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 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 = ~+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