$\left|{\mathrm{X}}^2-2\right|\le \mathrm{Y}\le \mathrm{X}$

As shown in figure find point of intersection
$$\begin{gathered} \text{ For }\mathrm{x}\in (0,\sqrt{2})2-{\mathrm{x}}^2=\mathrm{x}\Rightarrow \mathrm{x}=1 \\ \text{ For }\mathrm{x}\in (\sqrt{2},\mathrm{\infty }){\mathrm{x}}^2-2=\mathrm{x}\Rightarrow \mathrm{x}=2 \\ \begin{array}{l}\text{ Area }={\int }_1^{\sqrt{2}} \left(\mathrm{x}-2+{\mathrm{x}}^2\right)\mathrm{dx}+{\int }_{\sqrt{2}}^2 \left(\mathrm{x}-{\mathrm{x}}^2+2\right)\mathrm{dx}\\ =\frac{{\mathrm{x}}^2}{2}-2\mathrm{x}+{\left.\frac{{\mathrm{x}}^3}{3}\right|}_1^2+\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^3}{3}+{\left.2\mathrm{x}\right|}_{\sqrt{2}}^2\\ =\frac{9}{2}-\frac{8\sqrt{2}}{3}\\ \therefore 6\mathrm{A}+16\sqrt{2}=27\end{array} \end{gathered}$$