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 are of curve in all quadrants are equal

⇒ total area = 4 × area of curve in Ist quadrant

$⇒4\int_0^9y\,dx=4×\frac{6}{9}\int_0^9\sqrt{9^2-x^2}dx$

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

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