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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 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 sqrt [ 1 - x 2 / a 2 ] The upper part of the ellipse (y positive) is given by y = b sqrt [ 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 sqrt [ 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 ( sqrt [ 1 - sin2 t ] ) cos t dt
sqrt [ 1 - sin2 t ] = cos t since t varies from 0 to pi/2, hence (1 / 4) Area of ellipse = 0 pi/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 = 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. |
