Let the area enclosed between the curves $$|y|= 1-x^{2}$$ and $$x^{2}+y^{2}=1$$ be $$\alpha$$. If $$9\alpha$$ = $$\beta \pi + \gamma; \beta,\gamma$$ are integers, then the value of $$|\beta - \gamma |$$ equals.

Curves: 1. $$|y| = 1 - x^2 \implies y = \pm(1 - x^2)$$. These are two parabolas opening toward each other, intersecting the x-axis at $$(\pm1, 0)$$.

2. $$x^2 + y^2 = 1$$. This is a unit circle centered at the origin.

The area is symmetric across both axes. We can calculate the area in the first quadrant and multiply by 4.

In the first quadrant ($$x, y \ge 0$$), we find the area between the circle $$y = \sqrt{1-x^2}$$ and the parabola $$y = 1-x^2$$.

$$\frac{\alpha}{4} = \int_{0}^{1} (\sqrt{1-x^2} - (1-x^2)) \, dx$$

$$\frac{\alpha}{4} = \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}x - x + \frac{x^3}{3} \right]_0^1 = \frac{\pi}{4} - 1 + \frac{1}{3} = \frac{\pi}{4} - \frac{2}{3}$$

$$\alpha = \pi - \frac{8}{3} \implies 9\alpha = 9\pi - 24$$

 Comparing $$9\alpha = \beta\pi + \gamma$$, we get $$\beta = 9$$ and $$\gamma = -24$$.

$$|\beta - \gamma| = |9 - (-24)| = |9 + 24| = \mathbf{33}$$