The area that is enclosed in the circle `x^(2)+y^(2)=2` which is not common enclosed by `y=x" " & " "y^(2)=x` is

Options

A. `(1)/(12)(24pi-1)`
B. `(1)/(6)(12pi-1)`
C. `(1)/(12)(6pi-1)`
D. `(1)/(12)(12pi-1)`

Answer

To solve the problem of finding the area enclosed by the circle \( x^2 + y^2 = 2 \) that is not common to the area enclosed by the line \( y = x \) and the parabola \( y^2 = x \), we can follow these steps:

### Step 1: Understand the Shapes
The circle \( x^2 + y^2 = 2 \) has a radius of \( \sqrt{2} \) and is centered at the origin (0, 0). The line \( y = x \) passes through the origin at a 45-degree angle, while the parabola \( y^2 = x \) opens to the right with its vertex at the origin.

### Step 2: Find Points of Intersection
To find the points of intersection between the line \( y = x \) and the parabola \( y^2 = x \), we substitute \( y = x \) into the parabola's equation:

\[ x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0 \]

This gives us the points \( (0, 0) \) and \( (1, 1) \).

### Step 3: Determine the Area of the Circle
The area of the circle is given by the formula:

\[ \text{Area of Circle} = \pi r^2 = \pi \cdot 2 = 2\pi \]

### Step 4: Set Up the Integral for the Area Between the Line and the Parabola
The area we want to exclude is bounded by the line \( y = x \) and the parabola \( y^2 = x \) from \( x = 0 \) to \( x = 1 \). We need to find the area between these two curves:

- The upper curve is \( y = \sqrt{x} \) (from \( y^2 = x \)).
- The lower curve is \( y = x \).

The area between these curves can be computed using the integral:

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

### Step 5: Evaluate the Integral
Now we compute the integral:

\[ \int_0^1 \sqrt{x} \, dx - \int_0^1 x \, dx \]

Calculating each integral separately:

1. For \( \int_0^1 \sqrt{x} \, dx \):

\[ \int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \frac{2}{3} \]

2. For \( \int_0^1 x \, dx \):

\[ \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \]

Now, substituting back:

\[ \text{Area} = \frac{2}{3} - \frac{1}{2} \]

Finding a common denominator (which is 6):

\[ \text{Area} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \]

### Step 6: Calculate the Required Area
Now, we subtract the area we found from the area of the circle:

\[ \text{Required Area} = \text{Area of Circle} - \text{Area between curves} \]

\[ \text{Required Area} = 2\pi - \frac{1}{6} \]

### Step 7: Final Expression
To express this in a simplified form:

\[ \text{Required Area} = \frac{12\pi - 1}{6} \]

### Final Answer
Thus, the area that is enclosed in the circle \( x^2 + y^2 = 2 \) which is not common to the area enclosed by \( y = x \) and \( y^2 = x \) is:

\[ \frac{12\pi - 1}{6} \]