Let the area of the region $$\{(x, y) : x - 2y + 4 \geq 0, \; x + 2y^2 \geq 0, \; x + 4y^2 \leq 8, \; y \geq 0\}$$ be $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime numbers. Then $$m + n$$ is equal to _______.

We need the area of the region defined by:

1) $$x - 2y + 4 \geq 0$$ (i.e., $$x \geq 2y - 4$$)

2) $$x + 2y^2 \geq 0$$ (i.e., $$x \geq -2y^2$$)

3) $$x + 4y^2 \leq 8$$ (i.e., $$x \leq 8 - 4y^2$$)

4) $$y \geq 0$$

Let us integrate with respect to $$y$$. From condition 3: $$x \leq 8 - 4y^2$$, so $$y \leq \sqrt{2}$$ (for $$x \geq 0$$, max).

The right boundary is $$x = 8 - 4y^2$$.

The left boundary is $$x = \max(2y - 4, -2y^2)$$.

For $$y \geq 0$$: When is $$2y - 4 \geq -2y^2$$?

$$2y^2 + 2y - 4 \geq 0$$, i.e., $$y^2 + y - 2 \geq 0$$, i.e., $$(y+2)(y-1) \geq 0$$. Since $$y \geq 0$$: $$y \geq 1$$.

So left boundary = $$-2y^2$$ for $$0 \leq y \leq 1$$ and $$2y - 4$$ for $$y \geq 1$$.

Upper limit from right boundary: $$8 - 4y^2 \geq 2y - 4$$ gives $$4y^2 + 2y - 12 \leq 0$$, $$2y^2 + y - 6 \leq 0$$, $$(2y - 3)(y + 2) \leq 0$$, so $$y \leq 3/2$$.

Also $$8 - 4y^2 \geq -2y^2$$ gives $$2y^2 \leq 8$$, $$y \leq 2$$.

Area = $$\int_0^1 (8 - 4y^2 - (-2y^2)) dy + \int_1^{3/2} (8 - 4y^2 - (2y - 4)) dy$$

$$= \int_0^1 (8 - 2y^2) dy + \int_1^{3/2} (12 - 4y^2 - 2y) dy$$

First integral: $$\left[8y - \frac{2y^3}{3}\right]_0^1 = 8 - \frac{2}{3} = \frac{22}{3}$$

Second integral: $$\left[12y - \frac{4y^3}{3} - y^2\right]_1^{3/2}$$

At $$y = 3/2$$: $$18 - \frac{4 \cdot 27/8}{3} - \frac{9}{4} = 18 - \frac{9}{2} - \frac{9}{4} = 18 - \frac{27}{4} = \frac{45}{4}$$

At $$y = 1$$: $$12 - \frac{4}{3} - 1 = 11 - \frac{4}{3} = \frac{29}{3}$$

Second integral = $$\frac{45}{4} - \frac{29}{3} = \frac{135 - 116}{12} = \frac{19}{12}$$

Total area = $$\frac{22}{3} + \frac{19}{12} = \frac{88 + 19}{12} = \frac{107}{12}$$

$$m + n = 107 + 12 = 119$$.

The answer is $$\boxed{119}$$.