12.273. Point $A$ lies on the circumference of the upper base of the cylinder, point $B$ - on the circumference of the lower base. The line $A B$ forms an angle $\alpha$ with the base plane, and an angle $\beta$ with the plane of the axial section passing through point $B$. Find the volume of the cylinder if the length of segment $A B$ is $l$.
## Solution.

Let $O$ and $O_{1}$ be the centers of the upper and lower bases of the cylinder, and $CD$ be its generatrix (Fig. 12.137), then $BD$ is the projection of $BC$ on the plane of the lower base and $\angle CBD = \alpha$.

The plane of the axial section $BMNA$ is perpendicular to the planes of the bases. Draw the perpendicular $CK$ to $MN$, then $BK$ is the projection of $BC$ on the plane of the axial section and $\angle CBK = \beta$.

$MN$ is the diameter of the upper base of the cylinder, so $\angle MCN = 90^{\circ}$.

In $\triangle CDB\left(\angle CDB = 90^{\circ}\right): CD = l \sin \alpha; BD = l \cos \alpha$.

In $\triangle CKB\left(\angle CKB = 90^{\circ}\right): CK = l \sin \beta$.

$CM = BD = l \cos \alpha$.

In $\triangle CKM\left(\angle CKM = 90^{\circ}\right):$

$$
\begin{aligned}
& MK = \sqrt{CM^2 - CK^2} = l \sqrt{\cos^2 \alpha - \sin^2 \beta} = l \sqrt{\frac{1 + \cos 2\alpha}{2} - \frac{1 - \cos 2\beta}{2}} = \\
& = l \sqrt{\cos (\alpha + \beta) \cos (\alpha - \beta)}
\end{aligned}
$$

In $\triangle MCN\left(\angle MCN = 90^{\circ}, CK\right.$ is the height $)$:

$$
CM^2 = MK \cdot MN; MN = \frac{CM^2}{MK} = \frac{l \cos^2 \alpha}{\sqrt{\cos (\alpha + \beta)} \cos (\alpha - \beta)}
$$

The radius of the base of the cylinder

$$
R = \frac{1}{2} MN = \frac{l \cos^2 \alpha}{2 \sqrt{\cos (\alpha + \beta) \cos (\alpha - \beta)}}
$$

The height of the cylinder $H = CD = l \sin \alpha$.

The volume of the cylinder

$$
\begin{aligned}
& V = \pi R^2 H = \frac{\pi l^3 \cos^4 \alpha \sin \alpha}{4 \cos (\alpha + \beta) \cos (\alpha - \beta)} = \frac{\pi l^3 \sin 2\alpha \cos^3 \alpha}{8 \cos (\alpha + \beta) \cos (\alpha - \beta)} \\
& \text{Answer: } \frac{\pi l^3 \sin 2\alpha \cos^3 \alpha}{8 \cos (\alpha + \beta) \cos (\alpha - \beta)}
\end{aligned}
$$