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# 2.3 The Area Under a Parabola

If is the set of points in such that and , then we showed in (2.6) that

By (2.9)

Also
 (2.12)

so
 (2.13)

for all .

The right side of (2.13) is greater than and the left side is less than for all , but by taking large enough, both sides can be made as close to as we please. Hence we conclude that the ratio is equal to . Thus, we have proved the following theorem:

2.14   Theorem (Area Under a Parabola.) Let be a positive real number and let be the set of points in such that and . Then

i.e.

Remark: The last paragraph of the proof of theorem 2.14 is a little bit vague. How large is large enough'' and what does as close as we please'' mean? Archimedes and Euclid would not have considered such an argument to be a proof. We will reconsider the end of this proof after we have developed the language to complete it more carefully. (Cf Example 6.54.)

The first person to calculate the area of a parabolic segment was
Archimedes (287-212 B.C.). The parabolic segment considered by Archimedes corresponds to the set bounded by the parabola and the line joining to where .

2.15   Exercise. Show that the area of the parabolic segment just described is . Use theorem 2.14 and any results from Euclidean geometry that you need. You may assume that . The cases where and are all handled by similar arguments.

The result of this exercise shows that the area of a parabolic segment depends only on its width. Thus the segment determined by the points and has the same area as the segment determined by the points and , even though the second segment is 400 times as tall as the first, and both segments have the same width. Does this seem reasonable?

Remark: Archimedes stated his result about the area of a parabolic segment as follows. The area of the parabolic segment cut off by the line is four thirds of the area of the inscribed triangle , where is the point on the parabola at which the tangent line is parallel to . We cannot verify Archimedes formula at this time, because we do not know how to find the point .

2.16   Exercise. Verify Archimedes' formula as stated in the above remark for the parabolic segment . In this case you can use your intuition to find the tangent line.

The following definition is introduced as a hint for exercise 2.18

2.17   Definition (Reflection about the line ) If is a subset of , then the reflection of about the line is defined to be the set of all points such that .

If denotes the reflection of about the line , then and have the same area.

2.18   Exercise. A Let and let be the set of all points such that and . Sketch the set and find its area.

2.19   Exercise. In the first figure below, the square has been divided into two triangles and two trapezoids by means of the lines , and . In the second figure the four pieces have been rearranged to form an rectangle. The square has area , and the rectangle has area . Where did the extra unit of area come from? (This problem was taken from W. W. Rouse Ball's Mathematical Recreations [4, page 35]. Ball says that the earliest reference he could find for the problem is 1868.)

Next: 2.4 Finite Geometric Series Up: 2. Some Area Calculations Previous: 2.2 Some Summation Formulas   Index
Ray Mayer 2007-09-07