Proving the Area of a Circle and More

What is the circumference of a circle with radius $r$ ? It is $C(r) := 2 \pi r$ . What about its area? It is $A(r):= \pi r^2$ . Fair enough. Why?

For the first result, we find the arc length of the circle $x^2 + y^2 = r^2$ .

Theorem 1. $C(r) = 2\pi r$ .

Proof. By considering the upper-half semicircle, $y = \sqrt{r^2 - x^2}$ . The circumference is, by definition, given by

$\displaystyle C(r) = 2 \int_{-r}^r \sqrt{1 + \left( \frac{\mathrm dy}{\mathrm dx} \right)^2}\, \mathrm dx$

Making the substitution $x = ru$ yields $y = r \sqrt{1-u^2}$ , $\mathrm dx = r\, \mathrm du$ , and $\mathrm dy = -(ru)/\sqrt{1-u^2}\,\mathrm du$ . Doing the needful algebra simplifies the integral to

$\displaystyle 2 \int_{-r}^r \sqrt{1 + \left( \frac{\mathrm dy}{\mathrm dx} \right)^2}\, \mathrm dx = 2r \int_{-1}^1 \frac{1}{\sqrt{1-u^2}}\, \mathrm du = 2r \pi,$

where we have defined $\pi$ previously.

What about the area formula? One might be tempted to prove $A' = C$ , but that approach requires some logical contemplation.

Let’s first write down the computation for definite integrals.

Theorem 2. Let $f : [a, b] \to \mathbb R$ be continuously differentiable. Then

$\displaystyle \int_a^b f'(x)\,\mathrm dx = f(b) - f(a) \equiv [f(x)]_a^b.$

We will adopt the more conventional approach of using an integral to define the area of a circle with radius $r$ , then prove that this area really must be $\pi r^2$ .

Theorem 3. The area of a circle with radius $r$ is $\pi r^2$ .

Proof. By definition, the area of a circle with radius $r$ given by the integral

$\displaystyle 2 \int_{-r}^r \sqrt{r^2-x^2}\, \mathrm dx.$

Making the substitution $x = ru$ ,

$\displaystyle 2\int_{-r}^r \sqrt{r^2 - x^2}\,\mathrm dx = r^2 \cdot 2 \int_{-1}^1 \sqrt{1-u^2}\,\mathrm du.$

Integrating by parts and using the chain rule,

$\displaystyle \begin{aligned} \int_{-1}^1 \sqrt{1-u^2}\, \mathrm du &= [ \underbrace{u}_{\text I} \underbrace{\sqrt{1-u^2}}_{\text S} ]_{-1}^1 - \int_{-1}^1 \underbrace{u}_{\text I} \underbrace{\frac{1}{2\sqrt{1-u^2}} \cdot (-2u)}_{\text D}\, \mathrm du \\ &= (0 - 0) + \int_{-1}^1 \frac{u^2}{\sqrt{1-u^2}}\, \mathrm du \\ &= \int_{-1}^1 \frac{u^2-1}{\sqrt{1-u^2}}\, \mathrm du + \int_{-1}^1 \frac{1}{\sqrt{1-u^2}}\, \mathrm du \\ &= - \int_{-1}^1 \sqrt{1-u^2}\, \mathrm du + \pi.\end{aligned}$

Adding the integral on both sides,

$\displaystyle 2 \int_{-1}^1 \sqrt{1-x^2}\, \mathrm dx = \pi.$

Alternate Proof. I loved the proof given by Professor Francesco Calegari here. Evaluate the integral

$\begin{aligned}&2\int_{-1}^1 \sqrt{1-x^2}\, \mathrm dx - \int_{-1}^1 \frac{1}{\sqrt{1-x^2}}\, \mathrm dx \\ &= \int_{-1}^1\left( 2\sqrt{1-x^2} - \frac{1}{\sqrt{1-x^2}} \right)\, \mathrm dx \\ &= \left[ x \sqrt{1-x^2} \right]_{-1}^1 = 0. \end{aligned}$

Therefore,

$\displaystyle 2\int_{-1}^1 \sqrt{1-x^2}\, \mathrm dx = \int_{-1}^1 \frac{1}{\sqrt{1-x^2}}\, \mathrm dx = \pi.$

Thus, we have the striking observation that becomes a question of independent interest and exploration.

Corollary 2. $A' = C$ .

Our goal is to prove that $\sin' = \cos$ . We still need some tools to help us with this objective, namely, the area of a sector subtended by an angle of $\theta$ . Fortunately, with the area of a circle, this is not hard.