Area of the Ellipse

Derivation

Deriving the formula was done using area by integration. Say we have an ellipse and consider one-quarter of it. Let's solve the area of this section using integration.

We compute the area of a single segment using integration.

\(A_1=\int_{x_1}^{x_2} y d x\)

\(A_1=\int_0^a \frac{b}{a} \sqrt{a^2-x^2} d x\)

First, we apply trigonometric substitution to solve this integral, let:

\(x=a \sin \theta\)

\(d x=a \cos \theta d \theta\)

We also need to change the upper and lower limits of the definite integral. We find the limits \(\theta_1\) and \(\theta_2\) by substituting \(x_1\) and \(x_2\) to \(x = a\sin\theta\):

\(0=a \sin \theta_1\)

\(\theta_1=0\)

\(a=a \sin \theta_2\)

\(\theta_2=\frac{\pi}{2}\)

Next, we proceed to find the integral of the expression:

\(A_1=\frac{b}{a} \int_0^{\frac{\pi}{2}} \sqrt{a^2-a^2 \sin ^2 \theta}(a \cos \theta d \theta)\)

\(A_1=a b \int_0^{\frac{\pi}{2}} \sqrt{1-\sin ^2 \theta}(\cos \theta d \theta)\)

Apply Pythagorean identity:

\(A_1=a b \int_0^{\frac{\pi}{2}} \cos ^2 \theta d \theta\)

Apply double angle identity:

\(A_1=a b \int_0^{\frac{\pi}{2}} \frac{1+\cos 2 \theta}{2} d \theta\)

Split into two terms:

\(A_1=\frac{a b}{2}\left(\int_0^{\frac{\pi}{2}} d \theta+\int_0^{\frac{\pi}{2}} \cos 2 \theta d \theta\right)\)

Solve for the definite integral:

\(A_1=\frac{a b}{2}\left([\theta]_0^{\frac{\pi}{2}}+\left[\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{2}}\right)\)

\(A_1=\frac{a b}{2}\left(\frac{\pi}{2}\right)\)

\(A_1=\frac{\pi a b}{4}\)

We have computed the area of a single segment of the ellipse. Finally, we multiply it by four segments to get:

\(A=\pi{ab}\)

Summary

The area of an ellipse is \(A=\pi{ab}\)