The radii of the base circles of a truncated cone are: $R$ and $r$ (where $R > r$). Each base circle, together with the opposite circle's center as the vertex, defines a rotational cone. The parts of these two cone surfaces and the lateral surface of the given truncated cone bound a solid with a volume of $k$. Calculate:

a) the volume $K$ of the truncated cone,

b) the surface area $F$ of the lateral surface of the truncated cone, given $R$, $r$, and $k$.

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The cross-section of the truncated cone and the inscribed cones is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_6b295cff2330f5907712g-1.jpg?height=431&width=615&top_left_y=224&top_left_x=733)

Figure 1

Using the notation in the figure, from the two pairs of similar triangles, we have:

$$
\frac{x}{m}=\frac{\varrho}{R} \quad \text { and } \quad \frac{m-x}{m}=\frac{\varrho}{r}
$$

From the latter,

$$
1-\frac{x}{m}=\frac{\varrho}{r}
$$

Combining the two,

$$
\frac{\varrho}{r}+\frac{\varrho}{R}=1, \quad \text { hence } \quad \varrho=\frac{R r}{R+r}
$$

Thus,

$$
x=\frac{\varrho}{R} m=\frac{r m}{R+r}
$$

and

$$
m-x=\frac{\varrho}{r} m=\frac{R m}{R+r}
$$

Let the volumes of the truncated cones determined by the circles with radii $\varrho$ and $R$, and $\varrho$ and $r$ be $K_{1}$ and $K_{2}$, respectively. Clearly, $K_{1}+K_{2}+k=K$. Using the calculated heights $x$ and $m-x$,

$$
\begin{gathered}
K_{1}=\frac{m \pi}{3} \frac{R}{R+r}\left(R^{2}+R \frac{R r}{R+r}+\frac{R^{2} r^{2}}{(R+r)^{2}}\right)= \\
\left.=\frac{m \pi}{3} \frac{R^{3}}{(R+r)^{3}}\left[(R+r)^{2}\right)+r(R+r)+r^{2}\right]=\frac{m \pi}{3} \frac{R^{3}}{(R+r)^{3}}\left(R^{2}+3 R r+3 r^{2}\right) .
\end{gathered}
$$

Similarly, we get (by swapping the roles of $R$ and $r$),

$$
K_{2}=\frac{m \pi}{3} \frac{r^{3}}{(R+r)^{3}}\left(3 R^{2}+3 R r+r^{2}\right)
$$

Thus,

$$
\begin{aligned}
K_{1}+K_{2} & =\frac{m \pi}{3} \frac{R^{2}\left(R^{3}+3 R^{2} r+3 R r^{2}\right)+r^{2}\left(r^{3}+3 r^{2} R+3 r R^{2}\right)}{(R+r)^{3}}= \\
& =\frac{m \pi}{3} \frac{R^{2}(R+r)^{3}+r^{2}(R+r)^{3}-R^{2} r^{3}-R^{3} r^{2}}{(R+r)^{3}}=\frac{m \pi}{3}\left(R^{2}+r^{2}-\frac{R^{2} r^{2}(R+r)}{(R+r)^{3}}\right)
\end{aligned}
$$

We will calculate the unknown $m$ using the known data.

$$
\begin{gathered}
k=K-\left(K_{1}+K_{2}\right)=\frac{m \pi}{3}\left(R^{2}+R r+r^{2}\right)-\left(K_{1}+K_{2}\right)= \\
=\frac{m \pi}{3}\left(R^{2}+R r+r^{2}-R^{2}-r^{2}+\frac{R^{2} r^{2}}{(R+r)^{2}}\right)=\frac{m \pi}{3} \frac{R^{2}+3 R r+r^{2}}{(R+r)^{2}} R r
\end{gathered}
$$

From this,

$$
m=\frac{3 k(R+r)^{2}}{\pi R r\left(R^{2}+3 R r+r^{2}\right)}
$$

and thus

$$
K=\frac{m \pi}{3}\left(R^{2}+R r+r^{2}\right)=\frac{k(R+r)^{2}\left(R^{2}+R r+r^{2}\right)}{R r\left(R^{2}+3 R r+r^{2}\right)} .
$$

The lateral surface area $F$ of the frustum is

$$
F=\pi(R+r) \sqrt{(R-r)^{2}+m^{2}}=\frac{k(R+r) \sqrt{\left(R^{2}+3 R r+r^{2}\right)^{2} \pi^{2} R^{2} r^{2}(R-r)^{2}+9 k^{2}(R+r)^{4}}}{R r\left(R^{2}+3 R r+r^{2}\right)}
$$

With this, the problem is solved.