1. Write the curves in convenient form

   The given curves are: $y^2-2y=-x$ $x+y=0$

   Rewrite them as: $x=2y-y^2$ and $x=-y$

   Since both are expressed naturally as $x$ in terms of $y$, it is convenient to integrate with respect to $y$.

2. Find the points of intersection

   At intersection, $2y-y^2=-y$ $3y-y^2=0$ $y(3-y)=0$

   Hence, $y=0 \,\text{or}\, y=3$

   Corresponding $x$-coordinates:

   • For $y=0$, $x=-y=0$, so point is $(0,0)$.
   • For $y=3$, $x=-y=-3$, so point is $(-3,3)$.

3. Determine which curve is on the right and which is on the left

   For $0\le y\le 3$,

   \qquad x_{\text{line}}=-y$$ Compare: $$x_{\text{parabola}}-x_{\text{line}}=(2y-y^2)-(-y)=3y-y^2=y(3-y)\ge 0$$ So, in this interval, the parabola lies to the **right** and the line lies to the **left**.

4. Set up the area integral

   Therefore, $A=\int_0^3 \big(x_{\text{right}}-x_{\text{left}}\big)\,dy$ $A=\int_0^3 \big[(2y-y^2)-(-y)\big]dy$ $A=\int_0^3 (3y-y^2)\,dy$

5. Evaluate the integral

   $A=\left[\frac{3y^2}{2}-\frac{y^3}{3}\right]_0^3$ $A=\left(\frac{3\cdot 9}{2}-\frac{27}{3}\right)-0$ $A=\frac{27}{2}-9$ $A=\frac{27-18}{2}=\frac{9}{2}$

6. Compute $8A$

   $8A=8\cdot \frac{9}{2}=36$

Final Answer

$\boxed{36}$

The derived answer matches the stored correct answer.