Question

Find the area of the circle $x^2+y^2=4$ using integration.

Answer 1

Equation of the circle is $x^2+y^2=4$

$\therefore y=\sqrt{4-x^2}$

But the area element is given by

$\therefore ydx=\sqrt{4-x^2}dx$

Integrating both the sides with x ranging from -2 to 2

$\therefore A={\int }_{-2}^{2}ydx={\int }_{-2}^2\sqrt{4-x^2}dx$

Let $x=2\sin \theta$ $\Rightarrow dx=2\cos \theta d\theta$

$\therefore A={\int }_{-\pi }^{\pi }\sqrt{4-{4\sin }^2\theta }2\cos \theta d\theta ={\int }_{-\pi }^{\pi }4{\cos }^2\theta d\theta$

$=2{\int }_{-\pi }^{\pi }(1+\cos 2\theta )d\theta =4\pi$