The area of the region enclosed between the circles x^2+y^2=4 and x^2+(y-2)^2=4 is:

Question

The area of the region enclosed between the circles $x^2+y^2=4$ and $x^2+(y-2)^2=4$ is:

Accepted Answer

$\begin{gathered} {x}^{2}+{y}^{2}=4\implies {C}_{1}(0,0),r=2 \ {x}^{2}+(y-2{)}^{2}=4\implies {C}_{2}(0,2),r=2 \end{gathered}$

Intersection points:

$\begin{gathered} {x}^{2}+{y}^{2}={x}^{2}+(y-2{)}^{2} \ {y}^{2}={y}^{2}-4y+4\implies 4y=4\implies y=1 \ {x}^{2}+1=4\implies {x}^{2}=3\implies x=\pm \sqrt{3} \end{gathered}$

Points are $(\sqrt{3},1)$ and $(-\sqrt{3},1)$

For the region between $y=0$ and $y=2$ :

$\begin{gathered} {y}_{}=\sqrt{4-{x}^{2}}({x}^{2}+{y}^{2}=4) \ {y}_{}=2-\sqrt{4-{x}^{2}}({x}^{2}+(y-2{)}^{2}=4) \end{gathered}$

$\begin{gathered} ={\int }_{\sqrt{3}}^{-\sqrt{3}}({y}_{}-{y}_{})dx \ ={\int }_{\sqrt{3}}^{-\sqrt{3}}(\sqrt{4-{x}^{2}}-(2-\sqrt{4-{x}^{2}}))dx \ ={\int }_{\sqrt{3}}^{-\sqrt{3}}(2\sqrt{4-{x}^{2}}-2)dx \end{gathered}$

Property: ${\int }_{a}^{-a}f(x)dx=2{\int }_{a}^{0}f(x)dx$ (for even functions)

$\begin{gathered} =2{\int }_{\sqrt{3}}^{0}(2\sqrt{4-{x}^{2}}-2)dx \ =4{\int }_{\sqrt{3}}^{0}(\sqrt{4-{x}^{2}}-1)dx \ \int \sqrt{{a}^{2}-{x}^{2}}dx=\frac{x}{2}\sqrt{{a}^{2}-{x}^{2}}+\frac{{a}^{2}}{2}{\sin }^{-1}(\frac{x}{a}) \ =4{[\frac{x}{2}\sqrt{4-{x}^{2}}+2{\sin }^{-1}(\frac{x}{2})-x]}_{\sqrt{3}}^{0} \ =4[\frac{\sqrt{3}}{2}(1)+2(\frac{\pi }{3})-\sqrt{3}] \ =4[\frac{2\pi }{3}-\frac{\sqrt{3}}{2}] \ =\frac{8\pi }{3}-2\sqrt{3}=\frac{2}{3}(4\pi -3\sqrt{3}) \end{gathered}$

Solutions · 1