The area enclosed by $y^2=8x$ and $y=\sqrt{2}x$ that lies outside the triangle formed by $\mathrm{y}=\sqrt{2}\mathrm{x},\mathrm{x}=1,\mathrm{y}=2\sqrt{2}$, is equal to 

area of $\mathrm{\Delta BCD}=\frac{1}{2}\left(\sqrt{2}\right)\left(1\right)=\frac{1}{\sqrt{2}}$
area between curves = ${\int }_0^4 \left(2\sqrt{2}{\mathrm{x}}^{\frac{1}{2}}-\sqrt{2}\mathrm{x}\right)\mathrm{dx}$
$\begin{array}{l}={\left.\left(\frac{2\sqrt{2}{\mathrm{x}}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{{\mathrm{x}}^2}{\sqrt{2}}\right)\right|}_0^4={\left.\left(\frac{4}{3}\sqrt{2}{\mathrm{x}}^{\frac{3}{2}}-\frac{1}{\sqrt{2}}{\mathrm{x}}^2\right)\right|}_0^4\\ =\frac{32\sqrt{2}}{3}-8\sqrt{2}=\frac{8\sqrt{2}}{3}\end{array}$
Required area = $\left(\frac{8\sqrt{2}}{3}-\frac{1}{\sqrt{2}}\right)=\left(\frac{8\sqrt{2}}{3}-\frac{\sqrt{2}}{2}\right)=\frac{13\sqrt{2}}{6}$