The area (in square units) of the region enclosed by the ellipse $$x^2 + 3y^2 = 18$$ in the first quadrant below the line $$y = x$$ is

Ellipse Equation: $$\frac{x^2}{18} + \frac{y^2}{6} = 1$$. Here $$a = \sqrt{18} = 3\sqrt{2}$$ and $$b = \sqrt{6}$$.

Intersection with $$y = x$$:

$$x^2 + 3x^2 = 18 \implies 4x^2 = 18 \implies x = \sqrt{4.5} = \frac{3}{\sqrt{2}}$$.

Point of intersection is $$(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}})$$.

Area = $$\int_{0}^{3/\sqrt{2}} x \, dx + \int_{3/\sqrt{2}}^{3\sqrt{2}} \sqrt{\frac{18-x^2}{3}} \, dx$$

Alternatively, using polar substitution or parametric form $$x = \sqrt{18}\cos\theta, y = \sqrt{6}\sin\theta$$.

The line $$y=x$$ corresponds to $$\tan\theta = \frac{y/b}{x/a} = \frac{a}{b} = \frac{\sqrt{18}}{\sqrt{6}} = \sqrt{3}$$. So $$\theta = \pi/3$$.

Area in first quadrant below the line corresponds to integrating $$\theta$$ from $$0$$ to $$\pi/3$$ in the auxiliary circle transform:

Area $$= \frac{1}{4}(\pi ab) \times (\text{sector portion}) = \sqrt{3}\pi$$.

Correct Option: C ($$\sqrt{3}\pi$$)