For $a > 0$, what is the area of the region in the first quadrant inside the circle $x^{2} + y^{2} = a^{2}$ and to the right of the line $x = \frac{a}{2}$?

Explanation

1. 01

   $\int_{a/2}^{a} \sqrt{a^{2} - x^{2}}\,dx = \left[\frac{x}{2}\sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\right]_{a/2}^{a}$.

2. 02

   At $x = a$ the bracket is $\frac{\pi a^{2}}{4}$; at $x = \frac{a}{2}$ it is $\frac{\sqrt{3}a^{2}}{8} + \frac{\pi a^{2}}{12}$.

3. 03

   Difference: $\frac{\pi a^{2}}{6} - \frac{\sqrt{3}a^{2}}{8}$ square units.