AD
be parallel to the diameter. With some point E assumed, let EF be
drawn parallel to the diameter. I say that as FE is to EB, so is CA to
AE.

For DA is lengthwise to the FB as DC2 is to CF2,
or as DA2 is to FE2 (because it is a parabola).
Therefore, DA, FE, and FB are in a continuous ratio. Remember this.

Now as AC is to CE so is AD to EF, or EF to FB. By conversion of the
ratio, as CA is to AE so is FE to EB. Quod erat
ostendend. &c.

Lemma 33.

Any parabola is equal to two parabolas taken together which indeed
have a base equal to that one, a diameter in fact half, and which are
equally inclined.

Let ABC be a parabola whose diameter is BH. Let the two other
parabolas AEC and AGC be on the same base and indeed let the diameters
HE and HG both be 1/2 of the diameter HB, but inclined equally to the
base. I say that the parabola ABC is equal to the figure AECG.

For let some point be taken on the base AC, and let it be M. With PMN
drawn parallel to the diameter BH, BH will be to NM as the rectangle
AHC is to the rectangle AMC, or as the line HE is to the line MO. By
permutando, as BH is to HE so will NM be to MO. Wherefore, NM will be
twice the MO. It will be shown in completely the same manner that NM
is also twice the MP. Therefore, the entire NM is equal to OP
itself. And this always. On that account, all lines of the figure ABC
together (namely, the parabola ABC itself) will be equal to all the
lines of the figure AECG together (namely, to the two parabolas AEC
and AGC). Quod erat &c.

Proposition 20.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose diameter BE is conceived as perpendicular
to the horizontal, and let the parabola itself be inverted. Let CA be
extended to D, so that CA and AD are equal, and let DC be a balance
whose fulcrum is A. Let CF be drawn tangent to the parabola and AF
parallel to the diameter EB. Suppose GH is equal to AC itself, and let
GH be divided in half at I. Let IL and IM both be half of the line EB
and be inclined equally to the base as EB itself is to AC. Also, let
the two parabolas GLH and GMH be constructed, which (by the preceding
lemma) together will be equal to the parabola ABC, and let the figure
GLHM be suspended from the point D.

Now let the points O and N be taken equally in distance from the
points I and E, respectively. With NQ drawn parallel to EB and ROS to
LM, NP will be equal to RS itself, as in the preceding lemma.

Now QN is to RS as QN is to NP (because of equality), or reciprocally
as DA is to AN (by Lemma 32). Therefore, the lines QN and RS balance
at equilibrium, and thus always. Therefore, all the lines of the
triangle AFC together (namely, the triangle itself) balance at
equilibrium all the lines of the figure GLHM together (namely, the
figure GLHM itself).

Let AV be 1/3 of the entire AC. It is clear that if the line from V is
sent down parallel to AF itself, the center of gravity of the triangle
AFC will be on it, and the line will be perpendicular to the
horizontal. Therefore, the triangle AFC will be suspended centrally
from the point V, and the triangle AFC will reciprocally be to the
region GLHM as DA is to AV–namely, 3 to 1.

Moreover, since the region GLHM is equal to the parabola ABC, the
triangle AFC will also be three times the parabola ABC.

The remainder of the quadrature is solved as was done in Proposition
9. Quod &c.

Proposition 21.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a semiparabola whose diameter is CE and ordinate is AE, and
let CD be tangent. Let the parallelogram AECD be completed. It is
clear that all the lines of the mixed trilineum DABC which are in fact
parallel to the diameter are in the same ratio among themselves as all
of the circles of a cone which has axis DC and vertex C. Therefore,
the center of gravity of all the lines of the trilineum DABC, will be
on that line which divides the balance DC just as the center of
gravity of the cone divides the same balance–namely, so that the part
nearest to C is 3 times the remainder (by Lemma 22). Therefore, let
CF be made 3 times FD itself, and let FM be drawn parallel to CE. The
center of gravity of the trilineum DABC, wherever it may be, will be
on the line FM.

Likewise, all lines which are drawn parallel to the diameter of the
semiparabola ABCE are in the same ratio among themselves which are all
circles of some hemisphere whose axis is AE and vertex clearly is
A (by Lemma 22). Therefore, the center of gravity of all the lines
suspended from the balance AE, or from the semiparabola itself, will
be on that line which thus divides the balance AE as the center of
gravity of the hemisphere divides the balance. Namely, so that the
part terminated at A is to the remainder as 5 is to
3.11 Therefore, let
AI be to IE as 5 is to 3, and let IH be drawn parallel to CE. The
center of the parabola, wherever it is, will be on the line
IH. Finally, let GL be drawn which cuts in half the sides AE and DC.
The center of gravity of the parallelogram DE, which is O, will be on
GL. Suppose that the center of gravity of the semiparabola is some
point P, and let PO be extended onto N. N will be the center of
gravity of the trilineum DABC. Now the semiparabola is to the
trilineum as NO is to OP, or as ML is to LI–namely, as 2 is to 1.
(For by construction the whole AE is 8 of some kind of parts, AM is 2
of such parts, ML is 2, LI is 1, and the remaining IE will be 3.)
Therefore, the semiparabola will be to the parallelogram as 2 is to 3,
or as 4 is to 6, and the semiparabola is to its inscribed triangle as
4 is to 3–namely, 4/3. Quod &c.

FINIS.

Translation Notes

A word about typesetting: Torricelli uses a combination
of italics and Roman type, depending on the type of claim he is
making (e.g., “Lemma” or “Proposition”). We have followed him in
our typesetting of the translation, so, e.g., a figure ABCD may
sometimes be referred to as ABCD or ABCD depending on the
context.
A trilineum is a generalization of a triangle in which
one or more of the sides is curved instead of rectilinear. The
terminology goes back at least to Commandino.
Here Torricelli is using Euclid XII.10, which establishes
that a cone is 1/3 of the cylinder in the same base with the same
height.
In Proposition XXII of his Liber de Centro Gravitatis
Solidorum (1565), Federico Commandino established that the
center of gravity of a cone–i.e., the point at which it balances at
equilibrum–lies at the point on its axis which is 1/4 of the way
from the base to the vertex. In other words, the distance from the
center of gravity to the vertex is three times the distance from the
center of gravity to the base.
Proposition 14 of Archimedes’ On the Equilibrum of
Planes I establishes that the center of gravity of a triangle
lies at the intersection of any two median lines. It follows from
this that the center of gravity lies at a point on a given median
line that is 2/3 of the way from the vertex to the opposite side. A
well-known property of centers of gravity is that a figure will be
suspended at equilibrium from a point on its boundary if and only if
the line connecting that point to the center of gravity is vertical,
which means parallel to FA in this case. Since the triangle is
right, FI must be twice IC by similarity.
The Latin version of this sentence reads: Dico in
huiusmodi flexilineo esse omnes, & singulos ad unguem terminos qui
sunt in progressione proportionis AC ad DE. in infinitum
continuatae. “Ad unguem” means literally “to a fingernail” or
“exactly”.
This result is the main conclusion of Archimedes’ On
the Equilibrium of Planes II (see Proposition 8 of the work),
which was established with much effort on Archimedes’ part.
Torricelli is demonstrating how much easier it is to prove this
result using indivisibles.
Giannantonio Rocca (1607-1656) was a mathematical correspondent of Cavalieri
and Galileo, among others. His demonstration was apparently already established by 1628,
two years before the publication of Guldin’s result. (See p. 83 of
Seventeenth-Century Indivisibles Revisited, ed by Vincent Julien.)
He was instrumental in helping Cavalieri formulate a response to Guldin’s
criticism of his work in Cavalieri’s Exercitationes. (See pp.154-155
of Infinitesimal: How a Dangerous Mathematical Theory Shapred the Modern World
by Amir Alexander.) Letters between Cavalieri
and Rocca can be found in Lettere d’ uomini illustri del secolo XVII
a Giannantonio Rocca filosofo e matematico Reggiano con alcune del Rocca a’
medesimi (Moderno 1765).
For a discussion of what a “compounded” ratio is, see the
discussion of Euclid Definition V.9 on pages 132-133 of Heath,
Thomas trans., Euclid: The Thirteen Books of the Elements,
Vol. 2. Dover (1956). Also, the definition of
a moment stems from Archimedes’ Law of the Lever: Given two
masses m and M, the masses will balance on a lever when placed
distances d and D (respectively) from a fulcrum, where d and D
satisfy the reciprocal proportion m/M = D/d. By cross-multiplfying,
we have m d = M D. This product is called the moment of
the system. For, say, the line segment DH the center of gravity is
the midpoint L. Thus, the moment is (algebraically speaking) \( DH
\cdot {1 \over 2} DH = {1 \over 2} DH^2\). When taking the ratio of
the moment of DH to the moment of HF, the “1/2” terms will cancel.
Hence the ratio of the moments of the line segments is the same as
the ratios of the squares. Note that the conclusion of this Lemma is
a version of the Pappus-Guldin Theorem.
This follows from Proposition 21 of Archimedes’ On
Conoids and Spheroids, which states (in the Heath edition) that
“Any segment of a paraboloid of revolution is half as large as the
cone or segment of a cone which has the same base and the same
axis”, after applying Euclid XII.10.
In Proposition XXXI of De Centro Gravitatis
solidorum libri tres (1603), Luca Valerio established that the
center of gravity of a hemisphere was at the point on the axis which
is 3/8 of the way from the center to the circumference–i.e., which
divides the axis in a ratio of 5 to 3.

A Translation of Evangelista Torricelli’s Quadratura Parabolae per novam indivisibilium Geometriam pluribus modis absoluta.

A Translation of Evangelista Torricelli’s Quadratura Parabolae per novam indivisibilium Geometriam pluribus modis absoluta.

Introduction

Some historical background

Overview of the Translation

Some Prerequisites for Reading the Translation

Archimedes’ Quadrature of the Parabola

Torricelli’s Quadrature of the Parabola (One Version)

In the Classroom and Conclusion

In the Classroom

Conclusion

Acknowledgments and References

Acknowledgments

References

Appendix: The Quadrature of the Parabola solved by many methods through the new geometry of indivisibles

Translated by Andrew Leahy and Kasandara Sullivan

Proposition 11.

Lemma 20.

Lemma 21.

Proposition 12.

Proposition 13.

Lemma 22.

Lemma 23.

Proposition 14.

In another way.

Lemma 24.

Lemma 25.

Lemma 26.

Lemma 27.

Comment.

Proposition 15.

In Another Way.

Lemma 28.

Proposition 16.

Lemma 29.

Proposition 17.

Lemma 30.

Proposition 18.

Lemma 31.

Proposition 19.

Lemma 32.

Lemma 33.

Proposition 20.

Proposition 21.

Translation Notes

A Translation of Evangelista Torricelli’s Quadratura Parabolae per
novam indivisibilium Geometriam pluribus modis absoluta.

Appendix: The Quadrature of the Parabola
solved by many methods through the new geometry of indivisibles