If the area of the region $$\{(x,y) : |4 - x^2| \leq y \leq x^2, y \leq 4, x \geq 0\}$$ is $$\left(\frac{80\sqrt{2}}{\alpha} - \beta\right)$$, $$\alpha, \beta \in \mathbb{N}$$, then $$\alpha + \beta$$ is equal to _____.

The region is described by the simultaneous inequalities
  $$|4-x^2|\le y\le x^2,\; y\le 4,\; x\ge 0$$.

Combine the two upper bounds $$y\le x^2$$ and $$y\le 4$$ into a single bound:
  upper bound $$= \min\{x^2,\,4\}$$.
The lower bound remains $$|4-x^2|$$.

We now split the analysis according to whether $$x^2\lt 4$$ or $$x^2\gt 4$$.

Case 1: $$0\le x\le 2$$  ($$x^2\le 4$$)
  Upper bound $$=x^2$$, lower bound $$=4-x^2$$.
  The condition $$4-x^2\le x^2$$ gives $$x^2\ge 2\;\Longrightarrow\;x\ge\sqrt2$$.
  Hence this region exists for $$x\in[\sqrt2,\,2]$$ with
  $$4-x^2\le y\le x^2$$.

Area for Case 1:
$$ A_1=\int_{\sqrt2}^{2}\bigl[x^2-(4-x^2)\bigr]\,dx =\int_{\sqrt2}^{2}\bigl(2x^2-4\bigr)\,dx =2\int_{\sqrt2}^{2}(x^2-2)\,dx. $$
Using $$\int x^2\,dx=\dfrac{x^3}{3}$$:
$$ A_1=2\left[\frac{x^3}{3}-2x\right]_{\sqrt2}^{2} =2\left[\frac{8}{3}-4-\left(\frac{2\sqrt2}{3}-2\sqrt2\right)\right] =\frac{8\,(\,\sqrt2-1\,)}{3}. $$

Case 2: $$x\ge 2$$  ($$x^2\ge 4$$)
  Upper bound $$=4$$, lower bound $$=x^2-4$$.
  Require $$x^2-4\le 4\;\Longrightarrow\;x^2\le 8\;\Longrightarrow\;x\le 2\sqrt2.$$
  Thus this region exists for $$x\in[2,\,2\sqrt2]$$ with
  $$x^2-4\le y\le 4$$.

Area for Case 2:
$$ A_2=\int_{2}^{2\sqrt2}\bigl[4-(x^2-4)\bigr]\,dx =\int_{2}^{2\sqrt2}(8-x^2)\,dx =\left[8x-\frac{x^3}{3}\right]_{2}^{2\sqrt2}. $$
Compute the endpoints:
At $$x=2\sqrt2$$: $$8x-\dfrac{x^3}{3}=16\sqrt2-\dfrac{16\sqrt2}{3}=\dfrac{32\sqrt2}{3}.$$
At $$x=2$$: $$8x-\dfrac{x^3}{3}=16-\dfrac{8}{3}=\dfrac{40}{3}.$$
Therefore
$$ A_2=\frac{32\sqrt2}{3}-\frac{40}{3} =\frac{8\,(\,4\sqrt2-5\,)}{3}. $$

Total area:
$$ A=A_1+A_2 =\frac{8(\sqrt2-1)}{3}+\frac{8(4\sqrt2-5)}{3} =\frac{8}{3}\bigl[5\sqrt2-6\bigr] =\frac{40\sqrt2}{3}-16. $$

Express in the required form $$\dfrac{80\sqrt2}{\alpha}-\beta$$:
$$ \frac{40\sqrt2}{3}-16=\frac{80\sqrt2}{6}-16, $$
so $$\alpha=6,\;\beta=16$$ and $$\alpha+\beta=6+16=22.$$

Final Answer: $$\alpha+\beta=22$$.