Math 12 - 2 (Notes)
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Eudoxus' Method of Exhaustion

[Divided Pentagon]
In class we suggested that one way to compute the area of a circle or radius r is to estimate it by an inscribed polygon; the more sides the polygon has, the closer it is to the actual circle. Taking the limit as the number of sides goes to infinity should give the exact answer.

To carry out this process, we need a good way to calculate the area of a polygon with n sides for arbitrary n. A natural approach is to place a point at the center of the polygon and, for each edge of the polygon, look at the triangle that has that edge as its base and the central point as its third vertex. In a polygon with n sides (an n-gon), there are n such triangles. If we add up the areas of all n of these, we get the area of the entire n-gon.

[Right Triangle]
So our first task is to compute the area of one of these triangles. To do that, we need to get the length of its base and height. We can do that using some trigonometry. First, consider a line drawn perpendicular to the edge of the polygon; this is the height of the triangle. Notice that each triangle is an isosceles triangle, so this line divides the triangle into two congruent right triangles. The hypotenuse of one of these triangles is of length r (these lines are radii of the circle since the vertices of the polygon are on the circumference of the circle). We could compute the lengths of the other two sides of the right triangle using sine and cosine if we knew the measure of the angle that is at the center of the circle.

There are 360 degrees, or 2p radians, around the point at the center, and and we have divided this up among n isosceles triangles, so the angle in the isosceles triangle is 2p/n radians; each of these is divided in half to get the angle at the point of the right triangle, so the angle in the right triangle is half of that, or p/n. We know that the sine of an angle is the length of the opposite side divided by the length of the hypotenuse, so the opposite side equals the hypotenuse times the sine; i.e., the base of the right triangle is of length rsin(p/n). Similarly, using cosine, we can compute the length of the other side of the right triangle, which is rcos(p/n).

This tells us that the area of the right triangle is half the product of these two lengths, or 1/2[rsin(p/n)][rcos(p/n)] which is [(r2)/ 2]sin(p/n)cos(p/n). There are two such triangles in each isosceles triangle and n isosceles triangles, so the total area of the n-gon is nr2sin(p/n)cos(p/n). We can simplify this formula using the trigonometric identity sin2q = 2 sinq cosq, or sinq cosq = 1/2sin2q. This means that the area of the n-gon can be written as [(nr2)/ 2]sin(2p/n).

(Note: you can use this formula to get the area of the pentagon from last week's homework. You will have to find the correct radius in order to do it, as we wanted a pentagon with side length 1. This means we need rsin(p/5) to be equal to 1/2, since the base of the right triangle above must be half the side length of the pentagon. Now solve for r and plug it in with n = 5 in the formula above.)

At this point, we have estimated the area of a circle by an n-gon that has area [(nr2)/2]sin(2p/n). To get the exact area of the circle, we need to take the limit as n goes to infinity. I.e., the actual area is

sin( 2p
The next task is to evaluate this limit. Note that as n gets larger, 2p/n gets smaller, so sin(2p/n) goes to zero. Our hope is that this balances the n in nr2/2 so that the limit will exist.

One approach to computing the limit is to make a substitution: let x = 2p/n; then as n goes to infinity, x goes to 0, so we can look at a limit as x0 rather than n. Note that since x = 2p/n, we also have n = 2p/x, so nr2/2 becomes (2p/x)(r2/2) or pr2/x (this is where the pr2 shows up). So we have transformed the limit into

sin(x) = pr2
(since p and r are constants, they can be pulled outside the limit).

But we know from Math 10 that the limit of sin(x)/x as x goes to zero is 1 (graph it on your calculator if you need to see this for yourself), so that means the area is just pr2. Thus we have proven the fact that the area of a circle of radius r is pr2, a fact you were told back in grade school, but probably never knew why was true.

(Technically, we really need to be more careful about justifying the fact that the areas or the polygons really do go to the area of the circle in the limit, and that we aren't missing anything in the circle. Since the area of the inscribed polygon is always less than the area of the circle, the best we can say is that the limit is less than or equal to the area of the circle. One way to finish the proof is to do a similar limit using circumscribed polygons rather than inscribed ones and show that the limit of these areas as we increase the number of sides also goes to pr2. Since the area of the circumscribed polygons are always bigger than the area of the circle, their limit will be greater than or equal to the area of the circle. So we will have shown that the area of the circle is both less than or equal to pr2 and greater than or equal to it, hence it must be exactly equal to it. In other words, we will have squeezed the area of the circle between two limits that are equal, and so the area must be that value.)

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Created: 10 Jan 2000
Last modified: 10 Jan 2000 10:25:56
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