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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{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 \( \times  \) 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, \textit{etc}.). But it's not so simple to find the area
of a circle (those of you thinking ``It's just \( \pi r^{2} \)'' 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 \textit{in the limit} as the number of sides went to infinity, they could
find the area of the circle. 

\vspace{0.3cm}
{\par\centering \resizebox*{4in}{2in}{\includegraphics{exhaustion.eps}} \par}
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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 \( \pi r^{2} \) 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 \( x^{2}+y^{2}=a^{2} \),
or, since we have the top half and so \( y>0 \), 
\[
y=\sqrt{a^{2}-x^{2}}.\]
 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 \textit{inscribed rectangle}), and find the area
(width \( \times  \) height) of each rectangle. Add them up, and you get the
area of the semi-circle. 

\vspace{0.3cm}
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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.

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