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can give me the solution for a long time. You don't need to explain and clarify. Just write the solution to me in steps


To solve the problem of finding the area enclosed by the curves \( x - y^2 = 0 \) and \( x + 2y^2 = 3 \), here are the steps:

### Step 1: Express the curves
1. The first equation \( x - y^2 = 0 \) can be rewritten as \( x = y^2 \).
2. The second equation \( x + 2y^2 = 3 \) can be rewritten as \( x = 3 - 2y^2 \).

### Step 2: Find the points of intersection
To find the points where the curves intersect, set the two equations equal to each other:
\[
y^2 = 3 - 2y^2
\]
This simplifies to:
\[
3y^2 = 3
\]
\[
y^2 = 1 \quad \Rightarrow \quad y = \pm 1
\]

### Step 3: Determine the x-values of the intersection points
For \( y = 1 \):
\[
x = y^2 = 1
\]
For \( y = -1 \):
\[
x = y^2 = 1
\]
So, the points of intersection are \( (1, 1) \) and \( (1, -1) \).

### Step 4: Set up the integral for the area
The area is the region between the curves, so we will integrate the difference between the functions:
\[
\text{Area} = \int_{-1}^{1} \left[ (3 - 2y^2) - y^2 \right] \, dy
\]
Simplify the integrand:
\[
\text{Area} = \int_{-1}^{1} \left( 3 - 3y^2 \right) \, dy
\]

### Step 5: Compute the integral
\[
\int \left( 3 - 3y^2 \right) \, dy = 3y - y^3
\]
Now, evaluate the integral from \( y = -1 \) to \( y = 1 \):
\[
\text{Area} = \left[ 3(1) - (1)^3 \right] - \left[ 3(-1) - (-1)^3 \right]
\]
\[
\text{Area} = \left( 3 - 1 \right) - \left( -3 + 1 \right)
\]
\[
\text{Area} = 2 - (-2) = 4
\]

### Final Answer:
The area of the region enclosed by the curves is **4 square units**.

Let me know if you want more details or have further questions!

Here are 5 relative questions to expand on the problem:
1. How do we handle integrals for regions bounded by more complex curves?
2. What if one of the curves was a higher-degree polynomial, how would the area calculation change?
3. Could we use polar coordinates to calculate areas for different types of enclosed regions?
4. How would the solution change if the curves did not intersect symmetrically?
5. How can we verify the correctness of our area computation using numerical methods?

**Tip:** Always sketch the curves to visualize the enclosed area before setting up your integrals.

Find the areas of the regions enclosed by the lines and curves x - y^2 = 0 and x + 2y^2 = 3.

Learn how to find the area of the region enclosed by the curves x - y^2 = 0 and x + 2y^2 = 3. This detailed solution walks through using integral calculus and finding points of intersection. Suitable for students in Grades 11-12.

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