Example 3 As shown in Figure $2, \odot O$ has a radius of $2, \odot O$ contains a point $P$ that is 1 unit away from the center $O$, and a chord $AB$ passing through point $P$ forms a segment with the minor arc $\overparen{AB}$. Find the minimum value of the area of this segment. (13th Junior High School Mathematics Competition in Jiangsu Province)
Analysis: Under certain conditions of a circle, the shorter the chord, the smaller the central angle it subtends, and consequently, the smaller the area of the corresponding segment. Therefore, to find the minimum area of the segment, it is only necessary to determine the extreme position of the chord passing through point $P$. Clearly, the chord passing through point $P$ and perpendicular to $OP$ is the shortest.

Solution: As shown in Figure 2, connect $OP$. Draw the chord $A' B' \perp OP$ through point $P$, and draw the perpendicular distance $OQ$ from the center $O$ to the chord $AB$. Clearly, $OP > OQ$. Thus, $A' B' < AB$. Therefore, the segment with the smallest area should be segment $A' B B'$.
Since $O$ is the center, $OP = 1$, and $A' B' \perp OP$, then $A' B' = 2 B' P = 2 \sqrt{B' O^2 - PO^2} = 2 \sqrt{3}$.
And $\sin \angle B' OP = \frac{B' P}{B' O} = \frac{\sqrt{3}}{2}$,
Thus, $\angle B' OP = 60^\circ$, $\angle A' OB' = 120^\circ$.
Therefore, $S_{\text{segment } A' OB'} = \frac{120}{360} \pi \times 2^2 = \frac{4 \pi}{3}$,
$S_{\triangle A' OB'} = \frac{1}{2} A' B' \cdot OP = \sqrt{3}$.
Hence, $S_{\text{segment } C' BB} = \frac{4 \pi}{3} - \sqrt{3}$.