The area enclosed by the ellipse (x^2)/(9^2)+(y^2)/(6^2)=1 is:

$\frac{x^2}{9^2}+\frac{y^2}{6^2}=1$

so  $y = \frac{6}{9}\sqrt{9^2-x^2}$

so area of curve in all quadrants are equal 

⇒ total area = 4 × area of curve in 1^st quadrant

⇒ $4\int\limits_0^9 y~dx = 4 × \frac{6}{9} \int\limits_0^9 \sqrt{9^2-x^2}dx$

= $4 × \frac{6}{9} \left[\frac{x}{2} \sqrt{9^2-x^2} + \frac{9^2}{2} \sin \frac{x}{9} \right]_0^9$

$=4 \times \frac{6}{9} \times\left(\frac{9^2}{2} \times \frac{\pi}{2}\right)=54 \pi$