Archimedes and the parabolic segment area
How the Hellenistic genius computed the area of the parabolic segment without Calculus
Who does not know who Archimedes is. There are many anecdotes on him ranging from the famous “Eureka!” to the burning mirrors. We are in 300 BC, perhaps the most intellectually prosperous historical period in Europe until the end of the Middle Ages thanks to the Hellenistic culture. Archimedes was the prince of Hellenistic science, an undisputed genius able to range in every field of knowledge. For this reason, maybe, many of his works in the following centuries were lost because they were not understood by posterity and therefore not transcribed.
I will describe an Archimedes’ result in geometry: how he was able to compute the area of a parabolic segment without the mathematical equipment, the Calculus, born centuries later thanks to Newton and Leibtniz. I still remind when I was at the third year of high school and the math teacher shown the formula for computing such area, so simple and beauty! But he did not explain how the formula came from, and only two years later I was able to prove it by mean of Calculus but still it remained a mystery for me for many years how Archimedes did the computation.
The theorem
First of all, let’s clarify what is a parabolic segment. Given a parabola and a straight line intersecting it, the region of the plane between the parabola and the segment defined by the two intersection points is the parabolic segment.
The green region highlighted in the above picture is the parabolic segment.
The theorem states: the area of the parabolic segment is four over three times the area of the triangle having as vertices the two intersecting points described above and as third one the point of the parabola for which pass the tangent line to it parallel to the segment identified by the previous two points.
The Calculus proof
The way we can easier demonstrate such theorem is by using a bit of Calculus as taught at the end of high school. Let us start considering, without losing generality, a Cartesian reference frame and a parabola with its vertex at the origin. We can cast any other setting to this one via at most a couple of translations, one vertical and one horizontal (I’m assuming the axis of the parabola to be vertical, otherwise we should even use a rotation to report us to the simple setting).
The idea of the proof is to compute the area of the trapezoid defined by the AB segment and the X axis, and subtract from it the area between the parabola and the X axis again, obtaining the area of the parabolic segment (see the picture below). After that we will find the point C that identifies the triangle by imposing the tangent condition, and finally we can calculate the area of the triangle and show that it is four over three times the area of the parabolic segment.
The beautiful proof of Archimedes
We have seen how the Calculus proof is straightforward, but it requires several mathematical concepts absolutely not known to Archimedes. At his time many geometric theorems were demonstrated in the same way we still learn Euclidian geometry at school; Archimedes mastered them and in conjunction with the Method of Exaustion he was able to calculate the area of the parabolic segment. The beauty of his demonstration is in the way he applied the Method of Exaustion anticipating by a couple of millennia the idea of limit of a convergent infinite succession. By using the same approach Archimedes was able to also compute the area of the circle, of the sphere, of the spiral that takes his name and so on.
Archimedes first demonstrated that once the triangle of his theorem is constructed, two smaller parabolic segments will remain out of the original parabolic segment. Repeating the same triangle construction in such smaller parabolic segments, Archimedes shown that the new triangles have an area of one over eight the one of the triangles of the previous construction step. Using such relationship, Archimedes was able to proof by contradiction the theorem. I’ll provide in the first step the demonstration assuming known the relationship between the area of triangles of subsequent construction iterations, and only after I’ll give you the demonstration of this fact that relies on other theorems of Euclidean geometry known at that time.
We will deserve now a bit of Euclidean geometry to show you that the area of the triangle constructed at a given iteration is equal to one over eigth the area of the triangle constructed at the previous iteration. During the demonstration I’ll use some facts that were known to Archimedes and that I’ll fully demonstrated in a subsequent stage in order to do not break the logical reasoning.
In this last section I’m going to demonstrate the two facts used above, but for doing that we need to show a couple of lemmas before.
