The area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ is

1. Here $a^2 = 4$ and $b^2 = 9$, so $a = 2$ and $b = 3$. 2. Area of an ellipse $= \pi a b = \pi \cdot 2 \cdot 3$. 3. Therefore $A = 6\pi$. (Multiplying $a^2 b^2 = 36$ wrongly gives $36\pi$.) _Source: NCERT Class 12 Mathematics Ch 8 "Application of Integrals", p.4_