12.389. The larger base of an isosceles trapezoid is equal to $a$, and the acute angle is $\alpha$. The diagonal of the trapezoid is perpendicular to its lateral side. The trapezoid rotates around its larger base. Find the volume of the solid of revolution.
Solution.

Hexagon $A B E C D K$ (Fig. 12.238) is the axial section of a body obtained by rotating an isosceles trapezoid $E C D K$ about its larger base, $E K=a$,

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1152.jpg?height=543&width=555&top_left_y=338&top_left_x=691)

Fig. 12.238

$\angle K E C=\alpha, K C \perp E C$ (Fig. 12.238). $D L$ and $C F$ are the heights of trapezoid $E C D K$.

In $\triangle E C K\left(\angle E C K=\frac{\pi}{2}\right): E C=E K \cos \angle K E C=a \cos \alpha$.

In $\triangle E F C\left(\angle E F C=\frac{\pi}{2}\right): E F=E C \cos \angle K E C=a \cos ^{2} \alpha ; C F=$

$$
=E C \sin \angle K E C=a \sin \alpha \cos \alpha
$$

$V_{1}$ is the volume of the cylinder with axial section $A B C D$, and $V_{2}$ is the volume of the cone with axial section $B E C$.

The volume of the body of revolution is:

$$
\begin{aligned}
& V=V_{1}+2 V_{2}=\pi \cdot F C^{2} \cdot C D+\frac{2}{3} \pi \cdot F C^{2} \cdot E F=\pi \cdot F C^{2}\left(C D+\frac{2}{3} E F\right)= \\
& =\pi \cdot F C^{2}\left(C D+2 E F-2 E F+\frac{2}{3} E F\right)=\pi \cdot F C^{2}\left(E K-\frac{4}{3} E F\right)= \\
& =\pi a^{2} \cos ^{2} \alpha \sin ^{2} \alpha\left(a-\frac{4}{3} a \cos ^{2} \alpha\right)=\frac{\pi a^{3}}{4} \sin ^{2} 2 \alpha\left(1-\frac{4}{3} \cos ^{2} \alpha\right)= \\
& =\frac{\pi a^{3}}{3} \sin ^{2} 2 \alpha\left(\frac{3}{4}-\cos ^{2} \alpha\right)=\frac{\pi a^{3}}{3} \sin ^{2} 2 \alpha\left(\cos ^{2} \frac{\pi}{6}-\cos ^{2} \alpha\right)=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{\pi a^{3}}{3} \sin ^{2} 2 \alpha\left(\frac{1+\cos \frac{\pi}{3}}{2}-\frac{1+\cos 2 \alpha}{2}\right)=\frac{\pi a^{3}}{3} \sin ^{2} 2 \alpha \cdot \frac{\cos \frac{\pi}{3}-\cos 2 \alpha}{2}= \\
& =\frac{\pi a^{3}}{3} \sin ^{2} 2 \alpha \sin \left(\alpha+\frac{\pi}{6}\right) \sin \left(\alpha-\frac{\pi}{6}\right)
\end{aligned}
$$

Answer: $\frac{\pi a^{3}}{3} \sin ^{2} 2 \alpha \sin \left(\alpha+\frac{\pi}{6}\right) \sin \left(\alpha-\frac{\pi}{6}\right)$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1153.jpg?height=451&width=585&top_left_y=489&top_left_x=93)