If the area of the region bounded by the curves $$y^2 - 2y = -x$$ and $$x + y = 0$$ is $$A$$, then $$8A =$$

We need to find $$8A$$ where $$A$$ is the area bounded by $$y^2 - 2y = -x$$ and $$x + y = 0$$.

Parabola: $$x = 2y - y^2 = -(y - 1)^2 + 1$$ (opens left, vertex at $$(1, 1)$$).

Line: $$x = -y$$.

Setting $$-y = 2y - y^2$$ yields $$y^2 - 3y = 0 \implies y(y - 3) = 0$$, so the curves intersect at $$(0, 0)$$ and $$(-3, 3)$$.

For $$0 \leq y \leq 3$$, the parabola $$x = 2y - y^2$$ lies to the right of the line $$x = -y$$, and thus $$A = \int_0^3 \left[(2y - y^2) - (-y)\right] dy = \int_0^3 (3y - y^2)\,dy$$.

Evaluating, $$A = \left[\dfrac{3y^2}{2} - \dfrac{y^3}{3}\right]_0^3 = \dfrac{27}{2} - \dfrac{27}{3} = \dfrac{27}{2} - 9 = \dfrac{9}{2}$$.

Therefore, $$8A = 8 \times \dfrac{9}{2} = 36$$, and the answer is $$\boxed{36}$$.