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This discovery had a significant impact on the understanding of ancient Greek mathematics, for two reasons. The first is the aim of the paper, summarized by Archimedes[2] [54, Vol. 2, p. 221]:

Moreover, seeing in you, as I say, a zealous student and a man of considerable eminence in philosophy, who gives due honour to mathematical inquiries when they arise, I have thought fit to write out for you and explain in detail in the same book the peculiarity of a certain method, with which furnished you will be able to make a beginning in the investigation by mechanics of some of the problems in mathematics. I am persuaded that this method is no less useful even for the proofs the theorems themselves. For some things first became clear to me by mechanics, though they had later to be proved geometrically owing to the fact that investigation by this method does not amount to actual proof; but it is, of course, easier to provide the proof when some knowledge of the things sought has been acquired by this method rather than to seek it with no prior knowledge.

This is a radical divergence from all other extant Greek works, as T.L. Heath explains [6, Supplement, p. 6]:

Nothing is more characteristic of the classical works of the great geometers of Greece, or more tantalising, than the absolve of any indication of the steps by which they worked their ways to the discovery of their great theorems. As they have come down to us, these theorems are finished masterpieces which leave no traces of any rough-hewn stage, no hint of the method by which they were evolved ... A partial exception is now furnished by The Method; for here we have a sort of lifting of the veil, a glimpse of the interior of Archimedes' workshop as it were.

The other surprising aspect of The Method is the revelation that Archimedes worked with infinitesimals, for example, "The triangle GammaZA is composed of the straight lines drawn in GammaZA" [54, Vol. 2, p. 227], "The cylinder, the sphere and the cone being filled by circles thus taken." [8, vol. 3, p. 91], see [1] [42]. As every mathematician knows, infinitesimals were reinvented by mathematicians such as Cavalieri (1598-1647) and Leibniz (1646-1716), see [2] [21]. Archimedes used them to compute the area and volumes of various geometrical figures including what he considered his greatest achievement:[3]

Any cylinder having for its base the greatest of the circles in the sphere, and having its height equal to the diameter of the sphere, is one-and-a-half times the sphere, a result he subsequently proved, On the Sphere and Cylinder, I, Corollary to Proposition 34 [54, Vol. 2, p. 125]. Archimedes understood that his method does not produce valid proofs due to its use of infinitesimals, [4] though it is unclear if the same is true of his successors. In any case, it is easy to make the arguments rigorous, given present knowledge. The basic ideas of The Method are still presented in contemporary calculus courses [35] [51, p. 709], and a physical model of Archimedes's argument has been built [25].

On the other hand, Archimedes's On Spirals is a masterpiece of rigorous mathematics. In this paper, Archimedes computes the area and tangent of a spiral, and, in doing so, derives much of the Calculus I curriculum, including related rates, limits, tangents, and the evaluation of Riemann sums. This is reflected by the fact that a number of contemporary Calculus texts outline the basic idea of Archimedes's computation of the area of a spiral [3, p. 3] [12, p. 75], though both these works avoid technical difficulties by substituting a parabola, but then incorrectly imply that Archimedes used such an approach for the parabola [3, p. 8] [12, p. 75]. The considerable length of the paper is a consequence of proving these results from basic principles. Unfortunately, it does not yet have a faithful English translation [56]; Heath's intent in [6] was to capture the modern flavor of Archimedes's works in order to make them more accessible. A generally faithful French translation, including the Greek text, is available [8].

The mechanical method does not seem to produce directly the area of a spiral, or even the area of a circle (also computed by Archimedes), so one might wonder how he first derived them. W.R. Knorr [40] has suggested that the writings of Pappus of Alexandria (fourth century AD) indicate that Archimedes wrote an earlier version of On Spirals which used a different argument to compute the area of the spiral, but then rejected it as inelegant (this approach is developed in the solution to Exercise 1). The object of the next section is to show how a natural extension of the mechanical method easily produces these results.

Weighing a Spiral

The Method relies on a mechanical analogy by using a balance to compare objects. This requires a few simple assumptions and facts about the properties of a lever, which are developed (sometimes implicitly) in Archimedes's On the Equilibrium of Planes I, [6] [18, Chapter IX]. These can be summarized by

Assumption 1. Two objects will balance each other if the distances of their centers of gravity to the fulcrum are inversely proportional to their weights. The center of gravity of an object lies on an axis of symmetry.

When only the weight of an object is relevant to an argument, I will place it on a pan suspended from the balance. The object and any of its sections will then be assumed to have their centers of gravity at the point where the pan is suspended. I will also make extra assumptions not seen in Archimedes's works (however, see the solution to Exercise 4).

Assumption 2. A plane figure is composed of circular arcs with common center, and each circular arc weighs the same as a line segment of equal length.

Exercise 1. What happens if you instead decompose plane figures into radii with common center?

I will first show how the mechanical method can be used to derive Archimedes's formula for the area of a circle given in Measurement of the Circle, Proposition 1 [54, Vol. 1, p. 317].[5]

Proposition 1. Any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the base is equal to the circumference.

Exercise 2. Explain why Proposition 1 is equivalent to the familiar formula: Area of a circle = piR[sup 2].

Suspend two pans on opposite sides of a balance and at equal distances to the fulcrum. On one pan, place a circle with center at A and radius AB, on the other place a line segment CD of length AB. By Assumption 2, the circle is composed of circumferences with center A and radius AE for any E lying on AB. For each such circumference, place a line segment FG perpendicular to CD, of length the circumference through E, such that its endpoint F lies on CD and CF is equal to AE. By Assumption 2, the line segment FG is in equilibrium with the circumference through E. The resulting figure is a right triangle of height AB, base the circumference through B, and it balances a circle of radius AB, which is the statement of Proposition 1.

Exercise 3. Why is the resulting figure in this construction a triangle?

Exercise 4. Generalize the following heuristic from The Method [6, Supplement]: "... judging from the fact that any circle is equal to a triangle with base equal to the circumference and height equal to the radius of the circle, I apprehended that, in like manner, any sphere is equal to a cone with base equal to the surface of the sphere and height equal to the radius."

Archimedes's definition of a spiral and its relevant components is given by [54, Vol. 2, p. 183]:

1. If a straight line drawn in a plane revolve uniformly

any number of times about a fixed extremity until it

return to its original position, and if, at the same

time as the line revolves, a point move uniformly along

the straight line, beginning at the fixed extremity, the

point will describe a spiral in the plane.

2. Let the extremity of the straight line which remains

fixed while the straight line revolves be called the

origin of the spiral.

3. Let the position of the line, from which the straight

line began to revolve, be called the initial line of the

revolution.

Proposition 2. The area inside a spiral anywhere within its first revolution is one third the sector of a circle with center at the origin of the spiral, radius equal to the distance of the point describing the spiral to the origin, and angle equal to the angle between the line and the initial line. (Archimedes gave areas for complete revolutions only, but his proof also applies to this case.)

Consider a spiral with origin A, initial line AB, and C the position of the point describing the spiral. Consider also a balance arm DE of length twice AC and let the midpoint F of DE be the fulcrum. On this balance suspend a pan from D and place the spiral region in the pan.

By Assumption 2, the spiral region is composed of arcs GH for each G lying on AC, where H is the intersection of the circle with center A and radius AG and the spiral. Extend AH to intersect the circle of center A and radius AC at I. Consider a line segment JK of length equal to the arc CI and crossing DE at L such that JK and DE are perpendicular, L is the midpoint of JK, and FL is equal to AG. I claim that JK and the arc GH are in equilibrium. To see this note that, by Exercise 3, the length of an arc is proportional to its radius, so that

arc GH : arc CI :: AG : AC,

and the result follows from the assumption that the arc CI has its center of gravity at D and from Assumption 1. Now extend the arc CI to intersect AB at M; then the arc CI is equal to the arc CIM minus the arc IM, and by the definition of spiral, IM is proportional to AH. Since the arc CIM remains constant in this argument, the second part of Exercise 3 shows that the arc CIM minus JK is proportional to FL, which means that the resulting figure is an isosceles triangle which balances the inside of the spiral.

The exact same argument shows that the area between the spiral and the initial line that lies within the same sector balances the same isosceles triangle, but reversed so that, its vertex lies on the fulcrum. The crucial step is to recall the following

Fact: The center of gravity of a triangle lies at the intersection of the medians, and the medians of a triangle intersect each other in a ratio of 2:1.

The first part is suggested by the observation that a median divides a triangle into two triangles of equal weight. and its proof is one of the main results of On the Equilibrium of Planes L The second part is an easy exercise [15, Section/Sections1.4] and follows from On the Equilibrium of Planes I, Proposition 15, generalized to trapezoids.

This shows that reversing the first triangle places the center of gravity twice as far from the fulcrum, so the second triangle will balance twice the first. One concludes that the inside of the spiral weight one half the outside of the spiral and thus one third of the sector of the circle, which is the statement of Proposition 2.

Exercise 5. Evaluate the area of the spiral using the same procedure as for the circle, i.e., by only comparing weights placed on pans.

Exercise 6. Use the mechanical method to compute the Center of gravity of a spiral region.

A Modern Translation

The basic observation is that Assumption 2 extends Archimedes's method to polar coordinates. Consider a curve r = f(theta) in polar coordinates, where, for simplicity, f(r) is an increasing function, so there is an inverse function theta = g(r) (this notation is more convenient given the difficulties of Exercise 1). To compute the area of a region A lying inside the curve and having 0 </= theta </= Theta, one partitions A into thin circular shells of width h > 0, as in Figure 6. Using the formula thetar[sup 2]/2 for the area of a sector of angle theta and radius r, each shell has area (Theta - theta)rh + R(r, h), where the error R(r, h) is less than the area of the small shell element of area (r + h)h(g(r + h) - g(r)), see Figure 6, and this is less than Crh[sup 2], for some constant C, assuming that g(r) is well behaved. It follows that, ignoring terms of order h[sup 2], the area of each shell is (Theta - theta)rh, which is the length of the bottom arc of the shell multiplied by h. This shows why the first part of Assumption 2 holds. All these shells have area a linear function of h up to an error term of lower order, and form a disjoint union of A, which shows why the second part of Assumption 2 holds. Letting h arrow right 0, it follows that the area of A is

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The standard derivation of this formula uses the formula rdrd theta for the area element in polar coordinates

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A circle is simply g(r) = 0, which yields 2pi "[Greek text cannot be converted in ASCI text]"[sup R, sub 0] rdr = piR[sup 2].

A spiral, in polar coordinates, is given by the equation r = atheta, for some constant a, which can be written as theta = kr, where k = 1/a. By the above, the area of the spiral is

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