Find the area of an ellipse using integrals and calculus.

__Problem :__ Find the area of an ellipse with half axes \( a \) and \( b \).

The equation of the ellipse shown above may be written in the form

Since the ellipse is symmetric with respect to the \( x \) and \( y \) axes, we can find the area of one quarter and multiply by 4 in order to obtain the total area.

Solve the above equation for \( y \)

\( y = \pm b \sqrt{1 - \dfrac{x^2}{a^2}} \)

The upper part of the ellipse (\( y \) positive) is given by

\( y = b \sqrt{1 - \dfrac{x^2}{a^2}} \)

We now use integrals to find the area of the upper right quarter of the ellipse as follows

\( \dfrac{1}{4} \) Area of ellipse = \( \displaystyle \int_{0}^{a} b \sqrt{1 - \dfrac{x^2}{a^2}} \, dx \)

We now make the substitution \( \quad \sin t = \dfrac{x}{a} \) so that \( dx = a \cos t \, dt \) and the area is given by

\( \dfrac{1}{4} \) Area of ellipse = \( \displaystyle \int_{0}^{\dfrac{\pi}{2}} a b ( \sqrt{1 - \sin^2 t} ) \cos t \, dt \)

\( \sqrt{1 - \sin^2 t} = \cos t \) since \( t \) varies from \( 0 \) to \( \dfrac{\pi}{2} \), hence

\( \dfrac{1}{4} \) Area of ellipse = \( \displaystyle \int_{0}^{\dfrac{\pi}{2}} a b \cos^2 t \, dt \)

Use the trigonometric identity \( \cos^2 t = \dfrac{\cos 2t + 1}{2} \) to linearize the integrand;

\( \dfrac{1}{4} \) Area of ellipse = \( \displaystyle \int_{0}^{\dfrac{\pi}{2}} a b ( \dfrac{\cos 2t + 1}{2} ) \, dt \)

Evaluate the integral

\( \dfrac{1}{4} \) Area of ellipse = \( \dfrac{1}{2} b a [ \dfrac{1}{2} \sin 2t + t ] \bigg|_{0}^{\dfrac{\pi}{2}} \)

\( = \dfrac{1}{4} \pi a b \)

Obtain the total area of the ellipse by multiplying by 4

Area of ellipse = \( 4 \times \dfrac{1}{4} \pi a b = \pi a b \)

More references on integrals and their applications in calculus.