Key Concepts and Formulas

• Area between curves: $A = \int_a^b |f(x) - g(x)| dx$ where $f(x)$ and $g(x)$ are the functions bounding the region, and $a$ and $b$ are the intersection points.
• Equation of a parabola: $y = a(x-h)^2 + k$ (vertical axis) and $x = a(y-k)^2 + h$ (horizontal axis), where $(h,k)$ is the vertex.
• Solving polynomial equations to find intersection points.

Step-by-Step Solution

Step 1: Analyze the Equations and Identify the Curves

We are given the equations $y = x^2 - 4x + 4$ and $y^2 = 16 - 8x$. The first equation can be rewritten as $y = (x-2)^2$. This is a parabola opening upwards with vertex at $(2,0)$. The second equation can be rewritten as $8x = 16 - y^2$, or $x = 2 - \frac{y^2}{8}$. This is a parabola opening to the left with vertex at $(2,0)$.

Why: Identifying the type of curves and their orientation helps visualize the region and choose the appropriate integration strategy.

Step 2: Find the Intersection Points

To find the intersection points, we substitute $y = (x-2)^2$ into $y^2 = 16 - 8x$: $((x-2)^2)^2 = 16 - 8x$ $(x-2)^4 = 16 - 8x$ Let $u = x-2$, then $x = u+2$. Substituting, we get $u^4 = 16 - 8(u+2) = 16 - 8u - 16$ $u^4 = -8u$ $u^4 + 8u = 0$ $u(u^3 + 8) = 0$ So $u = 0$ or $u^3 = -8$, which means $u = -2$.

If $u = 0$, then $x = u+2 = 0+2 = 2$. Then $y = (x-2)^2 = (2-2)^2 = 0$. So one intersection point is $(2,0)$. If $u = -2$, then $x = u+2 = -2+2 = 0$. Then $y = (x-2)^2 = (0-2)^2 = 4$. So the other intersection point is $(0,4)$.

Why: The intersection points define the limits of integration.

Step 3: Choose the Integration Strategy and Set up the Integral

We can integrate with respect to $y$. The region is bounded by $x = 2 - \frac{y^2}{8}$ on the right and $y = (x-2)^2$ on the left. We need to express the left curve as $x$ in terms of $y$. $y = (x-2)^2$, so $x-2 = \pm \sqrt{y}$, and $x = 2 \pm \sqrt{y}$. Since we are in the region where $x \le 2$, we choose $x = 2 - \sqrt{y}$.

The limits of integration are $y=0$ to $y=4$. The area is $A = \int_0^4 \left[ \left(2 - \frac{y^2}{8}\right) - (2 - \sqrt{y}) \right] dy = \int_0^4 \left( -\frac{y^2}{8} + \sqrt{y} \right) dy$

Why: Integrating with respect to $y$ simplifies the calculation as it avoids dealing with square roots in the integrand.

Step 4: Evaluate the Integral

$A = \int_0^4 \left( -\frac{y^2}{8} + y^{1/2} \right) dy = \left[ -\frac{y^3}{24} + \frac{2}{3} y^{3/2} \right]_0^4$ $A = \left( -\frac{4^3}{24} + \frac{2}{3} (4)^{3/2} \right) - \left( -\frac{0^3}{24} + \frac{2}{3} (0)^{3/2} \right) = -\frac{64}{24} + \frac{2}{3} (8) = -\frac{8}{3} + \frac{16}{3} = \frac{8}{3}$

Why: Evaluating the definite integral gives the numerical value of the area.

Step 5: Check for Mistakes

We made an error. The problem statement says the answer is $8$. Let's go back to Step 3 and check our work. The right boundary is $x = 2 - \frac{y^2}{8}$. The left boundary is $x = 2 - \sqrt{y}$. The area is given by $A = \int_0^4 (x_{right} - x_{left}) dy = \int_0^4 (2 - \frac{y^2}{8} - (2 - \sqrt{y})) dy = \int_0^4 (-\frac{y^2}{8} + \sqrt{y}) dy = [-\frac{y^3}{24} + \frac{2}{3}y^{\frac{3}{2}}]_0^4 = -\frac{64}{24} + \frac{2}{3} \cdot 8 = -\frac{8}{3} + \frac{16}{3} = \frac{8}{3}$.

The error is that the problem statement answer is wrong. The correct answer is $\frac{8}{3}$.

Common Mistakes & Tips

• Sketching the curves is crucial for visualizing the region and choosing the correct integration strategy.
• Be careful when solving for $x$ or $y$ in terms of the other variable, especially when dealing with square roots.
• Double-check your integration and arithmetic to avoid errors.

Summary

We found the area of the region enclosed by the two parabolas by first identifying the curves and their intersection points. We then chose to integrate with respect to $y$ to simplify the calculation. We expressed both curves as functions of $y$, set up the definite integral, and evaluated it to find the area.

The final answer is $\boxed{\frac{8}{3}}$, which corresponds to option (A).