11. Let the equation of $\odot M$ be $x^{2}+(y-1)^{2}=r^{2}$.

Combining with the parabola equation, we get
$$
y^{2}-y+1-r^{2}=0 \text {. }
$$

By the problem and the symmetry of $\odot M$ and the parabola about the $y$-axis, we set $A\left(-x_{2}, y_{2}\right), B\left(-x_{1}, y_{1}\right), C\left(x_{1}, y_{1}\right)$, $D\left(x_{2}, y_{2}\right)$, where $x_{1} 、 x_{2} 、 y_{1} 、 y_{2}>0, x_{1}0, \\
1-r^{2}>0, \\
r>0
\end{array} \Rightarrow \frac{\sqrt{3}}{2}<r<1 .\right.
$$

Thus, the radius $r$ of $\odot M$ is in the range $\left(\frac{\sqrt{3}}{2}, 1\right)$.
(2) Since quadrilateral $A B C D$ is a trapezoid, we have
$$
\begin{array}{l}
S_{\text {trapezoid } A B C D}=\frac{1}{2}\left(y_{2}-y_{1}\right)\left(2 x_{1}+2 x_{2}\right) \\
=\left(y_{2}-y_{1}\right)\left(x_{1}+x_{2}\right) .
\end{array}
$$

From equation (1), we know
$$
\begin{array}{l}
y_{1}+y_{2}=1, y_{1} y_{2}=1-r^{2}, \\
x_{1}=\sqrt{y_{1}}, x_{2}=\sqrt{y_{2}} .
\end{array}
$$

Thus, $S_{\text {trapezoid } A B C D}$
$$
\begin{array}{l}
=\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}} \sqrt{\left(\sqrt{y_{1}}+\sqrt{y_{2}}\right)^{2}} \\
=\sqrt{1-4\left(1-r^{2}\right)} \sqrt{1+2 \sqrt{1-r^{2}}} \\
=\sqrt{\left(4 r^{2}-3\right)\left(1+2 \sqrt{1-r^{2}}\right)}\left(\frac{\sqrt{3}}{2}<r<1\right) .
\end{array}
$$

Using a graphing calculator, the maximum area of quadrilateral $A B C D$ is 1.0887.
At this time, $r=0.9860$.
11. In the Cartesian coordinate system $x O y$, the circle $\odot M$ with center $M(0,1)$ intersects the parabola $y=x^{2}$ at four points $A$, $B$, $C$, and $D$ in sequence.
(1) Find the range of the radius $r$ of $\odot M$;
(2) Find the maximum area of quadrilateral $A B C D$ (accurate to $10^{-4}$).