6. Given an equilateral $\triangle A B C$ with side length 2, if $\overrightarrow{A P}=\frac{1}{3}(\overrightarrow{A B}+\overrightarrow{A C}), \overrightarrow{A Q}=\overrightarrow{A P}+\frac{1}{2} \overrightarrow{B C}$, then the area of $\triangle A P Q$ is $\qquad$.
According to the problem, $P$ is the centroid of $\triangle A B C$, so $S_{\triangle A P Q}=\frac{1}{2} \cdot \frac{2 \sqrt{3}}{3} \cdot 1=\frac{\sqrt{3}}{3}$.