The area of the region $$A= \left\{(x,y): 4x^{2}+y^{2}\leq 8 \text{and } y^{2} \leq 4x \right\}$$ is :

Region: $$4x^2+y^2 \leq 8$$ (ellipse $$\frac{x^2}{2}+\frac{y^2}{8} \leq 1$$) and $$y^2 \leq 4x$$ (parabola).

Intersection: $$4x^2+4x = 8$$: $$x^2+x-2=0$$: $$(x+2)(x-1)=0$$: $$x=1$$ (taking $$x \geq 0$$).

At $$x=1$$: $$y = \pm 2$$. Using symmetry about x-axis:

Area $$= 2\int_0^1 2\sqrt{x} dx + 2\int_1^{\sqrt{2}} \sqrt{8-4x^2} dx$$.

$$= 4[\frac{2x^{3/2}}{3}]_0^1 + 2\int_1^{\sqrt{2}} 2\sqrt{2-x^2}dx = \frac{8}{3} + 4\int_1^{\sqrt{2}}\sqrt{2-x^2}dx$$.

$$\int_1^{\sqrt{2}}\sqrt{2-x^2}dx = [\frac{x\sqrt{2-x^2}}{2}+\frac{2}{2}\sin^{-1}\frac{x}{\sqrt{2}}]_1^{\sqrt{2}} = (0+\frac{\pi}{2})-(\frac{1}{2}+\frac{\pi}{4}) = \frac{\pi}{4}-\frac{1}{2}$$.

Area $$= \frac{8}{3}+4(\frac{\pi}{4}-\frac{1}{2}) = \frac{8}{3}+\pi-2 = \frac{2}{3}+\pi$$.

The answer is Option 1: $$\pi + 2/3$$.