What is the area of the part of the circle $x^{2} + y^{2} = 8$ between the lines $x = -2$ and $x = 2$?

Explanation

1. 01

   The part is symmetric about both axes: $4\int_0^{2} \sqrt{8 - x^{2}}\,dx$.

2. 02

   $\int_0^{2} \sqrt{8 - x^{2}}\,dx = \left[\frac{x}{2}\sqrt{8 - x^{2}} + 4\sin^{-1}\frac{x}{2\sqrt{2}}\right]_0^{2} = 2 + 4 \times \frac{\pi}{4} = 2 + \pi$.

3. 03

   Area $= 4(2 + \pi) = 8 + 4\pi$ square units.