3.54. A hemisphere is inscribed in a cone; the great circle of the hemisphere lies in the base plane of the cone, and the spherical surface touches the surface of the cone. Find the volume of the hemisphere if the slant height of the cone is $l$ and it makes an angle $\alpha$ with the base plane.
3.54. The axial section of the figure is depicted as a triangle $S A B$, in which a semicircle $M n K$ is inscribed; $S A$ and $S B$ are the generators of the cone, $S O \perp A B$ is the height of the cone; $A B$ is the diameter of the base of the cone and $M K$ is the diameter of the semicircle; $O$ is the center of the base of the cone and the semicircle (Fig. 3.56). According to the problem, $S A=l, \angle S A O=\alpha$. From $\triangle S O A$ we find $O A=l \cos \alpha$. Let $C-$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-534.jpg?height=314&width=341&top_left_y=1018&top_left_x=814)

Fig. 3.56 be the point of tangency of the hemisphere with the lateral surface of the cone; then $O C \perp S A$. From $\triangle O C A$ we obtain

$$
O C=O A \sin \alpha=l \cos \alpha \sin \alpha=\frac{1}{2} l \sin 2 \alpha
$$

Therefore,

$$
V=\frac{2}{3} \pi O C^{3}=\frac{2}{3} \pi \cdot \frac{1}{8} l^{3} \sin ^{3} 2 \alpha=\frac{1}{12} \pi l^{3} \sin ^{3} 2 \alpha
$$