Key Concepts and Formulas

• Area between curves: If $x = f(y)$ and $x = g(y)$ are continuous functions, and $f(y) \ge g(y)$ on the interval $[c, d]$, the area of the region bounded by the curves is given by $\int_c^d [f(y) - g(y)] \, dy$.
• Finding intersection points: To find the points where two curves intersect, set their equations equal to each other and solve for the variable.
• Even function integration: If $f(y)$ is an even function, then $\int_{-a}^a f(y) \, dy = 2 \int_0^a f(y) \, dy$.

Step-by-Step Solution

Step 1: Express the equations in terms of x = f(y)

We are given the equations $y^2 = 8x$ and $y^2 = 16(3 - x)$. To integrate with respect to $y$, we need to express $x$ as a function of $y$ for both equations. $x_1 = \frac{y^2}{8}$ $x_2 = 3 - \frac{y^2}{16}$ Why: This allows us to easily integrate with respect to y, as we will have expressions for x in terms of 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: $\frac{y^2}{8} = 3 - \frac{y^2}{16}$ Multiplying both sides by 16, we get: $2y^2 = 48 - y^2$ $3y^2 = 48$ $y^2 = 16$ $y = \pm 4$ Thus, the y-coordinates of the intersection points are $y = -4$ and $y = 4$.

Why: Finding the intersection points gives us the limits of integration for y.

Step 3: Determine which curve is to the right

We need to determine which function, $x_1(y)$ or $x_2(y)$, has larger values (i.e., is to the right) on the interval $[-4, 4]$. We can do this by subtracting the two expressions: $x_2 - x_1 = \left(3 - \frac{y^2}{16}\right) - \frac{y^2}{8} = 3 - \frac{y^2}{16} - \frac{2y^2}{16} = 3 - \frac{3y^2}{16}$ Since $y^2$ will always be positive on our interval, $3 - \frac{3y^2}{16}$ will always be less than or equal to $3$. More importantly, we need to know if it is positive. Consider the expression $3 - \frac{3y^2}{16}$. We know that $|y| \leq 4$. Thus $y^2 \le 16$. Therefore, $\frac{3y^2}{16} \le \frac{3(16)}{16}=3$. Therefore, $3 - \frac{3y^2}{16} \ge 0$. This means $x_2 - x_1 \ge 0$, so $x_2 \ge x_1$.

Why: This step is essential to ensure we set up the integral correctly.

Step 4: Set up the integral and evaluate

The area of the region is given by the integral: $A = \int_{-4}^{4} (x_2 - x_1) \, dy = \int_{-4}^{4} \left(3 - \frac{3y^2}{16}\right) \, dy$ Since the integrand is an even function, we can rewrite the integral as: $A = 2 \int_{0}^{4} \left(3 - \frac{3y^2}{16}\right) \, dy$ Now, we evaluate the integral: $A = 2 \left[3y - \frac{y^3}{16}\right]_0^4 = 2 \left[3(4) - \frac{4^3}{16} - (0)\right] = 2 \left[12 - \frac{64}{16}\right] = 2[12 - 4] = 2[8] = 16$

Why: This step applies the area between curves formula and the even function property to simplify the integration process.

Step 5: Re-evaluate the question

Let's reconsider the setup of the integral. $A = \int_{-4}^4 (3 - \frac{y^2}{16} - \frac{y^2}{8}) dy = \int_{-4}^4 (3 - \frac{3y^2}{16}) dy$ $A = [3y - \frac{y^3}{16}]_{-4}^4 = (12 - \frac{64}{16}) - (-12 + \frac{64}{16}) = 12 - 4 - (-12 + 4) = 8 - (-8) = 16$

Step 6: Check for errors

Let's check the integration limits and the functions again. The intersection points are at $y = \pm 4$. The functions are $x_1 = \frac{y^2}{8}$ and $x_2 = 3 - \frac{y^2}{16}$. $x_2$ is to the right of $x_1$. $A = \int_{-4}^4 (3 - \frac{y^2}{16} - \frac{y^2}{8}) dy = \int_{-4}^4 (3 - \frac{3y^2}{16}) dy = [3y - \frac{y^3}{16}]_{-4}^4 = (12 - 4) - (-12 + 4) = 8 - (-8) = 16$ The calculation seems correct.

Step 7: There seems to be an error in the question or provided answer.

If we assume the correct answer is $\frac{32}{3}$ and work backward, we would get: $\frac{32}{3} = 2 \int_0^4 (f(y)) dy$ $\frac{16}{3} = \int_0^4 (f(y)) dy$ $\frac{16}{3} = [3y - \frac{y^3}{16}]_0^4 = 12 - 4 = 8$ This is not equal to $\frac{16}{3}$.

There must be a mistake in either the question or the answer key. After careful examination of the steps, the area comes out to be 16. Let's assume there is a typo in the question. Let's assume the correct answer is 16.

Common Mistakes & Tips

• Always ensure you're subtracting the "leftmost" function from the "rightmost" function when integrating with respect to $y$.
• When integrating over symmetric intervals, check if the function is even or odd to simplify the integral.
• Double-check your algebra when finding intersection points and setting up the integral.

Summary

We found the area between the curves $y^2 = 8x$ and $y^2 = 16(3 - x)$ by expressing $x$ as a function of $y$, finding the intersection points, and integrating the difference between the rightmost and leftmost functions with respect to $y$. The calculated area is 16, but the provided answer is incorrect. We assume the question has a typo.

Final Answer

The final answer is \boxed{16}. The correct option is (C).