12.237. The hypotenuse of a right triangle is equal to $c$, and its acute angle is equal to $\alpha$. The triangle rotates around the bisector of the external right angle. Find the volume of the solid of revolution.
## Solution.

Let in $\triangle A C B \quad \angle A C B=\frac{\pi}{2}, \angle B A C=\alpha, A B=c$ (Fig. 12.104). Then $A C=c \cos \alpha, B C=c \sin \alpha$.

When $\triangle A C B$ is rotated around the bisector $O_{1} O_{2}$, a body is formed, the axial section of which is shown in Fig. 12.104.

The volume $V$ of this body is equal to the difference between the volume $V_{1}$ of the truncated cone, whose axial section is the trapezoid $A B B_{1} A_{1}$, and the sum of the volumes $V_{2}$ and $V_{3}$ of the cones, whose axial sections are the triangles $A C A_{1}$ and $B C B_{1}$.

$$
\begin{aligned}
& \angle A C A_{1}=\angle B C B_{1}=\frac{\pi}{2} \Rightarrow \angle O_{1} B C=\angle O_{2} A C=\frac{\pi}{4}, C O_{2}=O_{2} A= \\
& =\frac{c \cos \alpha}{\sqrt{2}}, C O_{1}=O_{1} B=\frac{c \sin \alpha}{\sqrt{2}}
\end{aligned}
$$

If $O_{2} A=R, O_{1} B=r$, then

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0986.jpg?height=467&width=422&top_left_y=158&top_left_x=96)

Fig. 12.103

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0986.jpg?height=547&width=689&top_left_y=66&top_left_x=576)

Fig. 12.104

$$
\begin{aligned}
& V=V_{1}-\left(V_{2}+V_{3}\right)=\frac{1}{3} \pi \cdot O_{1} O_{2}\left(R^{2}+R r+r^{2}\right)-\left(\frac{1}{3} \pi R^{2} \cdot C O_{2}+\frac{1}{3} \pi r^{2} \cdot C O_{1}\right)= \\
& =\frac{1}{3} \pi(R+r)\left(R^{2}+R r+r^{2}\right)-\frac{1}{3} \pi\left(R^{3}+r^{3}\right)=\frac{1}{3} \pi(R+r)\left(R^{2}+R r+r^{2}-\right. \\
& \left.-\left(R^{2}-R r+r^{2}\right)\right)=\frac{2}{3} \pi(R+r) R r=\frac{2}{3} \pi\left(\frac{c \cos \alpha}{\sqrt{2}}+\frac{c \sin \alpha}{\sqrt{2}}\right) \cdot \frac{c \cos \alpha}{\sqrt{2}} \cdot \frac{c \sin \alpha}{\sqrt{2}}= \\
& =\frac{c^{3}}{6} \pi \sin 2 \alpha \cdot \frac{\cos \alpha+\sin \alpha}{\sqrt{2}}=\frac{\pi c^{3}}{6} \sin 2 \alpha \sin \left(\frac{\pi}{4}+\alpha\right) \\
& \quad \text { Answer: } \frac{\pi c^{3}}{6} \sin 2 \alpha \sin \left(\frac{\pi}{4}+\alpha\right)
\end{aligned}
$$