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Deriving the Area of a Circle

During the 1998-99 school year I was teaching math at Ellsworth High School in Maine.  I got to talking with one of the other math teachers there about deriving the area of a circle.  We started approaching the idea from a similar perspective as Archimedes had 2300 years ago in his work with circles and $\pi$.  We inscribed a polygon in a circle and then took the limit of the area of the polygon as the number of sides ($n$) approached infinity.  For the area of the polygon, we divided it into $n$ triangles and used the area formula $A=\frac {1}{2} ab\sin C$.

In this diagram, both $a$ and $b$ are radii of the circle and are labeled as $R$.  The central angle $\theta$ which will be used to find the area of the triangle is actually equal to $\frac {360^\circ}{n}$ or (and this will be VERY important later) $\frac {2\pi}{n}$.

So the area of the polygon is $n$ times the area of the triangle, or:

$n*\frac {1}{2} R*R*\sin \frac{2 \pi}{n}$

For this to be the area of the circle we want to take the limit of the previous expression as $n \to \infty$.

$A=\lim_{n \to \infty} n*\frac {1}{2} R*R*\sin \frac {2 \pi}{n}$

Since the $\frac {1}{2}*R*R$ don’t involve $n$, we can pull them out from the limit.

$A=\frac {1}{2} R^2*\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$

Now comes the calculus – we were just talking about the derivative of the sine function in Calculus I last week, which is what made me think about this.  Remember that:

$\lim_{x \to 0} \frac{\sin x}{x}=1$

So, we need to establish this limit somewhere in our derivation and then replace it with $1$.  The most likely place for this would be in here:

$\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$

but we need to have:

$\lim_{\frac{2 \pi}{n} \to 0} \frac{\sin \frac {2 \pi}{n}}{\frac{2 \pi}{n}}$

OK – so, let’s multiply our origninal expression by $\frac{2 \pi}{2 \pi}$.

$\lim_{n \to \infty} \frac{2 \pi}{2 \pi}*n*\sin \frac {2 \pi}{n}$

or

$\lim_{n \to \infty} 2 \pi*\frac{n}{2 \pi}*\sin \frac {2 \pi}{n}$

which is equivalent to

$\lim_{n \to \infty} 2 \pi*\frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

Let’s bring back the whole expression:

$A=\frac {1}{2} R^2*\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$

$A=\frac {1}{2} R^2*\lim_{n \to \infty} 2 \pi*\frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

$A=\frac {1}{2} R^2*2 \pi \lim_{n \to \infty} \frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

as $n \to \infty$,

$\frac{2 \pi}{n} \to 0$

$A=\frac {1}{2} R^2*2 \pi \lim_{\frac{2 \pi}{n} \to 0} \frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

$A=\frac {1}{2} R^2*2 \pi *1$

or

$A=\pi R^2$

When we were working through this the first time, we kept the $360^\circ$ in the formula and ended up with $A=180^\circ R^2$, which didn’t make any sense until we realized that $180^\circ$ is $\pi$ radians!

Archimedes’ conclusion was that the area of the circle was equal to one-half the radius times the circumference:

$\frac{1}{2} R*2\pi R = \pi R^2$

or, since the mathematicians in the Greek tradition like Archimedes thought in terms of geometry – the area of a circle is the same as the area of triangle with height equal to the radius of the circle and base equal to the circumference of the circle!