The area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is

1. Compare with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: here $a^2 = 16$ and $b^2 = 9$, so $a = 4$, $b = 3$. 2. Area of an ellipse $= \pi a b$. 3. Substituting, $A = \pi \cdot 4 \cdot 3 = 12\pi$. (Using $a^2 b = 48$ instead of $ab$ gives the wrong $48\pi$.) _Source: NCERT Class 12 Mathematics Ch 8 "Application of Integrals", p.4_