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 $x = f(y)$, $x = g(y)$, $y = c$, and $y = d$ is given by $A = \int_c^d [f(y) - g(y)] \, dy$.
• Even Function: A function $f(y)$ is even if $f(-y) = f(y)$. The integral of an even function over a symmetric interval $[-a, a]$ is given by $\int_{-a}^a f(y) \, dy = 2 \int_0^a f(y) \, dy$.

Step-by-Step Solution

Step 1: Express the equations in terms of x as functions of y

We are given the equations $x + 2y^2 = 0$ and $x + 3y^2 = 1$. We need to rewrite these equations in the form $x = f(y)$.

• Equation 1: $x + 2y^2 = 0 \Rightarrow x = -2y^2$
• Equation 2: $x + 3y^2 = 1 \Rightarrow x = 1 - 3y^2$

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.

Set the two expressions for $x$ equal to each other: $-2y^2 = 1 - 3y^2$ Solve for $y$: $3y^2 - 2y^2 = 1$ $y^2 = 1$ $y = \pm 1$ So, the curves intersect at $y = -1$ and $y = 1$.

Step 3: Determine which curve is to the right

We need to determine which curve has the larger $x$-value for a given $y$ in the interval $[-1, 1]$. Let's test the value $y = 0$:

• For $x = -2y^2$: $x = -2(0)^2 = 0$
• For $x = 1 - 3y^2$: $x = 1 - 3(0)^2 = 1$

Since $1 > 0$, the curve $x = 1 - 3y^2$ is to the right of the curve $x = -2y^2$ in the region of interest. Therefore, $f(y) = 1 - 3y^2$ and $g(y) = -2y^2$.

Step 4: Set up the integral for the area

The area between the curves is given by the integral: $A = \int_{-1}^{1} [f(y) - g(y)] \, dy = \int_{-1}^{1} [(1 - 3y^2) - (-2y^2)] \, dy$ Simplify the integrand: $A = \int_{-1}^{1} (1 - 3y^2 + 2y^2) \, dy = \int_{-1}^{1} (1 - y^2) \, dy$

Step 5: Use symmetry to simplify the integral

Since the integrand $(1 - y^2)$ is an even function and the interval of integration is symmetric about $y = 0$, we can rewrite the integral as: $A = 2 \int_{0}^{1} (1 - y^2) \, dy$

Step 6: Evaluate the integral

Now, we evaluate the integral: $A = 2 \int_{0}^{1} (1 - y^2) \, dy = 2 \left[ y - \frac{y^3}{3} \right]_{0}^{1}$ Apply the limits of integration: $A = 2 \left[ \left( 1 - \frac{1^3}{3} \right) - \left( 0 - \frac{0^3}{3} \right) \right] = 2 \left[ 1 - \frac{1}{3} \right] = 2 \left[ \frac{2}{3} \right] = \frac{4}{3}$

Common Mistakes & Tips

• Incorrectly identifying the rightmost and leftmost curves: Always test a point within the interval to determine which curve is to the right.
• Sign errors: Be careful with signs when subtracting the two functions in the integral.
• Forgetting to multiply by 2 when using symmetry: If you use the symmetry property, remember to multiply the result by 2.

Summary

We found the area of the region bounded by the curves $x + 2y^2 = 0$ and $x + 3y^2 = 1$ by first expressing the equations in the form $x = f(y)$, then finding the points of intersection to determine the limits of integration. We then set up the integral $\int_{-1}^{1} [(1 - 3y^2) - (-2y^2)] \, dy$, simplified it using symmetry, and evaluated the integral to find the area.

Final Answer

The final answer is \boxed{\frac{4}{3}}, which corresponds to option (D).