The area enclosed by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is given by :
Answer & explanation
Correct answer: option 3
The correct answer is option (3) → $3\int\limits^{4}_{0}\sqrt{16-x^2}dx$
$\frac{x^2}{16}+\frac{y^2}{9}=1$
so $y=\frac{3}{4}\sqrt{16-x^2}$
area of $\frac{1}{4}th$ of ellipse in 1st quadrant
→ $\int\limits_0^4\frac{3}{4}\sqrt{16-x^2}dx$
$=3\int\limits_0^4\sqrt{16-x^2}dx$