Use integrals to establish the formula for the area of an ellipse

Denoting the unit circle, $x^2 + y^2 = 1$ , by $C$ we define an ellipse $E$ to be the set of points

$E = \{ (ax, by) \mid (x,y) \in C, a > 0, b > 0 \}.$

Show that the points on $E$ satisfy the equation:
$\left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1.$
Prove that the area of $E$ is measurable and that
$a(E) = \pi a b.$

Proof. If $(x,y)$ is a point on $E$ then $\left( \frac{x}{a}, \frac{y}{b} \right)$ is a point on $C$ (since all points of $E$ are obtained by taking a point of $C$ and multiplying the $x$ -coordinate by $a$ and the $y$ -coordinate by $b$ ). By definition of $C$ , we must then have,
$\left( \frac{x}{a} \right)^2 + \left(\frac{y}{b} \right)^2 = 1. \qquad \blacksquare$
Proof. From part (a) we know $E$ is the set of points $(x,y)$ such that $\left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1$ . But, this implies,
$y = b \cdot \sqrt{ 1 - \left( \frac{x}{a} \right)^2}, \qquad \text{or} \qquad y = -b \cdot \sqrt{1 - \left( \frac{x}{a} \right)^2}.$

Hence, the area of $E$ is the area enclosed from $-a$ to $a$ by the graphs of

$g(x) = b \cdot \sqrt{1 - \left( \frac{x}{a} \right)^2} \qquad \text{and} \qquad f(x) = -b \cdot \sqrt{1 - \left( \frac{x}{a} \right)^2}.$

To show this region is measurable and has area $\pi a b$ we begin with this identity (from Apostol, 2.4 Exercise 17),

$\begin{align*} \pi = 2 \int_{-1}^1 \sqrt{1-x^2} && \implies && \pi b &= 2 \int_{-1}^1 b \sqrt{1-x^2} \, dx \\ && \implies && \pi a b &= 2a \int_{-1}^1 b \sqrt{1-x^2} \, dx \\ && \implies && \pi a b &= 2 \int_{-a}^a b \sqrt{1 - \left( \frac{x}{a} \right)^2} \, dx \\ && \implies && \pi ab &= \int_{-a}^a \left( b \sqrt{1 - \left( \frac{x}{a} \right)} - \left(-b \sqrt{1 - \left( \frac{x}{a} \right)^2} \right)\right) \, dx. \end{align*}$

In the second to last line we have used the expansion/contraction of the interval of integration. Hence, we know the integral from $-a$ to $a$ of $g(x) - f(x)$ exists and has value $\pi a b$ . Thus, $E$ is measurable and $a(E) = \pi a b. \qquad \blacksquare$