Calculating Area $A_1$

Area $A_1$ is enclosed by $y = x^2 + 2$, $x + y = 8$ (or $y = 8 - x$), and the $y$-axis ($x=0$) in the first quadrant.

1. Find intersection points: Set $x^2 + 2 = 8 - x$. This gives $x^2 + x - 6 = 0$, or $(x+3)(x-2) = 0$. In the first quadrant, the intersection is at $x=2$.
2. Set up the integral: The upper curve is $y = 8 - x$ and the lower curve is $y = x^2 + 2$ between $x=0$ and $x=2$. $A_1 = \int_{0}^{2} [(8 - x) - (x^2 + 2)] dx$
3. Calculate the integral: $A_1 = \int_{0}^{2} (6 - x - x^2) dx = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{2}$ $A_1 = \left( 6(2) - \frac{2^2}{2} - \frac{2^3}{3} \right) - (0) = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{30 - 8}{3} = \frac{22}{3}$

Calculating Area $A_2$

Area $A_2$ is enclosed by $y = x^2 + 2$, $y^2 = x$ (or $y = \sqrt{x}$ in the first quadrant), $x = 2$, and the $y$-axis ($x=0$) in the first quadrant.

1. Determine the upper and lower curves: For $x \in [0, 2]$, $x^2+2 \ge \sqrt{x}$. Thus, $y = x^2 + 2$ is the upper curve and $y = \sqrt{x}$ is the lower curve.
2. Set up the integral: The area is between $x=0$ and $x=2$. $A_2 = \int_{0}^{2} [(x^2 + 2) - \sqrt{x}] dx$
3. Calculate the integral: $A_2 = \int_{0}^{2} (x^2 + 2 - x^{1/2}) dx = \left[ \frac{x^3}{3} + 2x - \frac{x^{3/2}}{3/2} \right]_{0}^{2}$ $A_2 = \left[ \frac{x^3}{3} + 2x - \frac{2}{3}x^{3/2} \right]_{0}^{2}$ $A_2 = \left( \frac{2^3}{3} + 2(2) - \frac{2}{3}(2^{3/2}) \right) - (0)$ $A_2 = \frac{8}{3} + 4 - \frac{2}{3}(2\sqrt{2}) = \frac{8}{3} + \frac{12}{3} - \frac{4\sqrt{2}}{3} = \frac{20 - 4\sqrt{2}}{3}$

Calculating $A_1 - A_2$

Subtract the calculated areas $A_1$ and $A_2$. $A_1 - A_2 = \frac{22}{3} - \frac{20 - 4\sqrt{2}}{3}$

1. Perform the subtraction: $A_1 - A_2 = \frac{22 - (20 - 4\sqrt{2})}{3} = \frac{22 - 20 + 4\sqrt{2}}{3} = \frac{2 + 4\sqrt{2}}{3}$
2. Factor the result: $A_1 - A_2 = \frac{2(1 + 2\sqrt{2})}{3} = \frac{2}{3}(1 + 2\sqrt{2})$

This result matches option C, which is $\frac{2}{3}(2\sqrt{2} + 1)$.