TITLE: Proving that the area of a circle is $\pi r^2$ without using $\lim\limits_{\theta \to 0}\frac{\sin\theta}{\theta}=1$ (or vice versa)
QUESTION [6 upvotes]: $\def\d{\mathrm{d}}$I've seen countless proofs for the area of a circle. There's the 'onion' proof, where we break up the circle into rings and integrate: 
$$\int_0^{r_0 }2\pi r\, \d r=\pi r_0^2.$$
Of course, this proof is made rigorous using multivariate substitution from Cartesian to polar coordinates: 
$$
\d x \d y =\begin{vmatrix}x_r & x_\theta\\ y_r& y_\theta\end{vmatrix}\,\d\theta\d r =r\,\d\theta \d r.
$$
Now we integrate over $D=\left\{(x,y)\,\big|\,x^2+y^2\leq r_0^2\right\}$:
$$
\iint_D\,\d x\d y =\int_0^{r_0}\int_0^{2\pi}r\, \d\theta\d r=\int_0^{r_0}2\pi r\, \d r.
$$
This is the same integral as the first one. Note that when making the substitution, we used the derivatives of $\sin\theta$ and $\cos\theta$, both of which rely on the following well known limit: 
$$\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1.$$
Instead of doing this, we can inscribe and circumscribe the circle with regular polygons. Doing this, we arrive at the following inequality: 
$$\frac{1}{2}nr_0^2\sin\frac{2\pi}{n}<A<nr_0^2\tan\frac{\pi}{n}.$$
In the above, $A$ denotes the area of the circle, and $n$ denotes the number of sides in the regular polygon. Taking the limit as $n$ goes to infinity: 
$$\lim_{n\to \infty}\frac{1}{2}nr_0^2\sin\frac{2\pi}{n}\leq \lim_{n\to \infty}A\leq\lim_{n\to \infty}nr_0^2\tan\frac{\pi}{n}$$
$$\pi r_0^2\leq   A\leq\pi r_0^2$$
Again, we've used the following well known limit: 
$$\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1.$$
Now, instead of doing this, we can integrate in Cartesian coordinates to find twice the area of a semicircle: 
\begin{align*}
&\mathrel{\phantom{=}} 2\int_{-1}^1\sqrt{1-x^2}\,\d x= 2\int_{-\pi/2}^{\pi/2}\cos\theta\sqrt{1-\sin\theta^2}\, \d\theta\\
&= 2\int_{-\pi/2}^{\pi/2}\cos^2\theta\, \d\theta = 2\int_{-\pi/2}^{\pi/2}\frac{1+\cos 2\theta}{2}\, \d\theta\\
&= 2\left.\left(\frac{\theta}{2}+\frac{\sin 2\theta}{4}\right)\right|_{-\pi/2}^{\pi/2} =\pi.
\end{align*}
Again, in the substitution we used the derivative of $\sin x$, which relies on the following well known limit:
$$\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1.$$
The problem I have with all of this is that the proof of the above limit relies on the area of a circle to obtain the following inequality: 
$$\frac{1}{2}\sin\theta<\frac{\theta}{2}<\frac{1}{2}\tan \theta.$$
After this, we can apply squeeze theorem. There are other proofs that rely on Taylor series or Euler's formula, but these are usually circular (pun intended). So, my question is, can we prove the area of a circle is $\pi r^2$ without using the above limit, or vice versa?

REPLY [8 votes]: In calculus terms, the formula for the circumference boils down to
$$\int_{-1}^1\frac{dt}{\sqrt{1-t^2}}=\pi$$ and for the area to
$$\int_{-1}^1\sqrt{1-t^2}\,dt=\frac\pi2.$$
So how to show that the first is twice the second without using trigonometric functions?
Consider
$$\frac d{dt}\left(t\sqrt{1-t^2}\right)
=\sqrt{1-t^2}-\frac{t^2}{\sqrt{1-t^2}}
=2\sqrt{1-t^2}-\frac1{\sqrt{1-t^2}}.$$
Integrate from $-1$ to $1$.