The area of the region $$S = \{(x,y) : y^2 \leq 8x, y \geq \sqrt{2}x, x \geq 1\}$$ is

We need to find the area of the region $$S = \{(x,y) : y^2 \leq 8x, \; y \geq \sqrt{2}x, \; x \geq 1\}$$.

The parabola $$y^2 = 8x$$ opens to the right with vertex at origin. The line $$y = \sqrt{2}x$$ passes through the origin.

First, find where the line $$y = \sqrt{2}x$$ intersects the parabola $$y^2 = 8x$$:

$$2x^2 = 8x \implies x = 0$$ or $$x = 4$$

At $$x = 4$$: $$y = 4\sqrt{2}$$.

For the region, we need $$y^2 \leq 8x$$ (inside the parabola), $$y \geq \sqrt{2}x$$ (above the line), and $$x \geq 1$$.

On the upper branch of the parabola, $$y = \sqrt{8x} = 2\sqrt{2x}$$. The condition $$y \geq \sqrt{2}x$$ combined with $$y \leq 2\sqrt{2x}$$ gives the region between the line and parabola for $$x \in [1, 4]$$.

At $$x = 1$$: the line gives $$y = \sqrt{2}$$, and the parabola gives $$y = 2\sqrt{2}$$. So both bounds are valid at $$x = 1$$.

The area is:

$$A = \int_1^4 \left(2\sqrt{2x} - \sqrt{2}x\right) dx$$

$$= \int_1^4 2\sqrt{2}\sqrt{x} \, dx - \int_1^4 \sqrt{2}x \, dx$$

$$= 2\sqrt{2} \cdot \frac{2}{3} x^{3/2} \Big|_1^4 - \sqrt{2} \cdot \frac{x^2}{2} \Big|_1^4$$

$$= \frac{4\sqrt{2}}{3} (4^{3/2} - 1) - \frac{\sqrt{2}}{2}(16 - 1)$$

$$= \frac{4\sqrt{2}}{3}(8 - 1) - \frac{15\sqrt{2}}{2}$$

$$= \frac{28\sqrt{2}}{3} - \frac{15\sqrt{2}}{2}$$

$$= \sqrt{2}\left(\frac{56 - 45}{6}\right) = \frac{11\sqrt{2}}{6}$$

Hence the correct answer is Option D: $$\dfrac{11\sqrt{2}}{6}$$.