6. As shown in the figure, in a cylindrical ball container with a height of 10 and a base radius of 1, there is a ball attached to both the top and bottom walls. The ball on the bottom wall is tangent to the bottom and the side of the container, and the ball on the top wall is tangent to the top and the side of the container. Ignoring the thickness of the container and the balls, a plane is tangent to both balls. It is known that the figure obtained by the intersection of this plane with the edge of the container is an ellipse. Then, the area of this ellipse is $\qquad$
6. $16 \pi$.

Analysis: When the cylinder is longitudinally cut along the diameter of the base, as shown in the figure, the cutting points are $A, A^{\prime}$, and it intersects the cylindrical surface at $C, C^{\prime}$. At this point, it is known that $C C^{\prime}$ is the major axis $2a$ of the ellipse. In the right triangle $ABO$, $AB=2$, $BO=\frac{20-2 \times 2}{2}=8$, so $\sin \angle AOB=\frac{AB}{OB}=\frac{2}{8}=\frac{1}{4}$. Also, $\sin \angle AOB=\frac{r}{OC}$, hence $a=OC=\frac{2}{\sin \angle AOB}=8$. The minor axis of the ellipse, which is the length of the diameter of the base of the cylinder, is $2b=4$, so $b=2, c=2\sqrt{15}$. The area of the ellipse is $S=ab\pi=16\pi$.