The area enclosed by \( y^{2}=8x \) and \( y=\sqrt{2x} \) that lies outside the triangle formed by \( y=\sqrt{2x}, \ x=1, \ y=2\sqrt{2} \) is equal to:
\( \dfrac{16\sqrt{2}}{6} \)
\( \dfrac{11\sqrt{2}}{6} \)
\( \dfrac{13\sqrt{2}}{6} \)
\( \dfrac{5\sqrt{2}}{6} \)