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 $x = f(y)$, $x = g(y)$, and the lines $y = c$ and $y = d$ is given by: $A = \int_c^d [f(y) - g(y)] \, dy$
• Finding Intersection Points: To find where curves intersect, set their equations equal to each other and solve for the variable(s).

Step-by-Step Solution

Step 1: Express the curves in terms of y.

We are given $y \leq 4x^2$, $x^2 \leq 9y$, and $y \leq 4$. We rewrite the inequalities as equations and express $x$ as a function of $y$.

• From $y = 4x^2$, we get $x^2 = \frac{y}{4}$, so $x = \pm \frac{\sqrt{y}}{2}$. Since we are looking for the area enclosed, we will focus on the positive root for the right boundary. Thus, $x = \frac{\sqrt{y}}{2}$.
• From $x^2 = 9y$, we get $x = \pm 3\sqrt{y}$. Again, we'll focus on the positive root for the right boundary. Thus, $x = 3\sqrt{y}$.
• The line $y = 4$ is already in the desired form.

Step 2: Find the intersection points.

We need to find the points where the curves intersect to determine the limits of integration.

• Intersection of $x^2 = 9y$ and $y = 4$: Substituting $y=4$ into $x^2 = 9y$, we have $x^2 = 9(4) = 36$, so $x = \pm 6$. Since $x = 3\sqrt{y}$, we have $x = 3\sqrt{4} = 3(2) = 6$.
• Intersection of $y = 4x^2$ and $y = 4$: Substituting $y=4$ into $y = 4x^2$, we have $4 = 4x^2$, so $x^2 = 1$, and $x = \pm 1$. Since $x = \frac{\sqrt{y}}{2}$, we have $x = \frac{\sqrt{4}}{2} = \frac{2}{2} = 1$.
• Intersection of $y = 4x^2$ and $x^2 = 9y$: Substituting $x^2 = 9y$ into $y = 4x^2$, we get $y = 4(9y) = 36y$. Thus, $35y = 0$, so $y = 0$. Then $x^2 = 9(0) = 0$, so $x = 0$.

The relevant intersection points in terms of y are at $y=0$ and $y=4$. Now we need to find the value of $y$ where $\frac{\sqrt{y}}{2} = 3\sqrt{y}$. $\frac{\sqrt{y}}{2} = 3\sqrt{y} \implies \sqrt{y} = 6\sqrt{y} \implies 5\sqrt{y} = 0 \implies y=0$.

Step 3: Determine which curve is on the right and which is on the left.

We are integrating with respect to $y$, so we need to determine which $x$ value is greater for a given $y$ in the region. From the equations $x = \frac{\sqrt{y}}{2}$ and $x = 3\sqrt{y}$, we see that $3\sqrt{y} > \frac{\sqrt{y}}{2}$ for $y > 0$. Thus, $x = 3\sqrt{y}$ is to the right of $x = \frac{\sqrt{y}}{2}$.

Step 4: Set up and evaluate the integral.

The area is given by the integral: $A = \int_0^4 \left(3\sqrt{y} - \frac{\sqrt{y}}{2}\right) \, dy = \int_0^4 \left(\frac{6\sqrt{y} - \sqrt{y}}{2}\right) \, dy = \int_0^4 \frac{5}{2}\sqrt{y} \, dy$ $A = \frac{5}{2} \int_0^4 y^{1/2} \, dy = \frac{5}{2} \left[\frac{2}{3} y^{3/2}\right]_0^4 = \frac{5}{2} \cdot \frac{2}{3} \left[4^{3/2} - 0^{3/2}\right] = \frac{5}{3} \left[(\sqrt{4})^3\right] = \frac{5}{3} (2^3) = \frac{5}{3}(8) = \frac{40}{3}$

Common Mistakes & Tips

• Sign Errors: Be careful with signs when subtracting the functions. Ensure you are subtracting the left boundary from the right boundary.
• Choosing the Correct Limits of Integration: The limits of integration must correspond to the range of $y$-values that define the region.
• Simplifying the Integrand: Simplify the integrand before integrating to avoid complicated calculations.

Summary

We found the area of the region enclosed by $y \leq 4x^2$, $x^2 \leq 9y$, and $y \leq 4$ by integrating with respect to $y$. First, we expressed the curves as functions of $y$. Then, we found the intersection points of the curves to determine the limits of integration. Finally, we set up and evaluated the integral to find the area, which is $\frac{40}{3}$.

Final Answer

The final answer is $\boxed{\frac{40}{3}}$, which corresponds to option (A).