Find the area of a circle of radius a using integrals in calculus.
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 circle and multiply by 4 in order to obtain the total area of the circle. Solve the above equation for y y = ~+mn~ √[ a 2 - x 2 ] The equation of the upper semi circle (y positive) is given by y = √[ a 2 - x 2 ] = a √ [ 1 - x 2 / a 2 ] We use integrals to find the area of the upper right quarter of the circle as follows (1 / 4) Area of circle =  0a a √ [ 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 circle = 0π/2 a 2 ( √ [ 1 - sin2 t ] ) cos t dt
We now use the trigonometric identity √ [ 1 - sin2 t ] = cos t since t varies from 0 to π/2, hence (1 / 4) Area of circle = 0π/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 = 0π/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 ]0π/2 = (1/4) π a 2 The total area of the circle is obtained by a multiplication by 4 Area of circle = 4 * (1/4) π a 2 = π a 2 More references on integrals and their applications in calculus. |
