Find the area of the region bounded by the curves `x^(2)+y^(2)=4` and `(x-2)^(2)+y^(2)=4.`

Answer

To find the area of the region bounded by the curves \(x^2 + y^2 = 4\) and \((x - 2)^2 + y^2 = 4\), we can follow these steps:

### Step 1: Identify the curves
The first equation \(x^2 + y^2 = 4\) represents a circle centered at the origin \((0, 0)\) with a radius of 2. The second equation \((x - 2)^2 + y^2 = 4\) represents a circle centered at \((2, 0)\) with a radius of 2.

### Step 2: Find the points of intersection
To find the area between the two circles, we first need to determine the points where they intersect. We can do this by solving the two equations simultaneously.

1. From the first equation, we can express \(y^2\):
\[ y^2 = 4 - x^2 \]

2. Substitute \(y^2\) into the second equation:
\[ (x - 2)^2 + (4 - x^2) = 4 \]
Expanding the left side:
\[ (x^2 - 4x + 4) + (4 - x^2) = 4 \]
Simplifying:
\[ -4x + 8 = 4 \]
Rearranging gives:
\[ -4x = -4 \implies x = 1 \]

3. Substitute \(x = 1\) back into the first equation to find \(y\):
\[ y^2 = 4 - 1^2 = 3 \implies y = \pm \sqrt{3} \]
Thus, the points of intersection are \((1, \sqrt{3})\) and \((1, -\sqrt{3})\).

### Step 3: Set up the area integral
The area between the two curves can be found by integrating the difference of the upper curve and the lower curve from the left intersection point to the right intersection point.

1. The left circle (centered at the origin) has the equation:
\[ y = \sqrt{4 - x^2} \]
The right circle (centered at \((2, 0)\)) has the equation:
\[ y = \sqrt{4 - (x - 2)^2} \]

2. The area \(A\) can be expressed as:
\[ A = 2 \int_{1}^{2} \left( \sqrt{4 - (x - 2)^2} - \sqrt{4 - x^2} \right) dx \]
The factor of 2 accounts for the symmetry of the area above and below the x-axis.

### Step 4: Evaluate the integral
1. Simplify the integrand:
\[ \sqrt{4 - (x - 2)^2} = \sqrt{4 - (x^2 - 4x + 4)} = \sqrt{4x - x^2} \]
Thus, the integral becomes:
\[ A = 2 \int_{1}^{2} \left( \sqrt{4x - x^2} - \sqrt{4 - x^2} \right) dx \]

2. We can evaluate each integral separately:
- For \(\int \sqrt{4x - x^2} \, dx\), we can use a trigonometric substitution or complete the square.
- For \(\int \sqrt{4 - x^2} \, dx\), this is a standard integral representing the area of a semicircle.

### Step 5: Calculate the area
After evaluating the integrals, we combine the results to find the total area.

### Final Result
The area of the region bounded by the two curves is:
\[ A = \frac{8\pi}{3} - 2\sqrt{3} \text{ square units.} \]