The bases of the trapezoid are 8 and 2. The angles adjacent to the larger base are each $45^{\circ}$. Find the volume of the solid formed by rotating the trapezoid around the larger base.
Let $C C_{1}$ and $D D_{1}$ be the heights of a given isosceles trapezoid $A B C D$, dropped from the vertices $C$ and $D$ of the smaller base. If $V_{1}$ is the volume of the cylinder obtained by rotating the rectangle $D C C_{1} D_{1}$ around the line $A B$, $V_{2}$ is the volume of the cone obtained by rotating the right triangle $A D D_{1}$ around the leg $A D_{1}$, and $V$ is the desired volume, then

$$
V=V_{1}+2 \cdot V_{2}
$$

In the isosceles trapezoid $A B C D$

$$
\begin{gathered}
D_{1} C_{1}=D C=2, A D_{1}=\frac{1}{2}(A B-C D)=\frac{1}{2}(8-2)=3, \\
D D_{1}=A D_{1}=3 .
\end{gathered}
$$

Therefore,

$$
\begin{gathered}
V_{1}=\pi \cdot D D_{1}^{2} \cdot D_{1} C_{1}=\pi \cdot 9 \cdot 2=18 \pi \\
V_{2}=\frac{1}{3} \pi \cdot D D_{1}^{2} \cdot A D_{1}=\frac{1}{3} \pi \cdot 9 \cdot 3=9 \pi
\end{gathered}
$$

Thus,

$$
V=V_{1}+2 \cdot V_{2}=18 \pi+18 \pi=36 \pi
$$

## Answer

## $36 \pi$.