The Method of Mechanical Theorems - Area of A Parabola

Area of A Parabola

As introduced by Aristotle, first consider balancing a triangle with the parabola. The triangle is the region in the x-y plane between the x-axis and the line y = x as x varies from 0 to 1. The parabola is the region in the x-y plane between the x-axis and y = x2, also as x varies from 0 to 1.

Slice the triangle and parabola into vertical slices, one for each value of x. Imagine that the x-axis is a lever, with a fulcrum at x=0. By the law of the lever—the mass times the distance to the fulcrum must be equal in order for two things to balance. The mass of the slice of the triangle at position x is equal to its height, to x, and it is at a distance x from the fulcrum, so that it will balance the corresponding slice of the parabola, of height x2, if it is placed at x = −1, at a distance of 1 on the other side of the fulcrum.

Since each slice balances, the whole parabola balances the whole triangle. This means that if the parabola is hung by a hook from the point x = −1, 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. The intersection of the two medians is at the point x = 2/3, so that the total mass of the triangle can be thought of as pushing down on this point. The total area of the triangle is 1/2, so the total torque exerted by the triangle is 1/2 times the distance to the center of mass, 2/3, which comes out to 1/3. The mass of the parabola, the area of the parabola, must then be 1/3.

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

which is 1/3 by elementary integral calculus.

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.