Find the area of the region bounded by the curve 4x^2 + y^2 = 36 usin…

The correct answer is Option (3) → $18\pi$ sq units

Given curve is $4x^2 + y^2 = 36$

or, $\frac{x^2}{9} + \frac{y^2}{36} = 1$

or, $\frac{x^2}{3^2} + \frac{y^2}{6^2} = 1$

since, ellipse is symmetrical along x-axis and y-axis.

$\text{Area of ellipse} = \text{Area of ABCD}$

$= 4 \times \text{Area of OBC}$

$= 4 \times \int\limits_{0}^{3} y dx$

$= 4 \times \int\limits_{0}^{3} (2 \sqrt{9 - x^2}) dx$

$= 8 \int\limits_{0}^{3} \sqrt{9 - x^2} dx$  [since, OBC is above x-axis]

$= 8 \int\limits_{0}^{3} \sqrt{3^{2} - x^{2}} \, dx$

$= 8 \left[ \frac{x}{2} \sqrt{3^{2} - x^{2}} + \frac{3^{2}}{2} \sin^{-1} \frac{x}{3} \right]_{0}^{3}$

$= 8 \left[ \frac{x}{2} \sqrt{9 - x^{2}} + \frac{9}{2} \sin^{-1} \frac{x}{3} \right]_{0}^{3}$

$= 8 \left[ 0 + \frac{9}{2} \sin^{-1}(1) - (0 + 0) \right]$

$= 8 \times \frac{9}{2} \times \frac{\pi}{2}$

$= 2 \times 9\pi$

$= 18\pi \text{ sq. units}$