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(A) The circle is $x^2 + y^2 = 2^2$, so the radius is $r = 2$.The area of the smaller region is given by $A = 2 \int_{1}^{2} \sqrt{4 - x^2} \, dx$.Usi

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$\frac{4\pi}{3} - \sqrt{3}$

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(A) The circle is $x^2 + y^2 = 2^2$, so the radius is $r = 2$.The area of the smaller region is given by $A = 2 \int_{1}^{2} \sqrt{4 - x^2} \, dx$.Using the formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})$, we get:$A = 2 \left[ \frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1}(\frac{x}{2}) \right]_{1}^{2}$.$A = 2 \left[ (0 + 2 \sin^{-1}(1)) - (\frac{1}{2} \sqrt{3} + 2 \sin^{-1}(\frac{1}{2})) \right]$.$A = 2 \left[ (2 \cdot \frac{\pi}{2}) - (\frac{\sqrt{3}}{2} + 2 \cdot \frac{\pi}{6}) \right]$.$A = 2 \left[ \pi - \frac{\sqrt{3}}{2} - \frac{\pi}{3} \right] = 2 \left[ \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right] = \frac{4\pi}{3} - \sqrt{3}$.

The area of the smaller region bounded by the circle $x^2 + y^2 = 4$ and the line $x = 1$ is:

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