22 Given the point set $A=\left\{(x, y) \mid x^{2}+(y-2)^{2} \leqslant 1\right\}, B_{m}=\{(x, y) \mid y=$ $-(x-m)^{2}+2 m_{i}^{\prime}, m \in \mathbf{R}$ Let the set $B$ be the union of all $B_{m}(m \in \mathbf{R})$, then the area of $A \cap B$ is $\qquad$ .
$22 \arctan 2-\frac{2}{5}$. First, prove that $2 x+1 \geqslant-(x-m)^{2}+2 m(1)$. Rearranging and simplifying, (1) becomes $(x+1-m)^{2} \geqslant 0$, which is clearly true.
Therefore, points above the line $y=2 x+1$ are not in $B$.
On the other hand, if $y_{0} \leqslant 2 x_{0}+1$, let $y_{0}=2 x_{0}+1-h$, where $h$ is a positive real number. Then from $2 x_{0}+1-h=-\left(x_{0}-m\right)^{2}+2 m$ we get $\left(x_{0}+1-m\right)^{2}=h$. Taking $m=\sqrt{h}+x_{0}+1$, we have $\left(x_{0}, y_{0}\right) \in B_{m}$.
Thus, the common part of circle $A$ and $B$ is the segment below the line $y=2 x+1$.
The central angle of the segment is $2 \alpha$, where $\alpha$ satisfies $\tan \alpha=2$. Therefore, the area of $A \cap B$ is
$$
\frac{1}{2} \times 1^{2} \times 2 \alpha-\frac{1}{2} \times 1^{2} \times \sin 2 \alpha=\alpha-\sin \alpha \cos \alpha=\alpha-\frac{2}{5} .
$$