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. `(1)/(3) (12pi-1)`
B. 2. `(1)/(6) (24pi-1)`
C. 3. `(1)/(6) (12pi-1)`
D. 4. `(1)/(3) (6pi-1)`

Answer

To find the area that is enclosed in the circle \(x^2 + y^2 = 2\) which is not common enclosed by the lines \(y = x\) and the parabola \(y^2 = x\), we can follow these steps:

### Step 1: Understand the shapes involved
The equation \(x^2 + y^2 = 2\) represents a circle centered at the origin (0,0) with a radius of \(\sqrt{2}\). The line \(y = x\) is a diagonal line passing through the origin, and the parabola \(y^2 = x\) opens to the right.

### Step 2: Find the points of intersection
To find the area that is not common, we first need to find the points where the line \(y = x\) intersects the parabola \(y^2 = x\).

Substituting \(y = x\) into the parabola's equation:
\[ x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0 \]
Thus, \(x = 0\) or \(x = 1\). The corresponding \(y\) values are also \(0\) and \(1\). Therefore, the points of intersection are \((0, 0)\) and \((1, 1)\).

### Step 3: Calculate the area of the circle
The area \(A\) of the circle is given by the formula:
\[ A = \pi r^2 \]
where \(r = \sqrt{2}\). Thus,
\[ A = \pi (\sqrt{2})^2 = 2\pi. \]

### Step 4: Set up the integral for the area between the curves
Next, we need to calculate the area between the curves \(y = \sqrt{x}\) (from \(y^2 = x\)) and \(y = x\) from \(x = 0\) to \(x = 1\). The area \(A_{common}\) can be calculated using the integral:
\[ A_{common} = \int_0^1 (\sqrt{x} - x) \, dx. \]

### Step 5: Evaluate the integral
Now we calculate the integral:
\[ A_{common} = \int_0^1 (\sqrt{x} - x) \, dx = \int_0^1 x^{1/2} \, dx - \int_0^1 x \, dx. \]
Calculating each integral separately:
1. \(\int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \frac{2}{3}\).
2. \(\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}\).

Thus,
\[ A_{common} = \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6}. \]

### Step 6: Calculate the required area
The required area that is enclosed in the circle but not common with the other shapes is:
\[ A_{required} = A_{circle} - A_{common} = 2\pi - \frac{1}{6}. \]

### Step 7: Finalize the answer
Thus, the final answer can be expressed as:
\[ A_{required} = 2\pi - \frac{1}{6}. \]