What is the area of the letter $O$ made by Dürer? The two circles have a unit radius. Their centers, or the angle of a triangle formed by an intersection point of the circles is $30^o$.
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1. **Identify the Geometry and Given Information:**
   - The two circles have a unit radius, i.e., \( r = 1 \).
   - The centers of the circles are \( O_1 \) and \( O_2 \).
   - The angle formed by the intersection point of the circles is \( 30^\circ \).

2. **Calculate the Distance Between the Centers:**
   - Since the angle at the intersection point is \( 30^\circ \), the angle at the center of each circle subtended by the chord is \( 2 \times 30^\circ = 60^\circ \).
   - The distance between the centers \( O_1 \) and \( O_2 \) can be calculated using the Law of Cosines in the triangle \( O_1AO_2 \):
     \[
     O_1O_2 = \sqrt{O_1A^2 + O_2A^2 - 2 \cdot O_1A \cdot O_2A \cdot \cos(60^\circ)}
     \]
     Since \( O_1A = O_2A = 1 \) and \( \cos(60^\circ) = \frac{1}{2} \):
     \[
     O_1O_2 = \sqrt{1^2 + 1^2 - 2 \cdot 1 \cdot 1 \cdot \frac{1}{2}} = \sqrt{2 - 1} = \sqrt{1} = 1
     \]

3. **Calculate the Area of the Sector:**
   - The area of each sector formed by the angle \( 60^\circ \) in each circle is:
     \[
     \text{Area of sector} = \frac{60^\circ}{360^\circ} \times \pi \times r^2 = \frac{1}{6} \pi \times 1^2 = \frac{\pi}{6}
     \]

4. **Calculate the Area of the Triangle:**
   - The area of the triangle \( O_1AO_2 \) can be calculated using the formula for the area of an equilateral triangle with side length \( 1 \):
     \[
     \text{Area of triangle} = \frac{\sqrt{3}}{4} \times 1^2 = \frac{\sqrt{3}}{4}
     \]

5. **Calculate the Area of the Intersection:**
   - The area of the intersection of the two circles is twice the area of the sector minus the area of the triangle:
     \[
     \text{Area of intersection} = 2 \times \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = \frac{\pi}{3} - \frac{\sqrt{3}}{2}
     \]

6. **Calculate the Area of the Letter \( O \):**
   - The area of the letter \( O \) is the area of one circle minus the area of the intersection:
     \[
     \text{Area of letter } O = \pi - \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) = \pi - \frac{\pi}{3} + \frac{\sqrt{3}}{2} = \frac{2\pi}{3} + \frac{\sqrt{3}}{2}
     \]

The final answer is \(\boxed{\frac{2\pi}{3} + \frac{\sqrt{3}}{2}}\)