The Method of Mechanical Theorems (Greek:
Περὶ μηχανικῶν θεωρημάτων
πρὸς Ἐρατοσθένη ἔφοδος),
also referred to as The Method, is considered
one of the major surviving works of the ancient
Greek polymath Archimedes. The Method takes
the form of a letter from Archimedes to Eratosthenes,
the chief librarian at the Library of Alexandria,
and contains the first attested explicit use
of indivisibles (sometimes referred to as
infinitesimals). The work was originally thought
to be lost, but in 1906 was rediscovered in
the celebrated Archimedes Palimpsest. The
palimpsest includes Archimedes' account of
the "mechanical method", so-called because
it relies on the law of the lever, which was
first demonstrated by Archimedes, and of the
center of mass (or centroid), which he had
found for many special shapes.
Archimedes did not admit the method of indivisibles
as part of rigorous mathematics, and therefore
did not publish his method in the formal treatises
that contain the results. In these treatises,
he proves the same theorems by exhaustion,
finding rigorous upper and lower bounds which
both converge to the answer required. Nevertheless,
the mechanical method was what he used to
discover the relations for which he later
gave rigorous proofs.
== Area of a parabola ==
To explain Archimedes' method today, it is
convenient to make use of a little bit of
Cartesian geometry, although this of course
was unavailable at the time. His idea is to
use the law of the lever to determine the
areas of figures from the known center of
mass of other figures. The simplest example
in modern language is the area of the parabola.
Archimedes uses a more elegant method, but
in Cartesian language, his method is calculating
the integral
∫
0
1
x
2
d
x
=
1
3
,
{\displaystyle \int _{0}^{1}x^{2}\,dx={\frac
{1}{3}},}
which can easily be checked nowadays using
elementary integral calculus.
The idea is to mechanically balance the parabola
(the curved region being integrated above)
with a certain triangle that is made of the
same material. The parabola is the region
in the x-y plane between the x-axis and y
= x2 as x varies from 0 to 1. The triangle
is the region in the x-y plane between the
x-axis and the line y = x, also as x varies
from 0 to 1.
Slice the parabola and triangle into vertical
slices, one for each value of x. Imagine that
the x-axis is a lever, with a fulcrum at x
= 0. The law of the lever states that two
objects on opposite sides of the fulcrum will
balance if each has the same torque, where
an object's torque equals its weight times
its distance to the fulcrum. For each value
of x, the slice of the triangle at position
x has a mass equal to its height x, and is
at a distance x from the fulcrum; so it would
balance the corresponding slice of the parabola,
of height x2, if the latter were moved to
x = −1, at a distance of 1 on the other
side of the fulcrum.
Since each pair of slices balances, moving
the whole parabola to x = −1 would balance
the whole triangle. This means that if the
original uncut parabola is hung by a hook
from the point x = −1 (so that the whole
mass of the parabola is attached to that point),
it will balance the triangle sitting between
x = 0 and x = 1.
The center of mass of a triangle can be easily
found by the following method, also due to
Archimedes. If a median line is drawn from
any one of the vertices of a triangle to the
opposite edge E, the triangle will balance
on the median, considered as a fulcrum. The
reason is that if the triangle is divided
into infinitesimal line segments parallel
to E, each segment has equal length on opposite
sides of the median, so balance follows by
symmetry. This argument can be easily made
rigorous by exhaustion by using little rectangles
instead of infinitesimal lines, and this is
what Archimedes does in On the Equilibrium
of Planes.
So the center of mass of a triangle must be
at the intersection point of the medians.
For the triangle in question, one median is
the line y = x/2, while a second median is
the line y = 1 − x. Solving these equations,
we see that the intersection of these two
medians is above the point x = 2/3, so that
the total effect of the triangle on the lever
is as if the total mass of the triangle were
pushing down on (or hanging from) this point.
The total torque exerted by the triangle is
its area, 1/2, times the distance 2/3 of its
center of mass from the fulcrum at x = 0.
This torque of 1/3 balances the parabola,
which is at a distance -1 from the fulcrum.
Hence, the area of the parabola must be 1/3
to give it the opposite torque.
This type of method can be used to find the
area of an arbitrary section of a parabola,
and similar arguments can be used to find
the integral of any power of x, although higher
powers become complicated without algebra.
Archimedes only went as far as the integral
of x3, which he used to find the center of
mass of a hemisphere, and in other work, the
center of mass of a parabola.
== First proposition in the palimpsest ==
Consider the parabola in the figure to the
right. Pick two points on the parabola and
call them A and B.
Suppose the line segment AC is parallel to
the axis of symmetry of the parabola. Further
suppose that the line segment BC lies on a
line that is tangent to the parabola at B.
The first proposition states:
The area of the triangle ABC is exactly three
times the area bounded by the parabola and
the secant line AB.Proof:Let D be the midpoint
of AC. Construct a line segment JB through
D, where the distance from J to D is equal
to the distance from B to D. We will think
of the segment JB as a "lever" with D as its
fulcrum. As Archimedes had previously shown,
the center of mass of the triangle is at the
point I on the "lever" where DI :DB = 1:3.
Therefore, it suffices to show that if the
whole weight of the interior of the triangle
rests at I, and the whole weight of the section
of the parabola at J, the lever is in equilibrium.
Consider an infinitely small cross-section
of the triangle given by the segment HE, where
the point H lies on BC, the point E lies on
AB, and HE is parallel to the axis of symmetry
of the parabola. Call the intersection of
HE and the parabola F and the intersection
of HE and the lever G. If the whole weight
of the triangle rests at I, it exerts the
same torque on the lever JB as it does on
HE. Thus, we wish to show that if the weight
of the cross-section HE rests at G and the
weight of the cross-section EF of the section
of the parabola rests at J, then the lever
is in equilibrium. In other words, it suffices
to show that EF :GD = EH :JD. But that is
a routine consequence of the equation of the
parabola. Q.E.D.
== Volume of a sphere ==
Again, to illuminate the mechanical method,
it is convenient to use a little bit of coordinate
geometry. If a sphere of radius 1 is placed
with its center at x = 1, the vertical cross
sectional radius
ρ
S
{\displaystyle \rho _{S}}
at any x between 0 and 2 is given by the following
formula:
ρ
S
(
x
)
=
x
(
2
−
x
)
.
{\displaystyle \rho _{S}(x)={\sqrt {x(2-x)}}.}
The mass of this cross section, for purposes
of balancing on a lever, is proportional to
the area:
π
ρ
S
(
x
)
2
=
2
π
x
−
π
x
2
.
{\displaystyle \pi \rho _{S}(x)^{2}=2\pi x-\pi
x^{2}.}
Archimedes then considered rotating the triangular
region between y = 0 and y = x and x = 2 on
the x-y plane around the x-axis, to form a
cone. The cross section of this cone is a
circle of radius
ρ
C
{\displaystyle \rho _{C}}
ρ
C
(
x
)
=
x
{\displaystyle \rho _{C}(x)=x}
and the area of this cross section 
is
π
ρ
C
2
=
π
x
2
.
{\displaystyle \pi \rho _{C}^{2}=\pi x^{2}.}
So if slices of the cone and the sphere both
are to be weighed together, the combined cross-sectional
area is:
M
(
x
)
=
2
π
x
.
{\displaystyle M(x)=2\pi x.}
If the two slices are placed together at distance
1 from the fulcrum, their total weight would
be exactly balanced by a circle of area
2
π
{\displaystyle 2\pi }
at a distance x from the fulcrum on the other
side. This means that the cone and the sphere
together, if all their material were moved
to x = 1, would balance a cylinder of base
radius 1 and length 2 on the other side.
As x ranges from 0 to 2, the cylinder will
have a center of gravity a distance 1 from
the fulcrum, so all the weight of the cylinder
can be considered to be at position 1. The
condition of balance ensures that the volume
of the cone plus the volume of the sphere
is equal to the volume of the cylinder.
The volume of the cylinder is the cross section
area,
2
π
{\displaystyle 2\pi }
times the height, which is 2, or
4
π
{\displaystyle 4\pi }
. Archimedes could also find the volume of
the cone using the mechanical method, since,
in modern terms, the integral involved is
exactly the same as the one for area of the
parabola. The volume of the cone is 1/3 its
base area times the height. The base of the
cone is a circle of radius 2, with area
4
π
{\displaystyle 4\pi }
, while the height is 2, so the area 
is
8
π
/
3
{\displaystyle 8\pi /3}
. Subtracting the volume of the cone from
the volume of the cylinder gives the volume
of the sphere:
V
S
=
4
π
−
8
3
π
=
4
3
π
.
{\displaystyle V_{S}=4\pi -{8 \over 3}\pi
={4 \over 3}\pi .}
The 
dependence of the volume of the sphere on
the radius is obvious from scaling, although
that also was not trivial to make rigorous
back then. The method then gives the familiar
formula for the volume of a sphere. By scaling
the dimensions linearly Archimedes easily
extended the volume result to spheroids.
Archimedes argument is nearly identical to
the argument above, but his cylinder had a
bigger radius, so that the cone and the cylinder
hung at a greater distance from the fulcrum.
He considered this argument to be his greatest
achievement, requesting that the accompanying
figure of the balanced sphere, cone, and cylinder
be engraved upon his tombstone.
== Surface area of a sphere ==
To find the surface area of the sphere, Archimedes
argued that just as the area of the circle
could be thought of as infinitely many infinitesimal
right triangles going around the circumference
(see Measurement of the Circle), the volume
of the sphere could be thought of as divided
into many cones with height equal to the radius
and base on the surface. The cones all have
the same height, so their volume is 1/3 the
base area times the height.
Archimedes states that the total volume of
the sphere is equal to the volume of a cone
whose base has the same surface area as the
sphere and whose height is the radius. There
are no details given for the argument, but
the obvious reason is that the cone can be
divided into infinitesimal cones by splitting
the base area up, and the each cone makes
a contribution according to its base area,
just the same as in the sphere.
Let the surface of the sphere be S. The volume
of the cone with base area S and height r
is
S
r
/
3
{\displaystyle \scriptstyle Sr/3}
, which must equal the volume of the sphere:
4
π
r
3
/
3
{\displaystyle \scriptstyle 4\pi r^{3}/3}
. Therefore, the surface area of the sphere
must be
4
π
r
2
{\displaystyle 4\pi r^{2}}
, or "four times its largest circle". Archimedes
proves this rigorously in On the Sphere and
Cylinder.
== Curvilinear shapes with rational volumes
==
One of the remarkable things about the Method
is that Archimedes finds two shapes defined
by sections of cylinders, whose volume does
not involve π, despite the shapes having
curvilinear boundaries. This is a central
point of the investigation—certain curvilinear
shapes could be rectified by ruler and compass,
so that there are nontrivial rational relations
between the volumes defined by the intersections
of geometrical solids.
Archimedes emphasizes this in the beginning
of the treatise, and invites the reader to
try to reproduce the results by some other
method. Unlike the other examples, the volume
of these shapes is not rigorously computed
in any of his other works. From fragments
in the palimpsest, it appears that Archimedes
did inscribe and circumscribe shapes to prove
rigorous bounds for the volume, although the
details have not been preserved.
The two shapes he considers are the intersection
of two cylinders at right angles, which is
the region of (x, y, z) obeying:
(2Cyl)
x
2
+
y
2
<
1
y
2
+
z
2
<
1
{\displaystyle x^{2}+y^{2}<1\;\;\;y^{2}+z^{2}<1}
and the circular prism, which is the region
obeying:
(CirP)
x
2
+
y
2
<
1
0
<
z
<
y
.
{\displaystyle x^{2}+y^{2}<1\;\;\;\;\;0<z<y.}
Both problems have a slicing which produces
an easy integral for the mechanical method.
For the circular prism, cut up the x-axis
into slices. The region in the y-z plane at
any x is the interior of a right triangle
of side length
1
−
x
2
{\displaystyle \scriptstyle {\sqrt {1-x^{2}}}}
whose area 
is
1
/
2
(
1
−
x
2
)
{\displaystyle \scriptstyle 1/2(1-x^{2})}
, so that the total volume is:
(CirP)
∫
−
1
1
1
2
(
1
−
x
2
)
d
x
{\displaystyle \displaystyle \int _{-1}^{1}{1
\over 2}(1-x^{2})\,dx}
which can be easily rectified using the mechanical
method. Adding to each triangular section
a section of a triangular pyramid with area
x
2
/
2
{\displaystyle \scriptstyle x^{2}/2}
balances a prism whose cross section is constant.
For the intersection of two cylinders, the
slicing is lost in the manuscript, but it
can be reconstructed in an obvious way in
parallel to the rest of the document: if the
x-z plane is the slice direction, the equations
for the cylinder give that
x
2
<
1
−
y
2
{\displaystyle \scriptstyle x^{2}\,<\,1-y^{2}}
while
z
2
<
1
−
y
2
{\displaystyle \scriptstyle z^{2}\,<\,1-y^{2}}
, which defines a region which is a square
in the x-z plane of side length
2
1
−
y
2
{\displaystyle \scriptstyle 2{\sqrt {1-y^{2}}}}
, so that the total volume is:
(2Cyl)
∫
−
1
1
4
(
1
−
y
2
)
d
y
.
{\displaystyle \displaystyle \int _{-1}^{1}4(1-y^{2})\,dy.}
And this is the same integral as for the previous
example.
== Other propositions in the palimpsest ==
A series of propositions of geometry are proved
in the palimpsest by similar arguments. One
theorem is that the location of a center of
mass of a hemisphere is located 5/8 of the
way from the pole to the center of the sphere.
This problem is notable, because it is evaluating
a cubic integral.
== See also ==
Archimedes Palimpsest
Method of indivisibles
Method of exhaustion
== Notes
