The area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is

1. The circle has radius $r = 2$, so its full area is $\pi r^2 = 4\pi$. 2. The bounds $x = 0$ to $x = 2$ in the first quadrant capture exactly one quarter of the disc. 3. Hence the required area $= \frac{1}{4}\cdot 4\pi = \pi$. 4. The fraction is a quarter, not a half, which rules out $\frac{\pi}{2}$. _Source: syllabus Mathematics Ch 8 "Application of Integrals", p.4_