What is the area of the part of the ellipse $9x^{2} + y^{2} = 36$ to the right of the $y$-axis?

Explanation

1. 01

   Divide by 36: $\frac{x^{2}}{4} + \frac{y^{2}}{36} = 1$, so $a = 2$, $b = 6$, and the ellipse encloses $\pi ab = 12\pi$.

2. 02

   The $y$-axis cuts it into two equal halves: $2\int_0^{2} 3\sqrt{4 - x^{2}}\,dx = 6\pi$ square units.