Key Concepts and Formulas

• Area Between Curves (Integration with respect to y): If $f(y) \ge g(y)$ for $c \le y \le d$, the area of the region bounded by $x=f(y)$, $x=g(y)$, $y=c$, and $y=d$ is given by $A = \int_{c}^{d} [f(y) - g(y)] \, dy$.
• Finding Intersection Points: To find where two curves intersect, set their equations equal to each other and solve for the variable (in this case, $y$).
• Basic Integration: $\int y^n \, dy = \frac{y^{n+1}}{n+1} + C$, where $n \neq -1$.

Step-by-Step Solution

Step 1: Express the curves in the form x = f(y)

We are given the equations $y^2 - 2y = -x$ and $x + y = 0$. We need to rewrite both equations in the form $x = f(y)$ to facilitate integration with respect to $y$.

• For the first equation, $y^2 - 2y = -x$, we multiply both sides by -1 to get: $x = 2y - y^2$

• For the second equation, $x + y = 0$, we subtract $y$ from both sides to get: $x = -y$

Step 2: Find the points of intersection

To find the points of intersection, we set the two expressions for $x$ equal to each other and solve for $y$:

$2y - y^2 = -y$

Adding $y$ to both sides, we get:

$3y - y^2 = 0$

Factoring out $y$, we have:

$y(3 - y) = 0$

This gives us two solutions for $y$:

$y = 0 \quad \text{or} \quad y = 3$

Now, we substitute these $y$ values back into either equation to find the corresponding $x$ values. Using $x = -y$:

• If $y = 0$, then $x = -0 = 0$. So, one intersection point is $(0, 0)$.

• If $y = 3$, then $x = -3$. So, the other intersection point is $(-3, 3)$.

Thus, the limits of integration will be from $y = 0$ to $y = 3$.

Step 3: Determine which curve is to the right

We need to determine which curve has the larger $x$-value for a given $y$ in the interval $[0, 3]$. Let's test $y = 1$:

• For $x = 2y - y^2$, when $y = 1$, $x = 2(1) - (1)^2 = 1$.

• For $x = -y$, when $y = 1$, $x = -1$.

Since $1 > -1$, the parabola $x = 2y - y^2$ is to the right of the line $x = -y$ in the interval $[0, 3]$.

Step 4: Set up the definite integral for the area A

The area A is given by the integral:

$A = \int_{0}^{3} [(2y - y^2) - (-y)] \, dy$

Simplifying the integrand:

$A = \int_{0}^{3} (3y - y^2) \, dy$

Step 5: Evaluate the definite integral

Integrating term by term, we have:

$A = \left[ \frac{3y^2}{2} - \frac{y^3}{3} \right]_{0}^{3}$

Evaluating at the limits of integration:

$A = \left( \frac{3(3)^2}{2} - \frac{(3)^3}{3} \right) - \left( \frac{3(0)^2}{2} - \frac{(0)^3}{3} \right)$

$A = \left( \frac{27}{2} - \frac{27}{3} \right) - (0 - 0)$

$A = \frac{27}{2} - 9 = \frac{27}{2} - \frac{18}{2} = \frac{9}{2}$

So, the area of the region is $A = \frac{9}{2}$.

Step 6: Calculate 8A

The question asks for the value of $8A$.

$8A = 8 \times \frac{9}{2} = 4 \times 9 = 36$

Common Mistakes & Tips

• Incorrect Order of Subtraction: Always ensure you are subtracting the 'left' function from the 'right' function when integrating with respect to $y$.
• Sign Errors: Pay close attention to signs during algebraic manipulations, especially when simplifying the integrand.
• Choosing the Right Variable: Consider integrating with respect to $x$ or $y$ and choose the one that simplifies the problem.

Summary

We found the area between the curves $y^2 - 2y = -x$ and $x+y=0$ by expressing both equations in the form $x = f(y)$, finding their intersection points, and integrating the difference of the functions with respect to $y$ from $y=0$ to $y=3$. The area was calculated as $A = \frac{9}{2}$, and the value of $8A$ is 36.

Final Answer

The final answer is \boxed{36}.