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

Options

A. `(3)/(2)`
B. 2
C. 3
D. 4

Answer

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

### Step 1: Find the points of intersection
To find the points where the curves intersect, we set the equations equal to each other:

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

Rearranging gives:

\[ 3y^2 + y^2 - 3 = 0 \implies 3y^2 - 3 = 0 \implies y^2 = 1 \implies y = 1 \text{ or } y = -1 \]

Now, substituting \( y = 1 \) and \( y = -1 \) back into either equation to find \( x \):

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 2: Set up the integral for the area
The area between the curves can be found by integrating the difference of the two functions from the lower intersection point to the upper intersection point. The area \( A \) is given by:

\[ A = 2 \int_0^1 \left( (3 - 2y^2) - y^2 \right) dy \]

This simplifies to:

\[ A = 2 \int_0^1 (3 - 3y^2) dy \]

### Step 3: Calculate the integral
Now we calculate the integral:

\[ A = 2 \left[ 3y - y^3 \right]_0^1 \]

Evaluating this from 0 to 1:

\[ A = 2 \left[ (3(1) - (1)^3) - (3(0) - (0)^3) \right] \]
\[ = 2 \left[ 3 - 1 - 0 \right] \]
\[ = 2 \cdot 2 = 4 \]

### Final Answer
The area bounded by the curves \( x = y^2 \) and \( x = 3 - 2y^2 \) is \( 4 \) square units.

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