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 \( x = y^2 \), which is a parabola that opens to the right with its vertex at the origin (0, 0).

The second curve is \( x = 3 - 2y^2 \), which is also a parabola that opens to the left with its vertex at (3, 0).

### Step 2: Find the points of intersection
To find the area between the two curves, we first need to determine where they intersect. We set the equations equal to each other:

\[ y^2 = 3 - 2y^2 \]

Rearranging gives:

\[ 3y^2 = 3 \]

\[ y^2 = 1 \]

Taking the square root of both sides, we find:

\[ y = 1 \quad \text{and} \quad y = -1 \]

### Step 3: Determine the corresponding x-values
Now, we can find the x-values at the points of intersection by substituting \( y = 1 \) and \( y = -1 \) back into either equation. Using \( x = y^2 \):

For \( y = 1 \):
\[ x = 1^2 = 1 \]

For \( y = -1 \):
\[ x = (-1)^2 = 1 \]

Thus, the points of intersection are \( (1, 1) \) and \( (1, -1) \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( y = -1 \) to \( y = 1 \) can be computed using the integral of the difference between the upper curve and the lower curve. The upper curve is \( x = 3 - 2y^2 \) and the lower curve is \( x = y^2 \).

The area \( A \) is given by:

\[ A = \int_{-1}^{1} \left( (3 - 2y^2) - (y^2) \right) dy \]

This simplifies to:

\[ A = \int_{-1}^{1} (3 - 3y^2) \, dy \]

### Step 5: Calculate the integral
Now we can calculate the integral:

\[ A = \int_{-1}^{1} (3 - 3y^2) \, dy \]

We can split this into two separate integrals:

\[ A = \int_{-1}^{1} 3 \, dy - \int_{-1}^{1} 3y^2 \, dy \]

Calculating the first integral:

\[ \int_{-1}^{1} 3 \, dy = 3[y]_{-1}^{1} = 3(1 - (-1)) = 3 \times 2 = 6 \]

Calculating 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:

\[ A = 6 - 2 = 4 \]

### Step 6: Final area
Since the area calculated is symmetric about the x-axis, we can conclude that the total area bounded by the curves is:

\[ \text{Total Area} = 4 \text{ square units} \]