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

The ellipse is $$x^{2}+3y^{2}=18$$.
Its first-quadrant part is bounded by the coordinate axes, the ellipse itself and, for our problem, the line $$y=x$$.

First find the intersection of the line and the ellipse:
Substitute $$y=x$$ in $$x^{2}+3y^{2}=18$$:

$$x^{2}+3x^{2}=18 \;\Longrightarrow\; 4x^{2}=18 \;\Longrightarrow\; x^{2}=4.5 \;\Longrightarrow\; x=\frac{3}{\sqrt{2}}$$
Hence the common point in the first quadrant is $$\left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)$$.

For $$0\le y\le\frac{3}{\sqrt{2}}$$ every vertical strip inside the required region starts at $$x=y$$ (the line) and ends at the ellipse, where

$$x=\sqrt{\,18-3y^{2}\,}$$

Therefore, the required area $$A$$ is

$$A=\int_{0}^{\frac{3}{\sqrt{2}}}\Bigl(\sqrt{18-3y^{2}}-y\Bigr)\,dy$$

Split the integral:

$$A=\underbrace{\int_{0}^{\frac{3}{\sqrt{2}}}\sqrt{18-3y^{2}}\;dy}_{I}\;-\;\underbrace{\int_{0}^{\frac{3}{\sqrt{2}}}y\,dy}_{J}$$

Evaluating $$I$$:
Put $$y=\sqrt{6}\sin\theta\;,\;0\le\theta\le\frac{\pi}{3}$$ (because $$y=\frac{3}{\sqrt{2}}\Rightarrow\sin\theta=\frac{\sqrt{3}}{2}$$).
Then $$dy=\sqrt{6}\cos\theta\,d\theta$$ and

$$\sqrt{18-3y^{2}}=\sqrt{18-18\sin^{2}\theta}=3\sqrt{2}\cos\theta$$

Thus

$$I=\int_{0}^{\pi/3}(3\sqrt{2}\cos\theta)(\sqrt{6}\cos\theta)\,d\theta =6\sqrt{3}\int_{0}^{\pi/3}\cos^{2}\theta\,d\theta$$

Use $$\cos^{2}\theta=\frac{1+\cos2\theta}{2}$$:

$$I=6\sqrt{3}\int_{0}^{\pi/3}\frac{1+\cos2\theta}{2}\,d\theta =3\sqrt{3}\Bigl[\theta+\tfrac{1}{2}\sin2\theta\Bigr]_{0}^{\pi/3}$$

At $$\theta=\pi/3$$, $$\sin2\theta=\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}$$.
Hence

$$I=3\sqrt{3}\left(\frac{\pi}{3}+\frac{\sqrt{3}}{4}\right) =\sqrt{3}\pi+\frac{9}{4}$$

Evaluating $$J$$:

$$J=\int_{0}^{\frac{3}{\sqrt{2}}}y\,dy =\frac{1}{2}y^{2}\Bigl|_{0}^{\frac{3}{\sqrt{2}}} =\frac{1}{2}\left(\frac{9}{2}\right)=\frac{9}{4}$$

Required area:

$$A=I-J=\left(\sqrt{3}\pi+\frac{9}{4}\right)-\frac{9}{4} =\sqrt{3}\pi$$

So, the area enclosed by the ellipse in the first quadrant below the line $$y=x$$ equals $$\sqrt{3}\pi$$ square units.

Option C is correct.