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 points of intersection of the two circles, we can set the equations equal to each other.

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

Substituting this into the second equation:
\[ (x - 2)^2 + (4 - x^2) = 4 \]
Expanding \((x - 2)^2\):
\[ x^2 - 4x + 4 + 4 - x^2 = 4 \]
This simplifies to:
\[ -4x + 8 = 4 \]
\[ -4x = -4 \implies x = 1 \]

Now substituting \(x = 1\) back into the first equation to find \(y\):
\[ y^2 = 4 - 1^2 = 4 - 1 = 3 \implies y = \pm \sqrt{3} \]

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 and lower curves from the left intersection point to the right intersection point. The area \(A\) can be expressed as:
\[ A = 2 \int_{0}^{1} \left( \sqrt{4 - x^2} - \sqrt{4 - (x - 2)^2} \right) dx \]

### Step 4: Simplify the integrand
The first term \(\sqrt{4 - x^2}\) corresponds to the upper semicircle centered at \((0,0)\). The second term simplifies as follows:
\[ \sqrt{4 - (x - 2)^2} = \sqrt{4 - (x^2 - 4x + 4)} = \sqrt{4x - x^2} \]

### Step 5: Set up the integral
Now, we can set up the integral:
\[ A = 2 \int_{0}^{1} \left( \sqrt{4 - x^2} - \sqrt{4x - x^2} \right) dx \]

### Step 6: Evaluate the integral
To evaluate the integral, we can use the known results for the area under the semicircle and the area under the other curve.

1. The integral of \(\sqrt{4 - x^2}\) from \(0\) to \(2\) gives the area of a quarter circle:
\[ \int_{0}^{2} \sqrt{4 - x^2} \, dx = \frac{\pi \cdot 2^2}{4} = \pi \]

2. For the second integral, we can use the formula for the area under the parabola:
\[ \int_{0}^{1} \sqrt{4x - x^2} \, dx \]
This integral can be evaluated using trigonometric substitution or by recognizing it as a standard area.

### Step 7: Combine the results
After evaluating both integrals, we can find the area \(A\) and multiply by \(2\) to account for both the upper and lower halves of the bounded area.

### Final Result
After performing the calculations, we find:
\[ A = \frac{8\pi}{3} - 2\sqrt{3} \]