This is my first attempt to actually go ahead and prove the area of a circle is $\pi r^2$ using the methods shown on calculus. I first used a coordinate system and placed the centre of my circle (with radius $r$) at the origin. Now, we increase the number of sides of the inscribed regular polygon; we note that the number of sides (call it $n$) is the same as the number of isosceles triangle “sectors” that the polygon consists of. If the angle of each of these triangles is $\theta$, then the base of each triangle (or equivalently, one side of the regular polygon) is easily seen to be $r\sin\theta \over \cos\frac{\theta}{2}$. The apothem of the regular polygon is easily seen to be $r\cos\frac{\theta}{2}$. So that the area of each isosceles triangle “sector” is $\frac{1}{2}r^2\sin\theta$. Since there are $n$ triangles, the area of the whole polygon $\frac{1}{2}r^2\sin\theta n$. This can be used as the function take the limit of as it’s variable approaches infinity, however, it has two variables $\theta$ and $n$. So we rewrite $\theta$ as a function of $n$, which is $\theta = –\frac{\pi}{8} n$. Hence our function is $f(n) = – \frac{r^2\sin(\frac{\pi}{8})n}{2}$. This is where I am stumped. I need to find $\lim_{x\to\infty} f(x)$. However, the limit of $\sin x$ as $x$ approaches infinity is undefined. How do you continue from this point on? Thank you in advance.

  • $\begingroup$ Please include some visuals for your proof attempt, as of right now it is very hard to follow your thought process without them. $\endgroup$
    – uriyabsc
    Jul 2, 2023 at 12:02
  • $\begingroup$ If your polygon is divided into $n$ triangles and the area of each of them is $\frac12r^2\sin\theta$, then the area of the polygon is $\frac n2r^2\sin\theta$, rather than $\frac12r^2\sin(n\theta)$. $\endgroup$ Jul 2, 2023 at 12:04
  • $\begingroup$ $\theta=\frac{2\pi}{n}$ $\endgroup$
    – Eric
    Jul 2, 2023 at 12:07
  • $\begingroup$ You are trying to find the area of the circle without using calculus, and using inscribed or circumscribed polygons instead. If, as your question states, you want to use the calculus, then you don't need the approximating polygons. The calculation is immediate in polar coordinates, and even in Cartesian coordinates it is significantly easier than what you've tried. $\endgroup$ Jul 2, 2023 at 13:10
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    $\begingroup$ Because\begin{align}\lim_{n\to\infty}n\sin\frac n\pi&=\pi\lim_{n\to\infty}\frac{\sin(\pi/n)}{\pi/n}\\&=\pi.\end{align} $\endgroup$ Jul 2, 2023 at 13:27