The area of the region $\mathrm{S}=\left\{(\mathrm{x},\mathrm{y}):{\mathrm{y}}^2\le 8\mathrm{x},\mathrm{y}\ge \sqrt{2}\mathrm{x},\mathrm{x}\ge 1\right\}$ is

$$\begin{gathered} \begin{array}{r}{y}^2=8x...\left(1\right), y=\sqrt{2}x.....\left(2\right)\\ \Rightarrow y^2=2x^2\end{array} \\ \begin{array}{l}\therefore 8x=2x^2\\ \Rightarrow x=0\mathrm{\&}4\\ \text{ Area }:={\int }_1^4 2\sqrt{2}\sqrt{x}-\sqrt{2}xdx\\ =2\sqrt{2}{\left(\frac{x^{\frac{3}{2}}}{3/2}\right)}_1^4-\sqrt{2}{\left(\frac{x^2}{2}\right)}_1^4\\ =\frac{4\sqrt{2}}{3}(8-1)-\frac{\sqrt{2}}{3}(16-1)\\ =\frac{28\sqrt{2}}{3}-\frac{15\sqrt{2}}{2}=\frac{11\sqrt{2}}{6}\end{array} \end{gathered}$$