Question

Find the area of the region enclosed between the two circles $x^2+y^2=4$ and ${(x-2)}^2+y^2=4.$

Answer 1

$f\left(x\right)=\{\cos x:0\le x\le \pi /4\sin x:\pi /4\le x\le 5\pi /4\cos x:5\pi /4\le x\le 2\pi \}A={\int }_0^1\sqrt{4-{(x-2)}^2}dx+{\int }_1^{24}\sqrt{4-x^2}dx+A={[\frac{\sqrt{4-{(x-2)}^2}(x-2)}{2}+\frac{4}{2}{\sin }^{-1}\frac{x-2}{2}]}_0^1+{[\frac{\sqrt{4-{\left(x\right)}^2}\left(x\right)}{2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_1^{2}A=2\left[\right(\frac{\sqrt{3}(-1)}{2}+2{\sin }^{-1}\frac{-1}{2})-\left(\right(-2\left)\right(0)+2{\sin }^{-1}(-1))+\left(\right(0)+2{\sin }^{-1}\left(1\right))-(\frac{\sqrt{3}\left(1\right)}{2}+2{\sin }^{-1}\frac{1}{2})]A=2[\frac{-\sqrt{3}}{2}-2{\sin }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(1\right)+2{\sin }^{-1}\left(1\right)-\frac{\sqrt{3}}{2}-2{\sin }^{-1}\left(\frac{1}{2}\right)]A=2[-\sqrt{3}-4{\sin }^{-1}\left(\frac{1}{2}\right)+4{\sin }^{-1}\left(1\right)]A=2[-\sqrt{3}-4\left(\pi /6\right)+4\left(\pi /2\right)]A=2[-\sqrt{3}+4\left(\pi /3\right)]A=-2\sqrt{3}+8(\pi /3)$