8. Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with left and right foci $F_{1}$ and $F_{2}$, a line $l$ is drawn through the right focus of the ellipse, intersecting the ellipse at points $P$ and $Q$. Then the maximum value of the area of the incircle of $\triangle F_{1} P Q$ is $\qquad$ .
8. $\frac{9 \pi}{16}$.

Since the product of the radius of the incircle and the perimeter of the triangle is twice the area of the triangle, and the perimeter of $\triangle F_{1} P Q$ is a constant 8, we only need to find the maximum value of the area of $\triangle F_{1} P Q$.
Let the line $l: x=m y+1$. Combining with the equation of the ellipse, we get $\left(3 m^{2}+4\right) y^{2}+6 m y-9=0$.
Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
y_{1}+y_{2}=-\frac{6 m}{3 m^{2}+4}, y_{1} y_{2}=-\frac{9}{3 m^{2}+4} \text {. }
$$

Thus, $S_{\triangle F_{1} P Q}=\frac{1}{2}\left|F_{1} F_{2}\right|\left|y_{1}-y_{2}\right|$
$=\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}}$
$=12 \sqrt{\frac{m^{2}+1}{\left(3 m^{2}+4\right)^{2}}}$.
Also, $\frac{m^{2}+1}{\left(3 m^{2}+4\right)^{2}}=\frac{1}{9 m^{2}+15+\frac{1}{m^{2}+1}}$
$$
=\frac{1}{9 m^{2}+9+\frac{1}{m^{2}+1}+6} \leqslant \frac{1}{16} \text {, }
$$

Therefore, the inradius $r=\frac{2 S_{\triangle F_{1} P Q}}{8} \leqslant \frac{3}{4}$.
Hence, the maximum value of the area sought is $\frac{9 \pi}{16}$.