Question

Using integration, find the area of the region enclosed between the two circles
$x^2+y^2=4$ and ${\left(x-2\right)}^2+y^2=4$

Answer 1

Clearly, given circles intersect at $(1,\sqrt{3})$ and $(1,-\sqrt{3})$
$\therefore Requiredarea=2\left(AreaOABO\right)$
$=2\left({\int }_0^1{y}_{1}dx+{\int }_1^2{y}_{1}dx\right)=2\{{\int }_1^2\sqrt{4-x^2}dx+{\int }_0^1\sqrt{4-{(x-2)}^2}\}$
$={[x\sqrt{4-x^2}+4{\sin }^{-1}\frac{x}{2}]}_1^2+{[(x-2)\sqrt{4-{(x-2)}^2}+4{\sin }^{-1}\frac{x-2}{2}]}_0^1$
$=4{\sin }^{-1}1-(\sqrt{3}+4\times \frac{\pi }{6})+\{-\sqrt{3}+4{\sin }^{-1}(-\frac{1}{2})\}-\{0+4{\sin }^{-1}(-1)\}$
$=4\times \frac{\pi }{2}-(\sqrt{3}+\frac{2\pi }{3})+(-\sqrt{3}-\frac{4\pi }{6})-(-\frac{4\pi }{2})=\frac{8\pi }{3}-2\sqrt{3}$