CUET Preparation Today
The area enclosed by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is given by : |
$3\int\limits^{4}_{0}\sqrt{9-x^2}dx$ $\frac{3}{4}\int\limits^{4}_{0}\sqrt{9-x^2}dx$ $3\int\limits^{4}_{0}\sqrt{16-x^2}dx$ $\frac{3}{4}\int\limits^{4}_{0}\sqrt{16-x^2}dx$ |
$3\int\limits^{4}_{0}\sqrt{16-x^2}dx$ |
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 → $\int\limits_0^4\frac{3}{4}\sqrt{16-x^2}dx$ $=3\int\limits_0^4\sqrt{16-x^2}dx$ |
