If is the set of points
in
such that
and
, then we showed in (2.6) that
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:
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
.
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
.
The following definition is introduced as a hint for exercise 2.18
If denotes the reflection of
about the line
, then
and
have the same area.