Area of a circle

Math

The area of a circle is &pi;r2, which can be found by integration (and the fact that the circumference of the circle is 2&pi;r.

Using integration
The equation of a circle is \(x^2+y^2=r^2\).

Then

\[{x^2\over r^2}+{y^2\over r^2}=1\]

We can find y in terms of x and r:

\[y=r\sqrt{1-{x^2\over r^2}}\]

Due to symmetry, integration can be used to find the area A:

\[A=4r\int^r_0\sqrt{1-{x^2\over r^2}}dx\]

Substitute:

\[x=r\sin\theta\]

\[\theta=\arctan\frac{x}{r}\]

\[dx=r\cos\theta d\theta\]

The integral becomes:

\[A=4r\int^{\pi/2}_0r\sqrt{1-\sin^2\theta}\cos\theta d\theta\].

The trigonometric identities \(1-\sin^2\theta=\cos^2\theta\) and \(\cos^2\theta=\frac12(1+\cos2\theta)\) can be used and finally,

\[4r^2\int^{\pi/2}_0\cos^2\theta d\theta=4r^2\int^{\pi/2}_0\frac12(1+cos2\theta)d\theta=2r^2\theta^{\pi/2}_0+2r^2\int^{\pi/2}_0\cos2\theta d\theta&=\pi r^2\]