Area of the region \ (x, y): x ^ 2 + (y - 2) ^ 2 ≤ 4, x ^ 2 ≥ 2y\ is

Given,

${x}^{2} + {\left(y - 2\right)}^{2} \le 4 \text{ and } {x}^{2} \ge 2y$

Now on solving, ${x}^{2} + {\left(y - 2\right)}^{2} = 4 \text{ and } {x}^{2} = 2y$ we get intersecting point of the curves as \left(\pm 2, 2\right)$,</p><p>Now plotting the diagram we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/9bd57b61-d058-4515-8af2-fcd621dfe142-image.png" style="width: 300px; height: 207px;" /></p><p>So, from above diagram area of required region will be,</p><p>$= 2\left[\dfrac{1}{4}\left(\pi \times {2}^{2}\right) - {∫}_{0}^{2}\sqrt{2} \cdot \sqrt{y}dy\right]$</p><p>$= 2\left[\pi - {\left[\dfrac{\sqrt{2} ⋅ {y}^{\dfrac{3}{2}}}{\dfrac{3}{2}}\right]}_{0}^{2}\right]$</p><p>$= 2\left[\pi - \dfrac{2\sqrt{2}}{3} \cdot 2\sqrt{2}\right]$</p><p>$= 2\left[\pi - \dfrac{8}{3}\right]$</p><p>$= 2\pi - \dfrac{16}{3}