When asked to find area of any given triangle, anyone will be able to solve it with the known formula for area of a triangle which is (1/2)(base)(height). Anyone can do the same with a square because the formula for the area of a square is (base)(height). This is basic geometry/algebra that students learn and is part of the curriculum. If you were asked to solve the area of a curved object, would you be able to solve for the area without using Calculus? Would you be able to find the area without using any type of formula? How can the area be solved for any object with curves, like a circle? Greek mathematicians used a technique called the method of exhaustion, which is precursor to Calculus. This method is no longer commonly used to solve problems but this method if very similar to Riemann Integration. The idea of this method is to take an approximation of an object repeatedly to end up with an area, which would otherwise be difficult to find.

The method of exhaustion was developed by Greek mathematicians in order to find the area of a shape. At the time this method was used, there was no known formula for the area of a circle. This method helped find a formula for the area of a circle that we use today. What was needed was to find an approximation for pi. They did not know the constant they needed to multiply by. The way the Greeks would use the method of approximation was to inscribe a polygon with a known area inside the circle. A square inside of a circle is how this might look like if you are trying to imagine it. The next step would be to add a side to the square, to create a pentagon inside the circle. With this process we could solve for the area of the pentagon and then we will know some of the area of the circle. But there is still the area of outside of the pentagon and inside of the circle that is still unknown so we would then repeat the process. The next step would be to add another length to the pentagon to create a hexagon inside of the circle. The Greeks would use this process over and over again until they got the n-polygon close enough to the circle. If this process is correctly constructed, the difference in area between the n-polygon inscribed and that of the circle will become really small. This would result in a more accurate estimation of the area of the circle. The possible values for the area of the circle are systematically exhausted by the polygon.

It is important to remember that Greeks did not have the same tools we have today. How were they able to construct such accurate polygons that were inscribed within the circles? Well the answer is that they used a straight edge and a compass. This would only work for specific polygons. “In 1837, mathematician Carl Gauss was able to prove that a n-polygon could be constructed with a compass and straight edge if and only if n is the product of a power of 2 and any number of the Fermat primes m = 2^{2k},” wrote Chelsea E. DeSouza in *The Greek Method of Exhaustion: Leading the Way to Modern Integration.* The first polygons that would able to be constructed would be 3=2+1, 4=2^{2}, 5=2^{2}+1, 6= 2(2+1), 8 =2^{3}, 10=2(2^{2}+1), 12=2^{2}(3), 16=2^{4}, 2^{4}+1. Not all polygons were able to be constructed with a compass and straight edge.

Though it is a long process, it is possible to find the area of a circle with polygons inscribed within. Can the same be done with other curved shapes, like parabolas? Parabolas can be inscribed with polygons like triangles to evaluate the area of the parabola. In the third century, Archimedes came up with 24 propositions that were about parabolas. At the end of these propositions, there is a proof that says, the region inside of the parabola and a line is 4/3 the area of a triangle inscribed inside the parabola. Archimedes split up the region inside the parabolas into triangles to solve for the area.

Using integration to find the area under a curve could be difficult at times but when you realize what the Greeks developed to find the area, integration seems like a quicker method. The method of exhaustion is a process that took some time to solve. This method is easy to learn because it is just using the area formulas for polygons.

Sources

http://www.math.ubc.ca/~cass/courses/m446-03/exhaustion.pdf

http://arxiv.org/pdf/math/0011078v1.pdf

https://etd.ohiolink.edu/!etd.send_file?accession=osu1338326658&disposition=inline