Which integral gives the area of the part of the circle $x^{2} + y^{2} = 16$ above the $x$-axis?

Explanation

1. 01

   Above the $x$-axis the circle is $y = \sqrt{16 - x^{2}}$, for $-4 \le x \le 4$.

2. 02

   Area $= \int_{-4}^{4} \sqrt{16 - x^{2}}\,dx = 8\pi$.