8. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A A_{1}=1, E, F$ are the midpoints of edges $C C_{1}, D D_{1}$, respectively. Then the area of the section formed by the plane $A E F$ cutting the circumscribed sphere of the cube is $\qquad$ .
As shown in the figure, $P$ is the center of the square $A A_{1} B_{1} B$, $M, N$ are the midpoints of $A B, E F$ respectively. It is easy to prove that plane $P M N \perp$ plane $A B E F$, and $O$ is the midpoint of $P N$. A line $O O_{1} \perp M N$ is drawn through $O$ at $O_{1}$, then $O_{1}$ is the center of the cross-sectional circle. In $\triangle P M N$, $P M=\frac{1}{2}$, $P N=1, M N=\frac{\sqrt{5}}{2} \Rightarrow O O_{1}=\frac{1}{2 \sqrt{5}}$, thus the radius of the cross-sectional circle $R^{2}=\frac{3}{4}-\frac{1}{20}=\frac{7}{10}$, hence the area of the required cross-sectional circle is $\frac{7 \pi}{10}$.