Key Concepts and Formulas

• Area Between Curves (Integration with respect to y): If $x = f(y)$ is the right boundary and $x = g(y)$ is the left boundary of a region between $y = c$ and $y = d$, then the area $A$ of the region is given by $A = \int_{c}^{d} [f(y) - g(y)] \, dy$.
• Intersection of Curves: To find the points of intersection of two curves, we set their equations equal to each other and solve for the variable(s).
• Solving Quadratic Equations: The quadratic formula is used to find the roots of a quadratic equation in the form $ax^2 + bx + c = 0$, given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Identify the curves and the region

We are given the following inequalities:

1. $x - 2y + 4 \geq 0 \Rightarrow x \geq 2y - 4$
2. $x + 2y^2 \geq 0 \Rightarrow x \geq -2y^2$
3. $x + 4y^2 \leq 8 \Rightarrow x \leq 8 - 4y^2$
4. $y \geq 0$

These inequalities define a region in the $xy$-plane. We will find the area of this region.

Step 2: Find the intersection points of the curves

We need to find where the boundary curves intersect to determine the limits of integration. We will integrate with respect to $y$ since the inequalities are given as $x$ in terms of $y$.

• Intersection of $x = 2y - 4$ and $x = -2y^2$: $2y - 4 = -2y^2 \Rightarrow 2y^2 + 2y - 4 = 0 \Rightarrow y^2 + y - 2 = 0 \Rightarrow (y+2)(y-1) = 0$. Thus, $y = -2$ or $y = 1$. Since $y \geq 0$, we consider $y = 1$. At $y = 1$, $x = 2(1) - 4 = -2$.
• Intersection of $x = 2y - 4$ and $x = 8 - 4y^2$: $2y - 4 = 8 - 4y^2 \Rightarrow 4y^2 + 2y - 12 = 0 \Rightarrow 2y^2 + y - 6 = 0 \Rightarrow (2y - 3)(y + 2) = 0$. Thus, $y = \frac{3}{2}$ or $y = -2$. Since $y \geq 0$, we consider $y = \frac{3}{2}$. At $y = \frac{3}{2}$, $x = 2(\frac{3}{2}) - 4 = 3 - 4 = -1$.
• Intersection of $x = -2y^2$ and $x = 8 - 4y^2$: $-2y^2 = 8 - 4y^2 \Rightarrow 2y^2 = 8 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$. Since $y \geq 0$, we consider $y = 2$. At $y = 2$, $x = -2(2^2) = -8$.

Step 3: Determine the correct order of integration

From the intersection points and the inequalities, we can see that the region is bounded by $x = 8 - 4y^2$ on the right and by $x = -2y^2$ and $x = 2y - 4$ on the left. We found that $2y-4 = -2y^2$ when $y = 1$. So we need to split the integral into two parts: from $y=0$ to $y=1$, the left boundary is $x=-2y^2$; and from $y=1$ to $y=3/2$, the left boundary is $x=2y-4$. The right boundary throughout is $x=8-4y^2$. The upper bound of the integral is $y=2$, where $x=-2y^2$ and $x=8-4y^2$ intersect. However, the region is bounded by $y = 3/2$, where $x = 2y - 4$ and $x = 8 - 4y^2$ intersect. Therefore, we need to integrate from $y=0$ to $y=1$ and from $y=1$ to $y=3/2$.

Step 4: Set up and evaluate the integrals

The area $A$ of the region is given by: $A = \int_{0}^{1} [(8 - 4y^2) - (-2y^2)] \, dy + \int_{1}^{3/2} [(8 - 4y^2) - (2y - 4)] \, dy$ $A = \int_{0}^{1} (8 - 2y^2) \, dy + \int_{1}^{3/2} (12 - 4y^2 - 2y) \, dy$ Now, we evaluate each integral separately:

$\int_{0}^{1} (8 - 2y^2) \, dy = \left[8y - \frac{2}{3}y^3\right]_{0}^{1} = 8(1) - \frac{2}{3}(1)^3 - (0) = 8 - \frac{2}{3} = \frac{24 - 2}{3} = \frac{22}{3}$

$\int_{1}^{3/2} (12 - 4y^2 - 2y) \, dy = \left[12y - \frac{4}{3}y^3 - y^2\right]_{1}^{3/2} = \left[12(\frac{3}{2}) - \frac{4}{3}(\frac{3}{2})^3 - (\frac{3}{2})^2\right] - \left[12(1) - \frac{4}{3}(1)^3 - (1)^2\right]$ $= \left[18 - \frac{4}{3}(\frac{27}{8}) - \frac{9}{4}\right] - \left[12 - \frac{4}{3} - 1\right] = \left[18 - \frac{9}{2} - \frac{9}{4}\right] - \left[11 - \frac{4}{3}\right] = 18 - \frac{18}{4} - \frac{9}{4} - 11 + \frac{4}{3} = 7 - \frac{27}{4} + \frac{4}{3} = \frac{84 - 81 + 16}{12} = \frac{19}{12}$

So, the total area is: $A = \frac{22}{3} + \frac{19}{12} = \frac{88 + 19}{12} = \frac{107}{12}$

Step 5: Find m + n

The area of the region is $\frac{107}{12}$. Since $m = 107$ and $n = 12$ are coprime, we have $m+n = 107 + 12 = 119$.

Common Mistakes & Tips

• Careful with Intersection Points: Always verify that the intersection points you find are within the region of interest, especially considering inequalities like $y \geq 0$.
• Choosing the Correct Order of Integration: If the curves are more easily expressed as $x = f(y)$, integrate with respect to $y$. Otherwise, integrate with respect to $x$.
• Splitting the Integral: If the bounding curves change within the region, you need to split the integral into multiple parts.

Summary

We found the area of the region by integrating with respect to $y$. We first identified the bounding curves and found their intersection points. Since the left boundary of the region was defined by two different curves, we split the integral into two parts. We then evaluated the integrals and added the results to find the total area of the region, which is $\frac{107}{12}$. Finally, we calculated $m+n = 107+12 = 119$.

Final Answer

The final answer is $\boxed{119}$.