6. As shown in Figure 2, given the ellipse $\frac{x^{2}}{2}+y^{2}=1, A$ and $B$ are the intersection points of the ellipse with the $x$-axis, $D A$ $\perp A B, C B \perp A B$, and $|D A|$ $=3 \sqrt{2},|C B|=\sqrt{2}$. A moving point $P$ is on the arc $\overparen{A B}$ above the $x$-axis. Then the minimum value of $S_{\triangle P C D}$ is
$6.4-\sqrt{6}$.
Given $C(\sqrt{2}, \sqrt{2}) 、 D(-\sqrt{2}, 3 \sqrt{2})$,
$$
|C D|=4, l_{C D}: y=-x+2 \sqrt{2} \text {. }
$$

Let $P(\sqrt{2} \cos \theta, \sin \theta)(0 \leqslant \theta \leqslant \pi)$.
Then the distance from point $P$ to line $C D$ is
$$
\begin{array}{l}
d=\frac{|\sqrt{2} \cos \theta+\sin \theta-2 \sqrt{2}|}{\sqrt{2}} \\
=\frac{|\sqrt{3} \sin (\theta+\varphi)-2 \sqrt{2}|}{\sqrt{2}},
\end{array}
$$

where, $\varphi=\arctan \sqrt{2}, \sin \varphi=\frac{\sqrt{2}}{\sqrt{3}}, \cos \varphi=\frac{\sqrt{3}}{3}$.
$$
\begin{array}{l}
\text { Also } \varphi \leqslant \theta+\varphi \leqslant \pi+\varphi \\
\Rightarrow \sin (\pi+\varphi) \leqslant \sin (\theta+\varphi) \leqslant 1 \\
\Rightarrow \sin \varphi \leqslant \sin (\theta+\varphi) \leqslant 1 \\
\Rightarrow-\frac{\sqrt{2}}{\sqrt{3}} \leqslant \sin (\theta+\varphi) \leqslant 1 \\
\Rightarrow-\sqrt{2} \leqslant \sqrt{3} \sin (\theta+\varphi) \leqslant \sqrt{3} \\
\Rightarrow-3 \sqrt{2} \leqslant \sqrt{3} \sin (\theta+\varphi)-2 \sqrt{2} \\
\quad \leqslant \sqrt{3}-2 \sqrt{2} \\
\Rightarrow 2 \sqrt{2}-\sqrt{3} \leqslant 2 \sqrt{2}-\sqrt{3} \sin (\theta+\varphi) \leqslant 3 \sqrt{2} \\
\Rightarrow 1 \sqrt{3} \sin (\theta+\varphi)-2 \sqrt{2} \mid \geqslant 2 \sqrt{2}-\sqrt{3} \\
\Rightarrow d \geqslant \frac{2 \sqrt{2}-\sqrt{3}}{\sqrt{2}}=2-\frac{\sqrt{6}}{2} \\
\Rightarrow S_{\triangle P C D}=\frac{1}{2} \times 4 \times d \\
\quad \geqslant 2\left(2-\frac{\sqrt{6}}{2}\right)=4-\sqrt{6} .
\end{array}
$$

Therefore, $\left(S_{\triangle P C D}\right)_{\min }=4-\sqrt{6}$.