The area of the smaller part of the circle `x^(2)+y^(2)=2` cut off by the line `x=1` is

Options

A. `(pi)/(2)` sq. units
B. `((pi)/(2)-1)` sq. units
C. `((pi)/(2)+1)` sq. units
D. `((pi)/(2)-(1)/(2))` sq. units

Answer

To find the area of the smaller part of the circle \(x^2 + y^2 = 2\) cut off by the line \(x = 1\), we can follow these steps:

### Step 1: Identify the Circle and the Line
The equation of the circle is given by:
\[ x^2 + y^2 = 2 \]
This represents a circle centered at the origin \((0, 0)\) with a radius of \(\sqrt{2}\).

The line \(x = 1\) is a vertical line that intersects the circle.

### Step 2: Find the Intersection Points
To find where the line intersects the circle, substitute \(x = 1\) into the circle's equation:
\[ 1^2 + y^2 = 2 \implies 1 + y^2 = 2 \implies y^2 = 1 \implies y = \pm 1 \]
Thus, the points of intersection are \((1, 1)\) and \((1, -1)\).

### Step 3: Set Up the Area Calculation
The area we need to calculate is the area of the smaller segment of the circle above the line \(x = 1\). We can find this area by integrating the upper half of the circle from \(x = 1\) to \(x = \sqrt{2}\).

The equation of the upper half of the circle can be expressed as:
\[ y = \sqrt{2 - x^2} \]

### Step 4: Calculate the Area
The area \(A\) can be calculated using the integral:
\[ A = \int_{1}^{\sqrt{2}} \sqrt{2 - x^2} \, dx \]

### Step 5: Use the Integral Formula
We can use the formula for the integral of \(\sqrt{a^2 - x^2}\):
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
In our case, \(a = \sqrt{2}\).

### Step 6: Apply the Limits
Now, we evaluate the definite integral:
\[ A = \left[ \frac{x}{2} \sqrt{2 - x^2} + \frac{2}{2} \sin^{-1}\left(\frac{x}{\sqrt{2}}\right) \right]_{1}^{\sqrt{2}} \]

Calculating the upper limit \(x = \sqrt{2}\):
\[ A(\sqrt{2}) = \frac{\sqrt{2}}{2} \cdot 0 + 1 \cdot \frac{\pi}{2} = \frac{\pi}{2} \]

Calculating the lower limit \(x = 1\):
\[ A(1) = \frac{1}{2} \cdot 1 + 1 \cdot \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2} + \frac{\pi}{4} \]

### Step 7: Subtract the Lower Limit from the Upper Limit
Now, we find the area:
\[ A = \left(\frac{\pi}{2}\right) - \left(\frac{1}{2} + \frac{\pi}{4}\right) \]
\[ A = \frac{\pi}{2} - \frac{1}{2} - \frac{\pi}{4} = \frac{2\pi}{4} - \frac{1}{2} - \frac{\pi}{4} = \frac{\pi}{4} - \frac{1}{2} \]

### Final Step: Multiply by 2 for the Total Area
Since we only calculated the area above the line, we need to multiply by 2 to account for both the upper and lower segments:
\[ \text{Total Area} = 2 \left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{\pi}{2} - 1 \]

### Conclusion
Thus, the area of the smaller part of the circle cut off by the line \(x = 1\) is:
\[ \boxed{\frac{\pi}{2} - 1} \]