Problem : Find the area of a circle with radius a.
Solution to the problem:
The equation of the circle shown above is given by
x^{ 2} + y^{ 2} = a^{ 2}
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 = + or  √[ 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} ^{pi/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 pi/2, hence
(1 / 4) Area of circle = _{0} ^{pi/2} a^{ 2} cos^{2} t dt
Use the trigonometric identity cos^{2} t = ( cos 2t + 1 ) / 2 to linearize the integrand;
(1 / 4) Area of circle = _{0} ^{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 circle is obtained by a multiplication by 4
Area of circle = 4 * (1/4) pi a^{ 2} = pi a^{ 2}
More references on
integrals and their applications in calculus.
