Key Concepts and Formulas

• Area Between Curves: If $x_1(y) \le x \le x_2(y)$ for $y \in [c, d]$, then the area of the region bounded by these curves is given by $\int_c^d (x_2(y) - x_1(y))\,dy$.
• Finding Intersection Points: To find where two curves intersect, set their equations equal to each other and solve for the variable.
• Fundamental Theorem of Calculus: $\int_a^b f(x) \, dx = F(b) - F(a)$, where $F(x)$ is an antiderivative of $f(x)$.

Step-by-Step Solution

Step 1: Identify the Bounding Curves

We are given the region $A = \{(x, y) : \frac{y^2}{2} \le x \le y + 4\}$. This tells us that the left boundary of the region is the curve $x = \frac{y^2}{2}$ and the right boundary is the curve $x = y + 4$. We will integrate with respect to $y$ since $x$ is given as a function of $y$.

Step 2: Find the Points of Intersection

To find the limits of integration, we need to find the points where the two curves intersect. This occurs when their $x$-values are equal, so we set up the equation: $\frac{y^2}{2} = y + 4$ Multiplying both sides by 2 gives: $y^2 = 2y + 8$ Rearranging into a quadratic equation: $y^2 - 2y - 8 = 0$ Factoring the quadratic gives: $(y - 4)(y + 2) = 0$ So, the solutions are $y = 4$ and $y = -2$. These are our limits of integration.

Step 3: Set Up the Integral

The area of the region is given by the integral of the difference between the right and left boundaries with respect to $y$, from the lower limit to the upper limit: $\text{Area} = \int_{-2}^{4} ( (y + 4) - \frac{y^2}{2} ) \, dy$

Step 4: Evaluate the Integral

First, find the antiderivative of the integrand: $\int (y + 4 - \frac{y^2}{2}) \, dy = \frac{y^2}{2} + 4y - \frac{y^3}{6} + C$ Now, evaluate the definite integral using the Fundamental Theorem of Calculus: $\text{Area} = \left[ \frac{y^2}{2} + 4y - \frac{y^3}{6} \right]_{-2}^{4}$ $\text{Area} = \left( \frac{4^2}{2} + 4(4) - \frac{4^3}{6} \right) - \left( \frac{(-2)^2}{2} + 4(-2) - \frac{(-2)^3}{6} \right)$ $\text{Area} = \left( \frac{16}{2} + 16 - \frac{64}{6} \right) - \left( \frac{4}{2} - 8 - \frac{-8}{6} \right)$ $\text{Area} = \left( 8 + 16 - \frac{32}{3} \right) - \left( 2 - 8 + \frac{4}{3} \right)$ $\text{Area} = \left( 24 - \frac{32}{3} \right) - \left( -6 + \frac{4}{3} \right)$ $\text{Area} = 24 - \frac{32}{3} + 6 - \frac{4}{3}$ $\text{Area} = 30 - \frac{36}{3}$ $\text{Area} = 30 - 12$ $\text{Area} = 18$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs when evaluating the antiderivative at the lower limit of integration.
• Order of Subtraction: Always subtract the left boundary from the right boundary. Reversing the order will result in a negative area.
• Sketching: A quick sketch of the region can help visualize the problem and ensure the correct setup of the integral.

Summary

We found the area of the region bounded by $x = \frac{y^2}{2}$ and $x = y + 4$ by integrating the difference between the right and left boundary curves with respect to $y$ from $y = -2$ to $y = 4$. The calculated area is 18 square units.

The final answer is \boxed{18}, which corresponds to option (B).