# Find The Area of a Circle Using Integrals in Calculus

 

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 $$y$$
$$y = \pm \sqrt{ a^2 - x^2 }$$
The equation of the upper semi circle (y positive) is given by
$$y = \sqrt { a^2 - x^2 }$$
Factor out $$a^2$$ inside the radicand
$$y = \sqrt { a^2(1 - x^2/a^2) }$$
Take $$a^2$$ from under the radicand and rewrite $$y$$ as follows
$$y = a \sqrt { 1 - x^2 / a^2 }$$
We use integrals to find the area of the upper right quarter of the circle as follows
(1 / 4) Area of circle = $$\displaystyle \int_0^a a \sqrt{1-x^2/a^2} dx$$
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 circle = $$\displaystyle \int_0^{\pi/2} a^2 \sqrt{1-\sin^2t} \cos t \; dt$$
We now use the trigonometric identity
$$\sin^2 t + \cos^2 t = 1$$
which gives
$$\sqrt{1-\sin^2t} = \cos t \quad$$ since t varies from 0 to $$\pi/2$$ hence
(1 / 4) Area of circle = $$\displaystyle \int_0^{\pi/2} a^2 \cos^2t \; dt$$
Use the trigonometric identity $$\; \cos^2 t = ( \cos 2t + 1 ) / 2 \;$$ to linearize the integrand;
(1 / 4) Area of circle = $$\displaystyle \int_0^{\pi/2} a^2 ( \cos 2t + 1 ) / 2 \; dt$$
Evaluate the integral
(1 / 4) Area of circle = $$\displaystyle (1/2)a^2 \left[(1/2) \sin 2t + t\right]_0^{\pi/2}$$
Simplify
(1 / 4) Area of circle = $$(1/4) \pi a^2$$
The total area of the circle is obtained by a multiplication by 4
Area of circle = $$4 \times (1/4) \pi a^2 = \pi a^2$$

## References

integrals and their applications in calculus.
Equation of Circle
Trigonometric Identities and Formulas