Find the area of an ellipse using integrals and calculus.

Solution to the problem:The equation of the ellipse shown above may be written in the form ^{ 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 = _{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}^{π/2} a b ( √ [ 1 - sin^{2} t ] ) cos t dt
√ [ 1 - sin ^{2} t ] = cos t since t varies from 0 to π/2, hence
(1 / 4) Area of ellipse = _{0}^{π/2} a b cos^{2} t dt
Use the trigonometric identity cos ^{2} t = ( cos 2t + 1 ) / 2 to linearise the integrand;
(1 / 4) Area of ellipse = _{0}^{π/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}^{π/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. |