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 cirle as follows (1 / 4) Area of cirle = _{0}^{a} 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 cirle = _{0}^{π/2} a^{ 2} ( √ [ 1 - sin^{2} t ] ) cos t dt We now use the trigonometric identity √ [ 1 - sin^{2} t ] = cos t since t varies from 0 to π/2, hence (1 / 4) Area of circle = _{0}^{π/2} a^{ 2} cos^{2} t dt Use the trigonometric identity cos^{2} t = ( cos 2t + 1 ) / 2 to linearise the integrand; (1 / 4) Area of circle = _{0}^{π/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}^{π/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. |