For the ellipse $9x^{2} + 4y^{2} = 36$, which of the following statements are true?

I. Its area is $36\pi$.

II. Its area is $4\int_0^{2} \frac{3}{2}\sqrt{4 - x^{2}}\,dx$.

III. Its area is $4\int_0^{3} \frac{2}{3}\sqrt{9 - y^{2}}\,dy$.

IV. Its part in the first quadrant has area $\frac{3\pi}{2}$.

Explanation

1. 01

   Standard form: $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$, so $a = 2$, $b = 3$ and the area is $6\pi$, not $36\pi$. I is false, and IV, a quarter of $6\pi$, is true.

2. 02

   II solves for $y = \frac{3}{2}\sqrt{4 - x^{2}}$ and III for $x = \frac{2}{3}\sqrt{9 - y^{2}}$; each gives $6\pi$, so both are true.