What is the area of the region $\{(x, y) : x^{2} + y^{2} \le 16,\ y \ge \lvert x \rvert\}$?

Explanation

1. 01

   $y = \lvert x \rvert$ meets the circle where $2x^{2} = 16$, at $x = \pm 2\sqrt{2}$; the region is symmetric about the $y$-axis.

2. 02

   Right half: $\int_0^{2\sqrt{2}} \left(\sqrt{16 - x^{2}} - x\right)dx = (4 + 2\pi) - 4 = 2\pi$.

3. 03

   Both halves: $4\pi$ square units, a quarter of the disc.