$A_1$: $y=x^2+2$ meets $y=8-x$ when $x^2+x-6=0\Rightarrow x=2$ (in Q1). The line lies above the parabola on $[0,2]$.

$A_1={\int }_0^2\left[(8-x)-(x^2+2)\right]dx={\int }_0^2(6-x-x^2)dx$

$={\left[6x-\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^2=12-2-\frac{8}{3}=\frac{22}{3}$.

$A_2$: Region between $y=x^2+2$ (top) and $y=\sqrt{x}$ (bottom, from $y^2=x$) on $[0,2]$.

$A_2={\int }_0^2\left[(x^2+2)-\sqrt{x}\right]dx={\left[\frac{x^3}{3}+2x-\frac{2}{3}{x}^{3/2}\right]}_0^2$

$=\frac{8}{3}+4-\frac{4\sqrt{2}}{3}=\frac{20}{3}-\frac{4\sqrt{2}}{3}$.

Difference:

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