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(D) The given circles are $x^{2}+y^{2}=4$ (center $(0,0)$,radius $2$) and $x^{2}+(y-2)^{2}=4$ (center $(0,2)$,radius $2$).Solving the equations: $x^{2

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$\frac{2}{3}(4\pi-3\sqrt{3})$

Vedclass · Answer

(D) The given circles are $x^{2}+y^{2}=4$ (center $(0,0)$,radius $2$) and $x^{2}+(y-2)^{2}=4$ (center $(0,2)$,radius $2$).Solving the equations: $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-4y+4=4$.Substituting $x^{2}+y^{2}=4$ into the second equation: $4-4y+4=4$,which gives $4y=4$,so $y=1$.Substituting $y=1$ into $x^{2}+y^{2}=4$,we get $x^{2}+1=4$,so $x^{2}=3$,$x=\pm\sqrt{3}$.The area of the intersection is symmetric about the $y$-axis.The area $A = 2 \int_{0}^{\sqrt{3}} [\sqrt{4-x^{2}} - (2 - \sqrt{4-x^{2}})] dx = 2 \int_{0}^{\sqrt{3}} (2\sqrt{4-x^{2}} - 2) dx = 4 \int_{0}^{\sqrt{3}} (\sqrt{4-x^{2}} - 1) dx$.Using the formula $\int \sqrt{a^{2}-x^{2}} dx = \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a})$:$A = 4 [\frac{x}{2}\sqrt{4-x^{2}} + 2\sin^{-1}(\frac{x}{2}) - x]_{0}^{\sqrt{3}}$$A = 4 [(\frac{\sqrt{3}}{2}\sqrt{4-3} + 2\sin^{-1}(\frac{\sqrt{3}}{2})) - \sqrt{3}]$$A = 4 [\frac{\sqrt{3}}{2} + 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})$.

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

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