3.93. The larger base of an isosceles trapezoid is equal to $a$, the acute angle is equal to $\alpha$. 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.
3.93. By the condition, \( A B \| D C, A D = B C, A B = a, \angle D A B = \alpha, B D \perp A D \) (Fig. 3.93). The volume of the solid of revolution of trapezoid \( A B C D \) around \( A B \) will be

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-560.jpg?height=507&width=433&top_left_y=108&top_left_x=120)

Fig. 3.92

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-560.jpg?height=294&width=525&top_left_y=322&top_left_x=630)

Fig. 3.93

sought using the formula \( V_{\text {solid of rev.}} = 2 V_{A E D} + V_{D E F C} \), where \( D E \perp A B, C F \perp A B \). We have

\[
\begin{aligned}
V_{\text {solid of rev.}} = & \frac{2}{3} \pi D E^{2} \cdot A E + \pi D E^{2} \cdot E F = \frac{1}{3} \pi D E^{2}(2 A E + 3 E F) = \\
& = \frac{1}{3} \pi D E^{2}(a + 2 E F) = \pi D E^{2}\left(\frac{1}{3} a + \frac{2}{3} E F\right) = \\
& \pi D E^{2}\left(\frac{1}{3} a + \frac{2}{3}(a - 2 A E)\right) = \pi D E^{2}\left(a - \frac{4}{3} A E\right)
\end{aligned}
\]

From \(\triangle A D B\) we find that \( A D = a \cos \alpha \), and from \(\triangle A D E\) - that \( D E = A D \sin \alpha = a \cos \alpha \sin \alpha = \frac{a}{2} \sin 2 \alpha, \quad A E = A D \cos \alpha = a \cos ^{2} \alpha \). Therefore,

\[
\begin{aligned}
& V_{\text {solid of rev.}} = \pi \frac{a^{2}}{4} \sin ^{2} 2 \alpha\left(a - \frac{4}{3} a \cos ^{2} \alpha\right) = \frac{\pi a^{3}}{3} \sin ^{2} 2 \alpha\left(\frac{3}{4} - \cos ^{2} \alpha\right) = \\
& = \frac{\pi a^{3}}{3} \sin ^{2} 2 \alpha \frac{1 - 2 \cos 2 \alpha}{4} = \frac{\pi a^{3}}{3} \sin ^{2} 2 \alpha \sin \left(\alpha - \frac{\pi}{6}\right) \sin \left(\alpha + \frac{\pi}{6}\right)
\end{aligned}
\]