Key Concepts and Formulas

• Area between curves (integration with respect to y): If a region is bounded by $x = f(y)$ (right curve) and $x = g(y)$ (left curve) between $y = c$ and $y = d$, the area is given by $A = \int_c^d [f(y) - g(y)] \, dy$.
• Power rule of integration: $\int y^n \, dy = \frac{y^{n+1}}{n+1} + C$, where $n \neq -1$.
• Symmetry: If the region is symmetric about the y-axis, the total area is twice the area in the first quadrant.

Step-by-Step Solution

Step 1: Analyze the given equations and visualize the region.

We are given the parabolas $x^2 = \frac{y}{4}$ and $x^2 = 9y$, and the line $y=2$. We want to find the area enclosed between these curves. Visualizing these curves helps to determine the limits of integration and which curve is to the right.

Step 2: Express the parabolas as functions of y.

Since the equations are given in terms of $x^2$, and one boundary is $y=2$, it's convenient to express $x$ as a function of $y$.

• From $x^2 = \frac{y}{4}$, we get $x = \pm \sqrt{\frac{y}{4}} = \pm \frac{\sqrt{y}}{2}$.
• From $x^2 = 9y$, we get $x = \pm \sqrt{9y} = \pm 3\sqrt{y}$.

Step 3: Determine the right and left boundaries in the first quadrant.

In the first quadrant, $x \ge 0$. Therefore, we consider the positive square roots:

• $x_1(y) = \frac{\sqrt{y}}{2}$
• $x_2(y) = 3\sqrt{y}$

For $y > 0$, we have $3\sqrt{y} > \frac{\sqrt{y}}{2}$. So, the right boundary is $x = 3\sqrt{y}$, and the left boundary is $x = \frac{\sqrt{y}}{2}$.

Step 4: Determine the limits of integration.

The region is bounded by the line $y=2$ above. The parabolas intersect at the origin $(0,0)$, so the lower limit of integration is $y=0$.

Step 5: Set up the integral for the area in the first quadrant.

The area in the first quadrant, $A_1$, is given by: $A_1 = \int_0^2 \left(3\sqrt{y} - \frac{\sqrt{y}}{2}\right) dy$

Step 6: Simplify the integrand and evaluate the integral.

Simplify the integrand: $3\sqrt{y} - \frac{\sqrt{y}}{2} = \frac{5}{2}\sqrt{y} = \frac{5}{2}y^{1/2}$

Now, evaluate the integral: $A_1 = \int_0^2 \frac{5}{2}y^{1/2} dy = \frac{5}{2} \left[\frac{y^{3/2}}{3/2}\right]_0^2 = \frac{5}{2} \cdot \frac{2}{3} \left[y^{3/2}\right]_0^2 = \frac{5}{3} \left[2^{3/2} - 0^{3/2}\right] = \frac{5}{3} (2\sqrt{2}) = \frac{10\sqrt{2}}{3}$

Step 7: Calculate the total area.

Since the region is symmetric about the $y$-axis, the total area $A$ is twice the area in the first quadrant: $A = 2 \times A_1 = 2 \times \frac{10\sqrt{2}}{3} = \frac{20\sqrt{2}}{3}$

Common Mistakes & Tips

• Incorrect order of subtraction: Ensure you subtract the left boundary from the right boundary (when integrating with respect to y) to get a positive area.
• Choosing the wrong integration variable: Integrating with respect to x would be more complicated in this case, as you would need to split the integral into multiple parts.
• Forgetting symmetry: Using symmetry simplifies the problem by allowing you to integrate over only half the region.

Summary

We calculated the area between the given parabolas and the line $y=2$ by integrating with respect to $y$. We exploited the symmetry of the region to simplify the calculation. The final area is $\frac{20\sqrt{2}}{3}$.

Final Answer

The final answer is \boxed{\frac{20\sqrt{2}}{3}}, which corresponds to option (C).