Question:

Step 1: Rewrite the equations of the given curves. The first curve is given by $y^2-2y=-x$. We can express $x$ in terms of $y$ to make integration with respect to $y$ easier. Completing the square for the $y$ terms: $$y^2 - 2y + 1 - 1 = -x$$ $$(y-1)^2 - 1 = -x$$ $$x = 1 - (y-1)^2$$ This is a parabola opening towards the negative x-axis with its vertex at $(1,1)$. The second curve is a straight line given by $x+y=0$. We can also express $x$ in terms of $y$: $$x = -y$$ Step 2: Find the points of intersection of the two curves. To find the intersection points, we set the $x$ values equal: $$1 - (y-1)^2 = -y$$ Expand the quadratic term: $$1 - (y^2 - 2y + 1) = -y$$ $$1 - y^2 + 2y - 1 = -y$$ $$-y^2 + 2y = -y$$ Move all terms to one side to form a quadratic equation: $$-y^2 + 3y = 0$$ Factor out $y$: $$y(-y + 3) = 0$$ This gives two possible values for $y$: $y=0$ or $y=3$. For $y=0$, $x = -0 = 0$. So, the first intersection point is $(0,0)$. For $y=3$, $x = -3$. So, the second intersection point is $(-3,3)$. Step 3: Determine the integration strategy and the limits of integration. Since both curves are easily expressed as $x$ in terms of $y$, it is convenient to integrate with respect to $y$. The limits of integration for $y$ are the $y$-coordinates of the intersection points, which are $y=0$ to $y=3$. We need to determine which curve has a larger $x$-value (is to the right) in the interval $y \in [0,3]$. Let's pick a test point, say $y=1$. For the parabola $x_p = 1 - (1-1)^2 = 1$. For the line $x_l = -1$. Since $x_p > x_l$ in this interval, the parabola $x = 1-(y-1)^2$ is the right curve, and the line $x=-y$ is the left curve. The area $A$ is given by the integral: $$A = \int_{y_1}^{y_2} (x_{right} - x_{left}) dy$$ Step 4: Set up and evaluate the definite integral for the area $A$. $$A = \int_{0}^{3} \left( (1 - (y-1)^2) - (-y) \right) dy$$ Expand the term $(y-1)^2$: $$A = \int_{0}^{3} \left( 1 - (y^2 - 2y + 1) + y \right) dy$$ $$A = \int_{0}^{3} \left( 1 - y^2 + 2y - 1 + y \right) dy$$ Simplify the integrand: $$A = \int_{0}^{3} (-y^2 + 3y) dy$$ Now, integrate term by term: $$A = \left[ -\frac{y^3}{3} + \frac{3y^2}{2} \right]_{0}^{3}$$ Evaluate the integral at the limits: $$A = \left( -\frac{(3)^3}{3} + \frac{3(3)^2}{2} \right) - \left( -\frac{(0)^3}{3} + \frac{3(0)^2}{2} \right)$$ $$A = \left( -\frac{27}{3} + \frac{3 \times 9}{2} \right) - (0)$$ $$A = \left( -9 + \frac{27}{2} \right)$$ To combine the terms, find a common denominator: $$A = \left( -\frac{18}{2} + \frac{27}{2} \right)$$ $$A = \frac{9}{2}$$ Step 5: Calculate the value of $8A$. Given that $A = \frac{9}{2}$, we need to find $8A$: $$8A = 8 \times \frac{9}{2}$$ $$8A = 4 \times 9$$ $$8A = 36$$ The final answer is $\boxed{36}$.