JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Area Under Curves

The region is bounded by the circle $x^2+(y-2{)}^2=4$ (center $(0,2)$, radius $2$) and the parabola $x^2=2y$. Solving simultaneously: $2y+y^2-4y+4=4\Rightarrow y^2-2y=0\Rightarrow y=0,2$. At $y=2$, $x=\pm 2$, giving intersection points $(2,2)$ and $(-2,2)$. The required area is twice the area from $x=0$ to $x=2$ of the region between the parabola and the circle. Using integration: Required area $=2\left[{\int }_0^2\frac{x^2}{2}dx-(4-\pi )\right]=2\left[\frac{x^3}{6}{|}_0^2-4+\pi \right]=2\left[\frac{4}{3}+\pi -4\right]=2\left[\pi -\frac{8}{3}\right]=2\pi -\frac{16}{3}$.