A) $\frac{2}{3} (4\pi – 3\sqrt{3})$ B) $\frac{4}{3} (2\pi – \sqrt{3})$ C) $\frac{4}{3} (2\pi – 3\sqrt{3})$ D) $\frac{2}{3} (2\pi – 3\sqrt{3})$
Equating $x^2 = 4 – y^2$ and $x^2 = 4 – (y – 2)^2$:
$4 – y^2 = 4 – (y^2 – 4y + 4) \implies -y^2 = -y^2 + 4y – 4 \implies 4y = 4 \implies y = 1$
Substituting $y = 1$ into $x^2 + y^2 = 4$: $x^2 = 3 \implies x = \pm\sqrt{3}$. Intersection points are $(\pm\sqrt{3}, 1)$.
The region is symmetric about the y-axis. The total area $A = 2 \int_{0}^{\sqrt{3}} (y_{upper} – y_{lower}) dx$.
For the region, $y_{upper} = \sqrt{4-x^2}$ (lower circle) and $y_{lower} = 2 – \sqrt{4-x^2}$ (upper circle). Wait, let’s look at the region: it is bounded between $y=1$ and $y=2$ by the upper circle and between $y=0$ and $y=1$ by the lower circle.
Total Area $A = 2 \int_{0}^{1} \sqrt{4 – (y-2)^2} dy + 2 \int_{1}^{2} \sqrt{4-y^2} dy$.
By symmetry, these two integrals are equal. $A = 4 \int_{1}^{2} \sqrt{4-y^2} dy$.
Use $\int \sqrt{a^2 – x^2} dx = \frac{x}{2}\sqrt{a^2 – x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$:
$A = 4 \left[ \frac{y}{2}\sqrt{4-y^2} + \frac{4}{2}\sin^{-1}(\frac{y}{2}) \right]_{1}^{2}$
$A = 4 \left[ (0 + 2\sin^{-1}1) – (\frac{1}{2}\sqrt{3} + 2\sin^{-1}\frac{1}{2}) \right]$
$A = 4 \left[ 2(\frac{\pi}{2}) – \frac{\sqrt{3}}{2} – 2(\frac{\pi}{6}) \right] = 4 \left[ \pi – \frac{\sqrt{3}}{2} – \frac{\pi}{3} \right] = 4 \left[ \frac{2\pi}{3} – \frac{\sqrt{3}}{2} \right]$
$A = \frac{8\pi}{3} – 2\sqrt{3} = \frac{8\pi – 6\sqrt{3}}{3} = \frac{2}{3}(4\pi – 3\sqrt{3})$