Quadrature of a Parabola

Hippocrates, the great Greek physician lived over 2300 years ago. He laid the foundation for constructing the quadrature of a lune. Another great mathematician and thinker, Archimedes came up with many new mathematical ideas that ahead of its time and so new to the public. 

For a lune to be produced, a parabola must be drawn.  The Ancient Greeks viewed Parabolas, circles, ellipses, and hyperbolas as conics. Archemedes had the brilliant idea of trying to figure out how to calculate the area of a curve. Finding the area under a curve had been a ongoing problem that had yet to be solved. Fair trade partly relied on being able to work out volumes of cylinders and spheres. Archimedes worked out some approximations for the area of a circle and the value of π (pi).Through this idea of finding the area of a parabola,  you can take any segment of a parabola cut off by a line AB, and if P is the point on the segment furthest from AB, then the area of the parabolic segment ABP is four-thirds the area of the triangle ABC. (shown below)

Source: http://www.math.mcgill.ca/rags/JAC/NYB/exhaustion2.pdf

This idea turned into Archimedes' proof. It states that the point P furthest from AB is  the vertex of the parabolic segment. Therefore, you can call the triangle ABP the vertex triangle. As stated earlier, Archimedes proved that the area of any parabolic segment is four-thirds the area of its vertex triangle.  

                            Source: http://www.math.mcgill.ca/rags/JAC/NYB/exhaustion2.pdf

The main insight is that when we remove the vertex triangle ABC from the parabolic segment we are left with two smaller parabolic segments which themselves have vertex triangles APP2 and BPP1. 

Removing this layer of two triangles we get four even smaller parabolic segments, whose four vertex triangles AP₂P₃, P₂P P₄, P P₁P_{5, and} P₁B P_6. After this "layer", it's onto the next, and so on. If we kept going and added the areas of an infinite number of such triangles, we would have an exact area for the parabolic segment. The idea of filling out a curved figure such as a parabolic segment with rectilinear figures such as triangles is called the “Method of Exhaustion”. It is the basis of what is now Integral Calculus, but  in the old days was called Quadrature. 

In trying to solve this mathematically, we can start with the area of each sea green triangle which is 1/8 the area of the largest triangle. This is because the green triangle has 1/2 the width of the largest triangle and 1/4 of the height. 

Then, considering the lime green triangles, they will be 1/8 the area of the sea green triangles.
So if we call the area of the largest triangle (ABP) x, then the area of all of the triangles is given by:


x + 2(x/8) + 4(x/64) + 8(x/512) + ...
= x + x/4 + x/16 + x/64 + ...
= x(1 + 1/4 + 1/16 + 1/64 + ...) and so on and so on...

We can use the expression in brackets as an Geometric Series with common ratio r = 1/4 and first term a = 1. The sum of this series is given by:

Sum = a/(1 − r) = 1/(1 − 1/4) = 1/(3/4) = 4/3

This goes back to Archimedes' Proof.

The total area of the triangles (which gives us the area of the parabolic segment) is 4x/3, which is 4/3 the area of the lime green triangle, as Archimedes claimed.

The ideas of rectification, quadrature, and the Method of Exhaustion are absolutely fascinating! Though a bit different from the rectification of a shape, parabolas still show the same general idea which shows it's close relation to an actual shape. I really enjoyed writing this blog!!