The area of the region enclosed between the circles
$x^2 + y^2 = 4$ and $x^2 + (y-2)^2 = 4$ is:
🎥 Video solution / Text Solution of this question is given below:

$A = 2\int_{0}^{\sqrt{3}} \left[\sqrt{4-x^2} - (2 - \sqrt{4-x^2})\right] dx$
$= 2\int_{0}^{\sqrt{3}} (2\sqrt{4-x^2} - 2),dx$
$= 4\int_{0}^{\sqrt{3}} (\sqrt{4-x^2} - 1),dx$
$= 4\left[\frac{1}{2}\left(x\sqrt{4-x^2} + 4\sin^{-1}\frac{x}{2}\right) - x\right]_{0}^{\sqrt{3}}$
$= \frac{8\pi}{3} - 2\sqrt{3}$