The area of the region in the first quadrant inside the circle $$x^2 + y^2 = 8$$ and outside the parabola $$y^2 = 2x$$ is equal to :

Find intersection: $$x^2 + 2x = 8 \implies x^2 + 2x - 8 = 0 \implies (x+4)(x-2) = 0$$.

In 1st quadrant, $$x = 2$$. At $$x=2, y=2$$.

2. The requested area is: (Area of circular sector) - (Area under parabola).

Area $$= \int_{0}^{2} (\sqrt{8-x^2} - \sqrt{2x}) dx$$.

Wait, the problem says "inside circle" and "outside parabola".

Total Area of circle in 1st quadrant is $$\frac{1}{4}\pi(2\sqrt{2})^2 = 2\pi$$.

Area $$= \int_{0}^{2} (\sqrt{8-x^2} - \sqrt{2x}) dx + \int_{2}^{2\sqrt{2}} \sqrt{8-x^2} dx$$.

Simplest way: Area = (Area of quadrant) - (Area under parabola from 0 to 2) - (Area under circle from 0 to 2).

Actually, it is simpler: Area $$= \text{Area of Sector} - \text{Area under Parabola}$$.

Looking at the options, the integral setup is:

Area $$= \int_{0}^{2} (\sqrt{8-y^2}/ \text{dy logic is messy}) \to$$ Let's use $$x$$:

$$\text{Area} = \int_{0}^{2} \sqrt{8-x^2} dx - \int_{0}^{2} \sqrt{2x} dx$$

$$= [\frac{x}{2}\sqrt{8-x^2} + \frac{8}{2}\sin^{-1}\frac{x}{2\sqrt{2}}]_{0}^{2} - [\frac{2\sqrt{2}}{3}x^{3/2}]_{0}^{2}$$

$$= (2 + 4 \cdot \frac{\pi}{4}) - \frac{2\sqrt{2} \cdot 2\sqrt{2}}{3} = 2 + \pi - \frac{8}{3} = \pi - \frac{2}{3}$$.