(a): Given that $x^2 + (y - 2)^2 \le 2^2$ and $x^2 \ge 2y$ Convert the given inequality into equation $x^2 + (y - 2)^2 = 4$ and $x^2 = 2y$Solving circle and parabola simultaneously $2y + y^2 - 4y + 4 = 4 \Rightarrow y^2 - 2y = 0 \Rightarrow y = 0, 2$ Put $y = 2$ in $x^2 = 2y$ $\Rightarrow x = \pm 2 \Rightarrow (2, 2)$ and $(-2, 2)$$A_1 = 2 \times 2 - \frac{1}{4} \cdot \pi \cdot 2^2 = 4 - \pi$Required area $= 2 \left[ \int_{0}^{2} \frac{x^2}{2} dx - (4 - \pi) \right]$$= 2 \left[ \frac{x^3}{6} \bigg|_{0}^{2} - 4 + \pi \right] = 2 \left[ \frac{4}{3} + \pi - 4 \right] = 2 \left[ \pi - \frac{8}{3} \right] = 2\pi - \frac{16}{3}$ sq. units