2.3 The Area Under a Parabola

If $S_a^2$ is the set of points $(x,y)$ in $\mbox{{\bf R}}^2$ such that $0\leq x\leq a$ and $0 \leq y \leq x^2$, then we showed in (2.6) that

$${ 1^2+2^2+\cdots +(n-1)^2 \over n^3} \leq {\mbox{\rm area}(S_... ...area}(\mbox{\rm cir}(S_a^2))} \leq {1^2+\cdots +n^2 \over n^3}.$$

By (2.9)

$$\begin{eqnarray*} {{1^2+2^2+ \cdots +n^2}\over {n^3}}&=&{{n(n+1)(2n+1)}\over {n^... ...over 3} \left( 1+{1\over n}\right) \left(1+{1\over {2n}}\right). \end{eqnarray*}$$

Also

$${{1^2+2^2+\cdots +(n-1)^2}\over {n^3}} = {{(n-1)n(\left( 2(n-1)+1\right)}\over {n^3\cdot 6}}= {1\over 3} \left( {{n-1}\over n}\right) \left( {{2n-1}\over 2n}\right)\mbox{{}}$$

$$= {1\over 3}\left( 1-{1\over n}\right) \left( 1-{1\over {2n}}\right),$$ (2.12)

so

$${1\over 3}\left( 1-{1\over n}\right) \left( 1-{1\over {2n}}\... ...over 3} \left( 1+{1\over n}\right)\left(1+{1\over {2n}}\right)$$ (2.13)

for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$.

The right side of (2.13) is greater than $${ {{1}\over 3}}$$ and the left side is less than $${ {{1}\over 3}}$$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$, but by taking $n$ large enough, both sides can be made as close to $${ {{1}\over 3}}$$ as we please. Hence we conclude that the ratio $${\mbox{\rm area}(S_a^2) \over \mbox{\rm area}(\mbox{\rm cir}(S_a^2))}$$ is equal to $${ 1\over 3}$$. Thus, we have proved the following theorem:

2.14   Theorem (Area Under a Parabola.) Let $a$ be a positive real number and let $S_a^2$ be the set of points $(x,y)$ in $\mbox{{\bf R}}^2$ such that $0\leq x\leq a$ and $0 \leq y \leq x^2$. Then

$${\mbox{\rm area}(S_a^2) \over \mbox{\rm area}(\mbox{\rm box circumscribed about }S_a^2)} = {1\over 3},$$

i.e.

$$\mbox{\rm area}(S_a^2) = {1\over 3}a^3.$$

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 $S(a,b)$ bounded by the parabola $y=x^2$ and the line joining $P(a) =(a,a^2)$ to $P(b)= (b,b^2)$ where $(a<b)$.

2.15   Exercise. Show that the area of the parabolic segment $S(a,b)$ just described is $${ {{(b-a)^3}\over 6}}$$. Use theorem 2.14 and any results from Euclidean geometry that you need. You may assume that $0<a\leq b$. The cases where $a<0<b$ and $a<b<0$ 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 $(-1,1)$ and $(1,1)$ has the same area as the segment determined by the points $(99,9801)$ and $(101,10201)$, 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 $AB$ is four thirds of the area of the inscribed triangle $ABC$, where $C$ is the point on the parabola at which the tangent line is parallel to $AB$. We cannot verify Archimedes formula at this time, because we do not know how to find the point $C$.

2.16   Exercise. Verify Archimedes' formula as stated in the above remark for the parabolic segment $S(-a,a)$. 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 $y=x$) If $S$ is a subset of $\mbox{{\bf R}}^2$, then the reflection of $S$ about the line $y=x$ is defined to be the set of all points $(x,y)$ such that $(y,x) \in S$.

If $S^\ast$ denotes the reflection of $S$ about the line $y=x$, then $S$ and $S^\ast$ have the same area.

2.18   Exercise. A Let $a \in \mbox{${\mbox{{\bf R}}}^{+}$}$ and let $T_a$ be the set of all points $(x,y)$ such that $0\leq x\leq a$ and $0 \leq y \leq \sqrt{x}$. Sketch the set $T_a$ and find its area.

2.19   Exercise. In the first figure below, the $8 \times 8$ square $ABCD$ has been divided into two $3 \times 8$ triangles and two trapezoids by means of the lines $EF$, $EB$ and $GH$. In the second figure the four pieces have been rearranged to form an $5 \times 13$ rectangle. The square has area $64$ , and the rectangle has area $65$. 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.)