4. A cone of height $H$ is circumscribed around a hemisphere of radius $r$, such that the bases of the hemisphere and the cone are concentric circles. Calculate the volume of the part of the cone that does not belong to the hemisphere, i.e., express that volume in terms of $r$ and $H$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_d793ff9f9da41a4dd093g-14.jpg?height=591&width=620&top_left_y=1201&top_left_x=841)

We use the labels as shown in the figure. From the similarity of triangles $A M N$ and $A N S$, we get

$$
\begin{aligned}
|A N|:|M N| & =|A S|:|N S| \\
R: r & =\sqrt{H^{2}+R^{2}}: H
\end{aligned}
$$

i.e.,

$$
R^{2}=\frac{r^{2} H^{2}}{H^{2}-r^{2}}
$$

The desired volume is the difference between the volume $V_{S}$ of the cone and the volume $V_{K}$ of the hemisphere, i.e.,

$$
\begin{aligned}
V & =V_{S}-V_{K}=\frac{1}{3} R^{2} \pi H-\frac{2}{3} r^{3} \pi \\
& \stackrel{(1)}{=} \frac{r^{2} \pi}{3}\left(\frac{H^{3}}{H^{2}-r^{2}}-2 r\right) .
\end{aligned}
$$