Key Concepts and Formulas

• Area between two 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 is given by $\int_{c}^{d} [f(y) - g(y)] \, dy$.
• Finding intersection points: To find where two curves intersect, set their equations equal to each other and solve for the variable.
• Symmetry: If the region is symmetric about the x-axis, the total area is twice the area above (or below) the x-axis.

Step-by-Step Solution

Step 1: Express the parabolas in terms of x as a function of y

We are given the equations $y^2 = 2x - 1$ and $y^2 = 4x - 3$. To integrate with respect to $y$, we need to solve for $x$ in terms of $y$ for both equations.

• For the first equation, $y^2 = 2x - 1$, we have $2x = y^2 + 1$, so $x_1 = \frac{y^2 + 1}{2}$.
• For the second equation, $y^2 = 4x - 3$, we have $4x = y^2 + 3$, so $x_2 = \frac{y^2 + 3}{4}$.

This step is crucial because it sets up the problem for integration with respect to $y$, which is the most convenient approach given the form of the equations.

Step 2: Find the points of intersection

To find the limits of integration, we need to find the $y$-values where the two parabolas intersect. We set $x_1 = x_2$:

$\frac{y^2 + 1}{2} = \frac{y^2 + 3}{4}$

Multiplying both sides by 4, we get:

$2(y^2 + 1) = y^2 + 3$

Expanding and simplifying:

$2y^2 + 2 = y^2 + 3$

$y^2 = 1$

$y = \pm 1$

The y-coordinates of the intersection points are $y = -1$ and $y = 1$. These will be our limits of integration.

Step 3: Determine which curve is to the right

We need to determine which parabola has larger $x$-values for a given $y$ in the interval $[-1, 1]$. Choose a test value, say $y = 0$.

• $x_1(0) = \frac{0^2 + 1}{2} = \frac{1}{2}$
• $x_2(0) = \frac{0^2 + 3}{4} = \frac{3}{4}$

Since $\frac{3}{4} > \frac{1}{2}$, the parabola $x_2 = \frac{y^2 + 3}{4}$ is to the right of $x_1 = \frac{y^2 + 1}{2}$ in the region of interest.

Step 4: Set up the integral

The area between the curves is given by the integral:

$\text{Area} = \int_{-1}^{1} (x_2 - x_1) \, dy = \int_{-1}^{1} \left( \frac{y^2 + 3}{4} - \frac{y^2 + 1}{2} \right) \, dy$

Step 5: Simplify the integrand

Simplify the expression inside the integral:

$\frac{y^2 + 3}{4} - \frac{y^2 + 1}{2} = \frac{y^2 + 3 - 2(y^2 + 1)}{4} = \frac{y^2 + 3 - 2y^2 - 2}{4} = \frac{-y^2 + 1}{4}$

So the integral becomes:

$\text{Area} = \int_{-1}^{1} \frac{1 - y^2}{4} \, dy$

Step 6: Evaluate the integral

Evaluate the definite integral:

$\text{Area} = \frac{1}{4} \int_{-1}^{1} (1 - y^2) \, dy = \frac{1}{4} \left[ y - \frac{y^3}{3} \right]_{-1}^{1}$

$\text{Area} = \frac{1}{4} \left[ \left( 1 - \frac{1}{3} \right) - \left( -1 - \frac{(-1)^3}{3} \right) \right] = \frac{1}{4} \left[ \frac{2}{3} - \left( -1 + \frac{1}{3} \right) \right] = \frac{1}{4} \left[ \frac{2}{3} - \left( -\frac{2}{3} \right) \right]$

$\text{Area} = \frac{1}{4} \left[ \frac{4}{3} \right] = \frac{1}{3}$

The area of the region enclosed between the two parabolas is $\frac{1}{3}$.

Common Mistakes & Tips

• Incorrectly identifying the "right" curve: Always test a point within the interval to determine which curve is greater (more to the right) for a given y.
• Forgetting the order of subtraction: Ensure you subtract the left curve from the right curve when integrating with respect to y.
• Algebra errors: Pay close attention to signs and distribution, especially when simplifying the integrand.

Summary

We found the area enclosed by the parabolas $y^2 = 2x - 1$ and $y^2 = 4x - 3$ by expressing the parabolas as functions of y, finding their intersection points, determining which parabola was to the right, setting up the definite integral, simplifying the integrand, and evaluating the integral. The area is $\frac{1}{3}$.

Final Answer

The final answer is \boxed{1/3}, which corresponds to option (A).