For $a > 0$, the part of the circle $x^{2} + y^{2} = a^{2}$ in the first quadrant has area $4\pi$. What is $a$?

Explanation

1. 01

   The quadrant is $\int_0^{a} \sqrt{a^{2} - x^{2}}\,dx = \frac{\pi a^{2}}{4}$.

2. 02

   $\frac{\pi a^{2}}{4} = 4\pi$ gives $a^{2} = 16$, so $a = 4$.