On-line Math 21

1.1 The area problem

Originally, the Greeks used this idea to find the area of a circle, called the method of exhaustion. It is geometrically clear how to find the area of a square (width × height), and if you think about it, by breaking it up into triangles, you can find the area of any regular polygon (like a pentagon, a hexagon, and octagon, etc.). But it's not so simple to find the area of a circle (those of you thinking ``It's just pr² '' are missing the problem of where that formula came from). What they did was to take regular polygons, inscribed in a circle, measure the area of each one, and take progressively more sides (which gives a polygon that looks progressively more like a circle), and in the limit as the number of sides went to infinity, they could find the area of the circle.

For now, that idea of ``in the limit'' should just mean you take more and more sides, finding the area of each polygon. Those areas are getting closer to pr² as the number of sides increases.

We don't have to deal with those hard-to compute areas of polygons; we can instead come up with an equivalent, but more modern, way to do the same measurement. We can use just the top half of the circle, and double it to get the whole area. If the circle has radius a , then the circles has equation x²+y² = a² , or, since we have the top half and so y > 0 ,

y =
Ö

a²-x²

.

We then chop up the x -axis, from -a to a , into a bunch of sections. Over each section we can construct the largest rectangle completely under the curve (called an inscribed rectangle), and find the area (width × height) of each rectangle. Add them up, and you get the area of the semi-circle.

Well, approximately. In order to get the actual area, precisely, and not an approximation, you need to repeat the process, over and over, with more and more rectangles that are skinnier and skinnier, and in the limit you will find the area. That is the idea of integration.

File translated from T_EX by T_TH, version 2.61.
On 2 May 2000, 11:53.