The given equation of curve is $9 x^{2}+4 y^{2}-36$
or it can written as $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
Required area = Area of curve $A D C B A$
$=4$ Area of curve $O C B A$
$=4 \int_{0}^{2} y\, d x$
$=4 \int_{0}^{2} 3 \sqrt{1-\frac{x^{2}}{4}} d x=6 \int_{0}^{2} \sqrt{4-x^{2}} d x$
$=4\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}$
$=6\left[0+2 \sin ^{-1} 1-(0+2(0))\right]$
$=6 \cdot 2 \cdot \frac{\pi}{2}=6 \pi$
