Problem : Find the area of an ellipse with half axes a and b.

Solution to the problem:

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

x^{ 2} / a^{ 2} + y^{ 2} / b^{ 2} = 1

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 = + or - b √ [ 1 - x^{ 2} / a^{ 2} ]

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

y = b √ [ 1 - x^{ 2} / a^{ 2} ]

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

(1 / 4) Area of ellipse = _{0}^{a} b √ [ 1 - x^{ 2} / a^{ 2} ] dx

We now make the substitution sin t = x / a so that dx = a cos t dt and the area is given by

(1 / 4) Area of ellipse = _{0}^{pi/2} a b ( √ [ 1 - sin^{2} t ] ) cos t dt

√ [ 1 - sin^{2} t ] = cos t since t varies from 0 to pi/2, hence

(1 / 4) Area of ellipse = _{0}^{pi/2} a b cos^{2} t dt

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

(1 / 4) Area of ellipse = _{0}^{pi/2} a b ( cos 2t + 1 ) / 2 dt

Evaluate the integral

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

= (1/4) pi a b

Obtain the total area of the ellipse by multiplying by 4

Area of ellipse = 4 * (1/4) pi a b = pi a b
More references on
integrals and their applications in calculus.