To find the area between the curves $x = 2y - y^2$ and $x = -12 + y^2$ by integrating with respect to $y$, we need to follow these steps:

Step 1: Set up the curves

We are given two functions of $y$:

1. $x = 2y - y^2$
2. $x = -12 + y^2$

Step 2: Find the points of intersection

To determine the limits of integration, we need to find where the two curves intersect. This occurs when $2y - y^2 = -12 + y^2$. Solving for $y$:

$2y - y^2 = -12 + y^2$

$2y - y^2 - y^2 = -12$

$2y - 2y^2 = -12$

Divide through by 2:

$y - y^2 = -6$

$y^2 - y - 6 = 0$

Now, solve the quadratic equation using the quadratic formula:

$y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)} = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm \sqrt{25}}{2} = \frac{1 \pm 5}{2}$

So, the two solutions are:

$y = \frac{1 + 5}{2} = 3, \quad y = \frac{1 - 5}{2} = -2$

Thus, the curves intersect at $y = -2$ and $y = 3$.

Step 3: Set up the integral

The area between the curves is given by the integral of the difference between the rightmost curve and the leftmost curve from $y = -2$ to $y = 3$.

For $x = 2y - y^2$ and $x = -12 + y^2$, the rightmost curve is $x = 2y - y^2$ and the leftmost curve is $x = -12 + y^2$.

The area $A$ is:

$A = \int_{-2}^{3} \left[ (2y - y^2) - (-12 + y^2) \right] \, dy$

Simplify the integrand:

$A = \int_{-2}^{3} \left( 2y - y^2 + 12 - y^2 \right) \, dy$

$A = \int_{-2}^{3} \left( 2y + 12 - 2y^2 \right) \, dy$

Step 4: Integrate

Now, we integrate term by term:

$A = \int_{-2}^{3} 2y \, dy + \int_{-2}^{3} 12 \, dy - \int_{-2}^{3} 2y^2 \, dy$

• For $\int 2y \, dy$:

$\int 2y \, dy = y^2$

• For $\int 12 \, dy$:

$\int 12 \, dy = 12y$

• For $\int 2y^2 \, dy$:

$\int 2y^2 \, dy = \frac{2y^3}{3}$

Now, evaluate these at the limits $y = -2$ and $y = 3$:

$A = \left[ y^2 + 12y - \frac{2y^3}{3} \right]_{-2}^{3}$

Step 5: Compute the result

Substitute $y = 3$ and $y = -2$ into the expression, then subtract the values:

For $y = 3$:

$A(3) = 3^2 + 12(3) - \frac{2(3)^3}{3} = 9 + 36 - \frac{54}{3} = 9 + 36 - 18 = 27$

For $y = -2$:

$A(-2) = (-2)^2 + 12(-2) - \frac{2(-2)^3}{3} = 4 - 24 - \frac{-16}{3} = 4 - 24 + \frac{16}{3} = -20 + \frac{16}{3} = \frac{-60 + 16}{3} = \frac{-44}{3}$

Now subtract:

$A = 27 - \left( \frac{-44}{3} \right) = 27 + \frac{44}{3} = \frac{81}{3} + \frac{44}{3} = \frac{125}{3}$

Thus, the area is $\frac{125}{3}$ square units.

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Would you like to dive into any particular detail or clarification? Here are 5 related questions you might find interesting:

1. How would the process change if integrating with respect to $x$?
2. What would the curves look like if graphed? How does the shape influence the solution?
3. How do you determine which function is the "rightmost" or "leftmost" in terms of integration?
4. Can the intersection points always be found algebraically? What if they can’t?
5. How would the answer differ if there were multiple regions of intersection?

Tip: Always check whether functions intersect more than once or whether a region needs to be broken into subregions for correct calculation.