Sketch the region enclosed between the circles `x ^(2) +y ^(2) =1 and (x -1) ^(2) +y ^(2) =1,` which lies in the first quadrant. Also, find the area of the region.

Answer

To solve the problem of finding the area of the region enclosed between the circles \(x^2 + y^2 = 1\) and \((x - 1)^2 + y^2 = 1\) that lies in the first quadrant, we will follow these steps:

### Step 1: Identify the equations of the circles
The first circle is given by:
\[ x^2 + y^2 = 1 \]
This circle has its center at \((0, 0)\) and a radius of \(1\).

The second circle is given by:
\[ (x - 1)^2 + y^2 = 1 \]
This circle has its center at \((1, 0)\) and a radius of \(1\).

### Step 2: Find the points of intersection
To find the points where these two circles intersect, we can set the equations equal to each other.

From the first circle:
\[ y^2 = 1 - x^2 \]

Substituting \(y^2\) into the second circle's equation:
\[ (x - 1)^2 + (1 - x^2) = 1 \]
Expanding this:
\[ x^2 - 2x + 1 + 1 - x^2 = 1 \]
Simplifying:
\[ -2x + 2 = 1 \implies -2x = -1 \implies x = \frac{1}{2} \]

Now substituting \(x = \frac{1}{2}\) back to find \(y\):
\[ y^2 = 1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \implies y = \frac{\sqrt{3}}{2} \]

Thus, the points of intersection are:
\[ \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \text{ and } \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \]

### Step 3: Sketch the region
In the first quadrant, we only consider the point \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\). The region enclosed between the two circles in the first quadrant is the area above the line segment connecting the two points of intersection and below the two circles.

### Step 4: Set up the integral for the area
To find the area of the region, we can set up the integral. The area can be calculated as:
\[ \text{Area} = \int_{0}^{\frac{1}{2}} (y_{\text{upper}} - y_{\text{lower}}) \, dx \]
Where:
- \(y_{\text{upper}} = \sqrt{1 - x^2}\) (from the first circle)
- \(y_{\text{lower}} = \sqrt{1 - (x - 1)^2} = \sqrt{2x - x^2}\) (from the second circle)

Thus, the area becomes:
\[ \text{Area} = \int_{0}^{\frac{1}{2}} \left(\sqrt{1 - x^2} - \sqrt{2x - x^2}\right) \, dx \]

### Step 5: Calculate the area
Now we compute the integral:
\[ \text{Area} = \int_{0}^{\frac{1}{2}} \sqrt{1 - x^2} \, dx - \int_{0}^{\frac{1}{2}} \sqrt{2x - x^2} \, dx \]

1. **First Integral**:
\[ \int_{0}^{\frac{1}{2}} \sqrt{1 - x^2} \, dx = \frac{\pi}{12} \quad \text{(Area of quarter circle)} \]

2. **Second Integral**:
\[ \int_{0}^{\frac{1}{2}} \sqrt{2x - x^2} \, dx = \frac{\pi}{12} - \frac{\sqrt{3}}{8} \]

Putting it all together:
\[ \text{Area} = \frac{\pi}{12} - \left(\frac{\pi}{12} - \frac{\sqrt{3}}{8}\right) = \frac{\sqrt{3}}{8} \]

### Final Result
The area of the region enclosed between the circles in the first quadrant is:
\[ \text{Area} = \frac{\sqrt{3}}{8} \]