6. In the figure, $P$ is a point inside square $A B C D$, $P D \doteq 1$, $P A=\sqrt{2}, P B=\sqrt{3}$. Then the area of the shaded region is $\qquad$
6. $\left(\frac{1}{2}(2+\sqrt{3})\right)$.

As shown in the figure, construct $\triangle A P^{\prime} B \cong \triangle A P D$,
connect $P P^{\prime}, \angle P A P^{\prime}=90^{\circ}$,
$$
A P=A P^{\prime}=\sqrt{2} \text {, }
$$

then $P P^{\prime}=2$.
In $\triangle P P^{\prime} B$, $P^{\prime} P^{2}=4=3+1=P B^{2}+P^{\prime} B^{2}$.
Thus, $\angle P B P^{\prime}=90^{\circ}$.
Therefore, $\angle A P^{\prime} B+\angle A P B=180^{\circ}$,

which means $\angle A P D+\angle A P B=180^{\circ}$.
Hence, $D, P, B$ are collinear, i.e., $P$ lies on the diagonal $D B$,
$$
B D=1+\sqrt{3} \text {. }
$$

Therefore, the area of the shaded region is half the area of the square $A B C D$, i.e., $\frac{1}{4}(1+\sqrt{3})^{2}=\frac{1}{2}(2+\sqrt{3})$.