Key Concepts and Formulas

• Area Between Curves (Integration with respect to y): If $x = f(y)$ and $x = g(y)$ are continuous functions and $f(y) \ge g(y)$ on the interval $[c, d]$, then the area of the region bounded by the curves is given by $A = \int_{c}^{d} [f(y) - g(y)] \, dy$.
• Finding Intersection Points: To find the points of intersection of two curves, set their equations equal to each other and solve for the variable.
• Quadratic Formula/Factoring: To solve a quadratic equation of the form $ax^2 + bx + c = 0$, either factor it or use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Express the equations in terms of y

We are given the parabola $y^2 = 4(x-2)$ and the line $y = 2x - 8$. We need to express both equations in the form $x = f(y)$.

• Parabola: $y^2 = 4(x-2)$ $\frac{y^2}{4} = x - 2$ $x = \frac{y^2}{4} + 2$ Let $x_P(y) = \frac{y^2}{4} + 2$.

• Line: $y = 2x - 8$ $y + 8 = 2x$ $x = \frac{y + 8}{2}$ Let $x_L(y) = \frac{y + 8}{2}$.

Step 2: Find the points of intersection

To find the points of intersection, we set the $x$-values equal to each other: $x_P(y) = x_L(y)$ $\frac{y^2}{4} + 2 = \frac{y + 8}{2}$

Multiply both sides by 4 to eliminate fractions: $y^2 + 8 = 2(y + 8)$ $y^2 + 8 = 2y + 16$ $y^2 - 2y - 8 = 0$

Factor the quadratic equation: $(y - 4)(y + 2) = 0$

So, $y = 4$ or $y = -2$. These are the y-coordinates of the intersection points, and will be our limits of integration.

Step 3: Determine which curve is to the right

We need to determine which curve has a larger $x$-value for a given $y$ in the interval $[-2, 4]$. Let's test $y = 0$:

• $x_P(0) = \frac{0^2}{4} + 2 = 2$
• $x_L(0) = \frac{0 + 8}{2} = 4$

Since $x_L(0) > x_P(0)$, the line is to the right of the parabola in the interval $[-2, 4]$. Therefore, $x_{right} = x_L(y)$ and $x_{left} = x_P(y)$.

Step 4: Set up and evaluate the integral

The area is given by the integral: $A = \int_{-2}^{4} [x_L(y) - x_P(y)] \, dy$ $A = \int_{-2}^{4} \left[ \frac{y + 8}{2} - \left( \frac{y^2}{4} + 2 \right) \right] \, dy$ $A = \int_{-2}^{4} \left[ \frac{y}{2} + 4 - \frac{y^2}{4} - 2 \right] \, dy$ $A = \int_{-2}^{4} \left[ -\frac{y^2}{4} + \frac{y}{2} + 2 \right] \, dy$

Now, integrate: $A = \left[ -\frac{y^3}{12} + \frac{y^2}{4} + 2y \right]_{-2}^{4}$

Evaluate at the limits: $A = \left( -\frac{4^3}{12} + \frac{4^2}{4} + 2(4) \right) - \left( -\frac{(-2)^3}{12} + \frac{(-2)^2}{4} + 2(-2) \right)$ $A = \left( -\frac{64}{12} + \frac{16}{4} + 8 \right) - \left( -\frac{-8}{12} + \frac{4}{4} - 4 \right)$ $A = \left( -\frac{16}{3} + 4 + 8 \right) - \left( \frac{2}{3} + 1 - 4 \right)$ $A = \left( -\frac{16}{3} + 12 \right) - \left( \frac{2}{3} - 3 \right)$ $A = \left( \frac{-16 + 36}{3} \right) - \left( \frac{2 - 9}{3} \right)$ $A = \frac{20}{3} - \left( -\frac{7}{3} \right)$ $A = \frac{20}{3} + \frac{7}{3} = \frac{27}{3} = 9$

Common Mistakes & Tips

• Incorrectly identifying the "right" and "left" curves: Always test a point within the interval to determine which curve has a larger x-value.
• Sign errors: Be extremely careful with signs, especially when evaluating the integral at the lower limit.
• Choosing the wrong variable of integration: Integrating with respect to y significantly simplifies this problem. Recognizing this is key.

Summary

We found the area of the region bounded by the given parabola and line by expressing both equations in terms of y, finding the intersection points to determine the limits of integration, determining which curve was to the right, and then evaluating the definite integral. This approach avoided the complexities of integrating with respect to x.

The final answer is \boxed{9}, which corresponds to option (C).