If the area of the region $$\{(x, y): |x^2 - 2| \leq y \leq x\}$$ is A, then $$6A + 16\sqrt{2}$$ is equal to _______.

1. Analyze the Inequalities

The region is defined by $$|x^2 - 2| \le y \le x$$.

This implies two conditions:

1. $$y \ge |x^2 - 2|$$: The region is above the absolute value of the parabola $$y = x^2 - 2$$.
2. $$y \le x$$: The region is below the line $$y = x$$.

2. Find Intersection Points

We need to find where $$x = |x^2 - 2|$$:

• Case 1 ($$x^2 - 2 \ge 0$$): $$x^2 - 2 = x \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0$$.

Since $$x^2 \ge 2$$, we take $$x = 2$$.

• Case 2 ($$x^2 - 2 < 0$$): $$-(x^2 - 2) = x \implies x^2 + x - 2 = 0 \implies (x+2)(x-1) = 0$$.

Since $$x^2 < 2$$, we take $$x = 1$$.

The region $$A$$ is bounded between $$x = 1$$ and $$x = 2$$.

3. Set up the Integral for Area $$A$$

The area $$A$$ is calculated by:

$$A = \int_{1}^{2} (x - |x^2 - 2|) \, dx$$

Since $$x^2 - 2$$ is positive for $$x \in [\sqrt{2}, 2]$$ and negative for $$x \in [1, \sqrt{2}]$$, we split the integral:

$$A = \int_{1}^{\sqrt{2}} (x - (2 - x^2)) \, dx + \int_{\sqrt{2}}^{2} (x - (x^2 - 2)) \, dx$$

$$A = \int_{1}^{\sqrt{2}} (x^2 + x - 2) \, dx + \int_{\sqrt{2}}^{2} (-x^2 + x + 2) \, dx$$

4. Calculate the Integrals

1. First part: $$\left[ \frac{x^3}{3} + \frac{x^2}{2} - 2x \right]_{1}^{\sqrt{2}} = \left( \frac{2\sqrt{2}}{3} + 1 - 2\sqrt{2} \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) = \frac{-4\sqrt{2}}{3} + 1 + \frac{7}{6} = \frac{13}{6} - \frac{4\sqrt{2}}{3}$$
2. Second part: $$\left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{\sqrt{2}}^{2} = \left( -\frac{8}{3} + 2 + 4 \right) - \left( -\frac{2\sqrt{2}}{3} + 1 + 2\sqrt{2} \right) = \frac{10}{3} - \frac{4\sqrt{2}}{3} - 1 = \frac{7}{3} - \frac{4\sqrt{2}}{3}$$

Summing them:

$$A = \left( \frac{13}{6} - \frac{4\sqrt{2}}{3} \right) + \left( \frac{14}{6} - \frac{4\sqrt{2}}{3} \right) = \frac{27}{6} - \frac{8\sqrt{2}}{3} = \frac{9}{2} - \frac{8\sqrt{2}}{3}$$

5. Final Calculation

We need to find the value of $$6A + 16\sqrt{2}$$:

$$6 \left( \frac{9}{2} - \frac{8\sqrt{2}}{3} \right) + 16\sqrt{2}$$

$$27 - 16\sqrt{2} + 16\sqrt{2} = \mathbf{27}$$

Final Answer: 27