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Problem : Find the area of a circle with radius a.
Solution to the problem: The equation of the circle shown above is given by The circle is symmetric with respect to the x and y axes, hence we can find the area of one quarter of a cicle and multiply by 4 in order to obtain the total area of the circle. Solve the above equation for y y = + or - sqrt [ a 2 - x 2 ] The equation of the upper semi cirle (y positive) is given by y = sqrt [ a 2 - x 2 ] = a sqrt [ 1 - x 2 / a 2 ] We use integrals to find the area of the upper right quarter of the cirle as follows (1 / 4) Area of cirle = ò0 a a sqrt [ 1 - x 2 / a 2 ] dx Let us substitute x / a by sin t so that sin t = x / a and dx = a cos t dt and the area is given by (1 / 4) Area of cirle = ò0 pi/2 a 2 ( sqrt [ 1 - sin2 t ] ) cos t dt We now use the trigonometric identity sqrt [ 1 - sin2 t ] = cos t since t varies from 0 to pi/2, hence (1 / 4) Area of circle = ò0 pi/2 a 2 cos2 t dt Use the trigonometric identity cos2 t = ( cos 2t + 1 ) / 2 to linearize the integrand; (1 / 4) Area of circle = ò0 pi/2 a 2 ( cos 2t + 1 ) / 2 dt Evaluate the integral (1 / 4) Area of circle = (1/2) a 2 [ (1/2) sin 2t + t ]0 pi/2 = (1/4) pi a 2 The total area of the cicle is obtained by a multiplcation by 4 Area of circle = 4 * (1/4) pi a 2 = pi a 2 More references on integrals and their applications in calculus. |