Let $$A_{1}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,x+Y=8$$ and y-axis that lies in the first quadrant. Let $$A_{2}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,y^{2}=x,x=2$$ and y-axis that lies in the first quadrant. Then $$A_{1}-A_{2}$$ is equal to

To find $$A_1 - A_2$$, we first compute the areas $$A_1$$ and $$A_2$$ separately.

The bounded area $$A_1$$ is enclosed by the curves $$y = x^2 + 2$$, $$x + y = 8$$, and the y-axis in the first quadrant.

The curves intersect when $$x + (x^2 + 2) = 8$$, simplifying to $$x^2 + x - 6 = 0$$. Solving this quadratic equation gives a discriminant of $$1 + 24 = 25$$, so $$x = \frac{-1 \pm 5}{2}$$. Thus, $$x = 2$$ (since $$x \geq 0$$ in the first quadrant) and $$y = 2^2 + 2 = 6$$ at the intersection point $$(2, 6)$$.

The line $$x + y = 8$$ intersects the y-axis at $$(0, 8)$$, and the parabola $$y = x^2 + 2$$ intersects the y-axis at $$(0, 2)$$. Between $$x = 0$$ and $$x = 2$$, the line $$y = 8 - x$$ is above the parabola $$y = x^2 + 2$$, so the area $$A_1$$ is given by

$$ A_1 = \int_{0}^{2} \bigl((8 - x) - (x^2 + 2)\bigr)\,dx = \int_{0}^{2} (6 - x - x^2)\,dx $$

Computing this integral, the antiderivative is $$6x - \frac{1}{2}x^2 - \frac{1}{3}x^3$$. Evaluating at $$x = 2$$ yields $$6(2) - \frac{1}{2}(4) - \frac{1}{3}(8) = 12 - 2 - \frac{8}{3} = \frac{22}{3}$$, and at $$x = 0$$ it is $$0$$. Therefore, $$A_1 = \frac{22}{3}$$.

The bounded area $$A_2$$ is enclosed by the curves $$y = x^2 + 2$$, $$y^2 = x$$, the line $$x = 2$$, and the y-axis in the first quadrant.

Rewriting $$y^2 = x$$ as $$y = \sqrt{x}$$ in the first quadrant, the relevant points are:

Intersection of $$y^2 = x$$ and $$x = 2$$: $$(2, \sqrt{2})$$

Intersection of $$y = x^2 + 2$$ and $$x = 2$$: $$(2, 6)$$

Intersection of $$y = x^2 + 2$$ and the y-axis: $$(0, 2)$$

Intersection of $$y^2 = x$$ and the y-axis: $$(0, 0)$$

Between $$x = 0$$ and $$x = 2$$, the parabola $$y = x^2 + 2$$ lies above $$y = \sqrt{x}$$, so the area $$A_2$$ is given by

$$ A_2 = \int_{0}^{2} \bigl((x^2 + 2) - \sqrt{x}\bigr)\,dx = \int_{0}^{2} \bigl(x^2 + 2 - x^{1/2}\bigr)\,dx $$

Here the antiderivative is $$\frac{1}{3}x^3 + 2x - \frac{2}{3}x^{3/2}$$. At $$x = 2$$ this becomes $$\frac{1}{3}(8) + 2(2) - \frac{2}{3}(2^{3/2}) = \frac{8}{3} + 4 - \frac{2}{3}(2\sqrt{2}) = \frac{8}{3} + 4 - \frac{4\sqrt{2}}{3} = \frac{20 - 4\sqrt{2}}{3}$$, and at $$x = 0$$ it is $$0$$. Hence, $$A_2 = \frac{20 - 4\sqrt{2}}{3}$$.

Subtracting these results gives

$$ A_1 - A_2 = \frac{22}{3} - \frac{20 - 4\sqrt{2}}{3} = \frac{22 - 20 + 4\sqrt{2}}{3} = \frac{2 + 4\sqrt{2}}{3} = \frac{2}{3}(2\sqrt{2} + 1) $$

Comparing with the options shows that $$\frac{2}{3}(2\sqrt{2} + 1)$$ corresponds to option A. Thus, $$A_1 - A_2 = \boxed{\frac{2}{3}(2\sqrt{2} + 1)}$$.