The area of the region enclosed by ` x ^ 2 + y ^ 2 = 2 , y ^ 2 = x ` and y - axis is

Options

A. ` ( pi ) /(4 ) + (1)/(3) `
B. ` ( pi ) /(2) + (1 ) /(3) `
C. ` ( pi ) /(4 ) - (1)/(3) `
D. ` ( pi ) /(2) - (1 ) /(3) `

Explanation

To find the area of the region enclosed by the curves \( x^2 + y^2 = 2 \), \( y^2 = x \), and the y-axis, we will follow these steps:

### Step 1: Identify the curves
The equation \( x^2 + y^2 = 2 \) represents a circle centered at the origin with a radius of \( \sqrt{2} \). The equation \( y^2 = x \) represents a parabola that opens to the right.

### Step 2: Find the points of intersection
To find the points where the curves intersect, we substitute \( y^2 = x \) into the circle's equation:

\[ x^2 + y^2 = 2 \implies x^2 + x = 2 \]

This simplifies to:

\[ x^2 + x - 2 = 0 \]

Factoring the quadratic:

\[ (x - 1)(x + 2) = 0 \]

Thus, the solutions are:

\[ x = 1 \quad \text{and} \quad x = -2 \]

Now, substituting \( x = 1 \) back into \( y^2 = x \):

\[ y^2 = 1 \implies y = 1 \quad \text{or} \quad y = -1 \]

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

### Step 3: Set up the area calculation
The area we want to find is symmetric about the x-axis. Therefore, we can calculate the area in the first quadrant and then double it.

The area can be calculated as:

\[ \text{Area} = 2 \left( \int_0^1 (y_{\text{circle}} - y_{\text{parabola}}) \, dx \right) \]

Where \( y_{\text{circle}} = \sqrt{2 - x^2} \) (from the circle) and \( y_{\text{parabola}} = \sqrt{x} \) (from the parabola).

### Step 4: Calculate the area
We need to compute:

\[ \text{Area} = 2 \left( \int_0^1 \left( \sqrt{2 - x^2} - \sqrt{x} \right) \, dx \right) \]

### Step 5: Evaluate the integral
1. **Calculate \( \int_0^1 \sqrt{2 - x^2} \, dx \)**:
- This can be solved using trigonometric substitution or recognizing it as a quarter circle area.
- The area of a circle with radius \( \sqrt{2} \) is \( \frac{1}{4} \pi (2) = \frac{\pi}{2} \).

2. **Calculate \( \int_0^1 \sqrt{x} \, dx \)**:
- This integral evaluates to \( \frac{2}{3} x^{3/2} \bigg|_0^1 = \frac{2}{3} \).

### Step 6: Combine results
Putting it all together:

\[ \text{Area} = 2 \left( \frac{\pi}{2} - \frac{2}{3} \right) = \pi - \frac{4}{3} \]

### Final Result
The area of the region enclosed by the curves is:

\[ \text{Area} = \pi - \frac{4}{3} \]