The figure shows the smaller part of the circle x^2 + y^2 = 16 cut off by the line x = 2√(2).

Explanation

1. 01

   The part is symmetric about the $x$-axis: $2\int_{2\sqrt{2}}^{4} \sqrt{16 - x^{2}}\,dx$.

2. 02

   $\int \sqrt{16 - x^{2}}\,dx = \frac{x}{2}\sqrt{16 - x^{2}} + 8\sin^{-1}\frac{x}{4}$; between the limits this gives $2\pi - 4$, so the area is $4\pi - 8$.