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

A

$$\frac{16\sqrt{2}}{6}$$

B

$$\frac{11\sqrt{2}}{6}$$

C

$$\frac{13\sqrt{2}}{6}$$

D

$$\frac{5\sqrt{2}}{6}$$