What is the area of the region bounded by the arc of $x^{2} + y^{2} = 16$ in the first quadrant and the lines $x = 4$ and $y = 4$?

Explanation

1. 01

   Each vertical strip runs from the arc $y = \sqrt{16 - x^{2}}$ up to $y = 4$, for $0 \le x \le 4$.

2. 02

   $\int_0^{4} \left(4 - \sqrt{16 - x^{2}}\right)dx = 16 - 4\pi$ square units: the square $4 \times 4$ minus the quarter circle.