$S_{\triangle F_{1} P Q}=\frac{1}{2} \cdot 2 c \cdot\left|y_{P}-y_{Q}\right| \leqslant \frac{2 b^{2} c}{a}$, equality holds when $P Q \perp x$ axis.
While the perimeter of $\triangle F_{1} P Q$ is $4 a \Rightarrow S_{\triangle F_{1} P Q}=2 a \cdot r$, thus $2 a \cdot r \leqslant \frac{2 b^{2} c}{a} \Rightarrow r \leqslant \frac{b^{2} c}{a^{2}}=\frac{3}{4}$. Therefore, the maximum value of the area of the incircle of $\triangle F_{1} P Q$ is $\frac{9 \pi}{16}$.
8. Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with left and right foci $F_{1}, F_{2}$, a line $l$ passing through the right focus intersects the ellipse at points $P, Q$. Then the maximum value of the area of the incircle of $\triangle F_{1} P Q$ is $\qquad$