What is the area of the part of the circle $x^{2} + y^{2} = 36$ in the first quadrant?

Explanation

1. 01

   $\int_0^{6} \sqrt{36 - x^{2}}\,dx = \frac{\pi a^{2}}{4}$ with $a = 6$.

2. 02

   $9\pi$ square units.