Two disks of radius 1 are drawn so that each disk's circumference passes through the center of the other disk. What is the circumference of the region in which they overlap?
1. **Understanding the Geometry**:
   - Each disk has a radius of 1.
   - The circumference of each disk passes through the center of the other disk.
   - This implies that the distance between the centers of the two disks is equal to the radius, which is 1.

2. **Forming the Equilateral Triangle**:
   - The distance between the centers of the disks is 1.
   - The radius of each disk is also 1.
   - Therefore, the line segment connecting the centers of the disks and the radii to the intersection points form an equilateral triangle with each side of length 1.

3. **Calculating the Central Angle**:
   - In an equilateral triangle, each internal angle is \(60^\circ\).
   - The central angle subtended by the arc of the circle at the center of the disk is \(120^\circ\) (since the angle at the center is twice the angle at the circumference).

4. **Calculating the Length of One Arc**:
   - The circumference of a full circle is \(2\pi \times \text{radius} = 2\pi \times 1 = 2\pi\).
   - The length of the arc corresponding to \(120^\circ\) is:
     \[
     \text{Arc length} = \frac{120^\circ}{360^\circ} \times 2\pi = \frac{1}{3} \times 2\pi = \frac{2\pi}{3}
     \]

5. **Calculating the Combined Length of Both Arcs**:
   - Since there are two such arcs (one from each disk), the combined length is:
     \[
     \text{Combined arc length} = 2 \times \frac{2\pi}{3} = \frac{4\pi}{3}
     \]

The final answer is \(\boxed{\frac{4\pi}{3}}\).