$$\begin{aligned} & \text { Area }=\int_\limits0^{3 \sqrt{2}} x d x+\int_\limits{3 / \sqrt{2}}^{3 \sqrt{2}} \sqrt{\frac{18-x^2}{3}} d x \\ & =\frac{1}{2}\left(x^2\right)_0^{3 \sqrt{2}}+\frac{1}{\sqrt{3}}\left[\frac{x}{2} \sqrt{18-x^2}+9 \sin ^{-1}\left(\frac{x}{3 \sqrt{2}}\right)\right]_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \\ & =\frac{1}{2}\left(\frac{9}{2}\right)+\frac{1}{\sqrt{3}}\left[9 \sin ^{-1}(1)-\frac{3}{2 \sqrt{2}} \cdot \frac{3 \sqrt{3}}{\sqrt{2}}-9 \sin ^{-1}\left(\frac{1}{2}\right)\right] \\ & =\frac{9}{4}+\frac{1}{\sqrt{3}}\left(\frac{9 \pi}{2}-\frac{9 \sqrt{3}}{4}-\frac{9 \pi}{6}\right)=\sqrt{3} \pi \end{aligned}$$