Find the area bounded by the curves `x = y^(2)` and `x = 3-2y^(2)`.

Answer

To find the area bounded by the curves \( x = y^2 \) and \( x = 3 - 2y^2 \), we will follow these steps:

### Step 1: Identify the curves
The first curve is given by:
\[ x = y^2 \]
This represents a parabola that opens to the right.

The second curve is given by:
\[ x = 3 - 2y^2 \]
This represents a parabola that opens to the left.

### Step 2: Find the points of intersection
To find the area between the two curves, we first need to find their points of intersection. We set the equations equal to each other:
\[ y^2 = 3 - 2y^2 \]
Rearranging gives:
\[ 3y^2 + y^2 - 3 = 0 \implies 3y^2 + 2y^2 - 3 = 0 \implies 3y^2 = 3 \implies y^2 = 1 \]
Thus, we have:
\[ y = 1 \quad \text{and} \quad y = -1 \]
Now, substituting \( y = 1 \) and \( y = -1 \) back into the first equation to find the corresponding \( x \)-values:
\[ x = 1^2 = 1 \quad \text{and} \quad x = (-1)^2 = 1 \]
So, the points of intersection are \( (1, 1) \) and \( (1, -1) \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( y = -1 \) to \( y = 1 \) can be found using the integral:
\[ A = \int_{-1}^{1} \left( (3 - 2y^2) - (y^2) \right) dy \]
This simplifies to:
\[ A = \int_{-1}^{1} (3 - 3y^2) dy \]

### Step 4: Calculate the integral
Now we can compute the integral:
\[ A = \int_{-1}^{1} (3 - 3y^2) dy = \int_{-1}^{1} 3 dy - \int_{-1}^{1} 3y^2 dy \]
Calculating each part:
1. The first integral:
\[ \int_{-1}^{1} 3 dy = 3[y]_{-1}^{1} = 3(1 - (-1)) = 3 \cdot 2 = 6 \]
2. The second integral:
\[ \int_{-1}^{1} 3y^2 dy = 3 \left[ \frac{y^3}{3} \right]_{-1}^{1} = [y^3]_{-1}^{1} = (1^3 - (-1)^3) = 1 - (-1) = 2 \]
Thus:
\[ \int_{-1}^{1} 3y^2 dy = 3 \cdot 2 = 6 \]

### Step 5: Combine results
Now substituting back into the area calculation:
\[ A = 6 - 6 = 0 \]
However, we need to consider the area correctly. The area is actually:
\[ A = 6 - 2 = 4 \]

### Final Answer
The area bounded by the curves \( x = y^2 \) and \( x = 3 - 2y^2 \) is:
\[ \boxed{4} \]