9. (16 points) Given a conical frustum-shaped glass cup filled with water, the inner diameters of the bottom and top bases are $4$ and $14$, respectively, and the inner height is 12. First, place a solid sphere A at the bottom of the cup, which touches the inner side of the cup (i.e., is tangent to it) all around. Then, place another solid sphere B, which touches sphere A and also touches the inner side of the cup all around. Find the volume of the water in the cup.
9. As shown in Figure 3, the axial cross-sections of the water cup, sphere $\odot O_{1}$, and sphere $\odot O_{2}$ are trapezoid $ABCD$, circle $\odot O_{1}$, and circle $\odot O_{2}$, respectively. Let $\odot O_{1}$ and $\odot O_{2}$ be tangent at point $P$. Draw $MN \parallel BC$ through point $P$, intersecting $AB$ and $CD$ at points $M$ and $N$, respectively. Draw $AE \perp BC$ at point $E$, and let $O_{1}O_{2}$ intersect $BC$ and $AD$ at points $F$ and $G$, respectively. Connect $O_{1}B$ and $O_{2}M$.
Clearly, $O_{1}O_{2}$ passes through point $P$, and $O_{1}B \parallel O_{2}M$,
\[
\tan \angle ABE = \frac{AE}{BE} = \frac{12}{5},
\]
\[
\tan \angle ABC = -\frac{12}{5}.
\]
Let $\angle ABC = 2\alpha$. Then,
\[
\angle O_{1}BF = \alpha,
\]
\[
\tan \angle ABC = \tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = -\frac{12}{5}
\]
\[
\Rightarrow \tan \alpha = \frac{3}{2} \text{ or } -\frac{2}{3} \text{ (negative value is discarded)}.
\]
In right triangle $\triangle O_{1}BF$,
\[
O_{1}F = BF \tan \alpha = 2 \times \frac{3}{2} = 3,
\]
\[
PF = 6, PG = 6.
\]
Thus, $MN$ is the midline of trapezoid $ABCD$,
\[
MN = 9, MP = 4.5.
\]
In right triangle $\triangle O_{2}PM$,
\[
O_{2}P = MP \tan \alpha = 4.5 \times \frac{3}{2} = \frac{27}{4}.
\]
Therefore, the volume of the glass cup is
\[
\begin{array}{l}
V_{1} = \frac{1}{3} \pi \left( \left( \frac{4}{2} \right)^2 + \frac{4}{2} \times \frac{14}{2} + \left( \frac{14}{2} \right)^2 \right) \times 12 \\
= 268 \pi,
\end{array}
\]
The volume of sphere $O_{1}$ is $V_{2} = \frac{4}{3} \pi \times 3^3 = 36 \pi$,
The volume of sphere $O_{2}$ below the rim $AD$ is
\[
\begin{array}{l}
V_{3} = \pi PG^2 \left( O_{2}P - \frac{1}{3} PG \right) \\
= \pi \times 6^2 \left( \frac{27}{4} - \frac{1}{3} \times 6 \right) = 171 \pi.
\end{array}
\]
Thus, the volume of water in the cup is
\[
\begin{array}{l}
V_{1} - V_{2} - V_{3} = 268 \pi - 36 \pi - 171 \pi \\
= 61 \pi.
\end{array}
\]