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 = ~+mn~ 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 = 0a 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 = 0π/2 a b ( √ [ 1 - sin2 t ] ) cos t dt
√ [ 1 - sin2 t ] = cos t since t varies from 0 to π/2, hence
(1 / 4) Area of ellipse = 0π/2 a b cos2 t dt
Use the trigonometric identity cos2 t = ( cos 2t + 1 ) / 2 to linearize the integrand;
(1 / 4) Area of ellipse = 0π/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 ]0π/2
= (1/4) π a b
Obtain the total area of the ellipse by multiplying by 4
Area of ellipse = 4 * (1/4) π a b = π a b
More references on
integrals and their applications in calculus.