To calculate the area of the shaded region between the two curves, we need to:

1. Determine the equations of the curves:

   • The red curve: $x = y^2 - 4y$
   • The blue curve: $x = 2y - y^2$

2. Find the points of intersection:
   Solve $y^2 - 4y = 2y - y^2$. Rearrange and simplify: $y^2 - 4y - 2y + y^2 = 0 \quad \Rightarrow \quad 2y^2 - 6y = 0 \quad \Rightarrow \quad 2y(y - 3) = 0$ Thus, $y = 0$ and $y = 3$. The points of intersection are $y = 0$ and $y = 3$.

3. Set up the integral for the area:
   The area is the integral of the difference between the curves: $\text{Area} = \int_{y=0}^{y=3} \big[(2y - y^2) - (y^2 - 4y)\big] \, dy$ Simplify the integrand: $(2y - y^2) - (y^2 - 4y) = 2y - y^2 - y^2 + 4y = 6y - 2y^2$

4. Integrate:
   $\text{Area} = \int_{0}^{3} (6y - 2y^2) \, dy$ Split and calculate: $\int_{0}^{3} 6y \, dy = \left[3y^2\right]_0^3 = 3(3^2) - 3(0^2) = 27$ $\int_{0}^{3} 2y^2 \, dy = \left[\frac{2y^3}{3}\right]_0^3 = \frac{2(3^3)}{3} - \frac{2(0^3)}{3} = \frac{54}{3} = 18$

   Subtract the results: $\text{Area} = 27 - 18 = 9$

Thus, the area of the shaded region is 9 square units.

Let me know if you need further explanation!

Expanding Questions:

1. What would change if the intersection points were different?
2. How can we verify the setup of the integrals graphically?
3. How does the orientation of the curves affect the integrand?
4. Could this integral be solved numerically instead? How?
5. How does symmetry play a role in simplifying such problems?

Tip:

Always check your curve equations and their limits of integration to ensure the area calculation is accurate!