Let $$S$$ be the region bounded by the curves $$y = x^3$$ and $$y^2 = x$$. The curve $$y = 2|x|$$ divides $$S$$ into two regions of areas $$R_1$$ and $$R_2$$. If $$ |R_1, R_2 | = R_2$$, then $$\frac{R_2}{R_1}$$ is equal to ______

The region $$S$$ is bounded by $$y = x^3$$ and $$y^2 = x$$ (i.e., $$y = \sqrt{x}$$) in the first quadrant, from $$x = 0$$ to $$x = 1$$.

First, the total area of $$S$$ is given by the integral $$\text{Area}(S) = \int_0^1 (\sqrt{x} - x^3)\,dx = \left[\frac{2x^{3/2}}{3} - \frac{x^4}{4}\right]_0^1 = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}.$$

Next, we determine where the line $$y = 2x$$ intersects the boundary curves. Substituting into $$y = \sqrt{x}$$ gives $$2x = \sqrt{x} \implies 4x^2 = x \implies x(4x - 1) = 0 \implies x = \frac{1}{4},$$ whereas substituting into $$y = x^3$$ leads to $$2x = x^3 \implies x^2 = 2 \implies x = \sqrt{2} > 1,$$ so the latter intersection lies outside $$S$$. Thus the line $$y = 2x$$ enters $$S$$ at the origin and exits through $$y = \sqrt{x}$$ at $$\left(\frac{1}{4}, \frac{1}{2}\right)$$.

We then split $$S$$ into two regions: $$R_1$$ is the portion above $$y = 2x$$ for $$x \in [0, 1/4]$$ (between $$y = \sqrt{x}$$ and $$y = 2x$$), and $$R_2$$ is the remainder of $$S$$, namely the region below $$y = 2x$$ for $$x \in [0, 1/4]$$ together with the entire region for $$x \in [1/4, 1]$$.

To compute $$R_1$$ we evaluate $$R_1 = \int_0^{1/4} (\sqrt{x} - 2x)\,dx = \left[\frac{2x^{3/2}}{3} - x^2\right]_0^{1/4} = \frac{2 \cdot \frac{1}{8}}{3} - \frac{1}{16} = \frac{1}{12} - \frac{1}{16} = \frac{1}{48}.$$

Since the total area of $$S$$ is $$\frac{5}{12}$$, the area of the remaining region is $$R_2 = \frac{5}{12} - \frac{1}{48} = \frac{20}{48} - \frac{1}{48} = \frac{19}{48}.$$

Because $$R_2 = \frac{19}{48} > R_1 = \frac{1}{48}$$, the desired ratio is $$\frac{R_2}{R_1} = \frac{19/48}{1/48} = 19.$$

Therefore, the correct answer is $$19$$.