3.49. The side of the rhombus is $a$, and its acute angle is $\alpha$. The rhombus rotates around a line passing through its vertex and parallel to the larger diagonal. Find the volume of the solid of revolution.
3.49. By the condition, $ABCD$ is a rhombus, $AB=a, \angle BAD=\alpha\left(\alpha<90^{\circ}\right)$, $MN$ is the axis of rotation, $MN \| AC, B \in MN$ (Fig. 3.51); it is required to find $V_{\text {rot }}=2\left(V_{B E C D}-V_{B E C}\right)$. The trapezoid $B E C D$ forms a frustum of a cone upon rotation, and the triangle $B E C$ forms a cone. We have

$$
\begin{gathered}
V_{\mathrm{rot}}=2\left(\frac{1}{3} \pi \cdot BE\left(BD^{2}+BD \cdot EC+EC^{2}-EC^{2}\right)\right)= \\
=\frac{2}{3} \pi BE \cdot BD(BD+EC)
\end{gathered}
$$

But $BD=2BO=2EC$; therefore, $V_{\text {rot }}=4 \pi BE \cdot EC^{2}$. Considering that $\angle CBE=\frac{\alpha}{2}$, from $\triangle BEC$ we find $EC=a \sin \frac{\alpha}{2}, BE=a \cos \frac{\alpha}{2}$. Thus,

$$
V_{\text {rot }}=4 \pi a \cos \frac{\alpha}{2} \cdot a^{2} \sin ^{2} \frac{\alpha}{2}=2 \pi a^{3} \sin \alpha \sin \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-530.jpg?height=596&width=527&top_left_y=235&top_left_x=58)

Fig. 3.51

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-530.jpg?height=513&width=527&top_left_y=255&top_left_x=627)

Fig. 3.52