How much water can be poured into a straight cylindrical, liter-sized container that has been tilted from a vertical position by $37^{\circ} 42^{\prime}$, if the height of the container is $15.3 \mathrm{~cm}$?
The diagram (from which the lettering can be read) shows the cross-section of the water in a cylindrical container rotated by an angle $\alpha$ from the vertical position, when $\alpha$ is not so large that the water level intersects the base circle.

![](https://cdn.mathpix.com/cropped/2024_05_02_2b94296b385bb05840dcg-1.jpg?height=718&width=724&top_left_y=297&top_left_x=690)

Let $O_{0}$ be the intersection point of the water level with the axis of the cylinder. The water body above the plane passing through $O_{0}$ and parallel to the base of the cylinder is centrally symmetric to the water-free body below the plane, bounded by the cutting plane, the cylinder wall, and the water level. Therefore, the volumes of these two parts are equal, and thus the volume of the water in the tilted cylinder is equal to the volume of a cylinder whose base is the base circle of the given cylinder and whose height is: $O O_{0}=m_{0}$.

Let the sought volume of water be $k$ liters, then

$$
k: 1=m_{0}: m
$$

or

$$
\begin{gathered}
k=\frac{m_{0}}{m} \\
m_{0}=m-x=m-r \operatorname{tg} \alpha
\end{gathered}
$$

where $x=r \operatorname{tg} \alpha<\frac{m}{2}$.

Since (measuring in cm)

$$
r^{2} \pi m=1000
$$

therefore

$$
r=\sqrt{\frac{1000}{m \pi}}
$$

and thus

$$
m_{0}=m-\operatorname{tg} \alpha \sqrt{\frac{1000}{m \pi}}=m-\sqrt{\frac{1000 \operatorname{tg}^{2} \alpha}{m \pi}}=m-x
$$

In this case, $\alpha=37^{\circ} 42^{\prime}$ and $m=15.3 \mathrm{~cm}$. Therefore, it is necessary that $x<\frac{m}{2}=7.65$.

$$
\begin{aligned}
& 2 \lg \operatorname{tg} 37^{\circ} 42^{\prime}=19.7762-20 \\
&+\lg \quad 1000=3.0000 \\
& \hline 2.7762 \\
&-\left\{\begin{aligned}
\lg 15.3 & =1.1847 \\
\lg \pi & =0.4971
\end{aligned}\right. \\
& \hline 2 \lg x=1.0944 \\
& \lg x=0.5472 \\
& x=3.525<7.65 .
\end{aligned}
$$

Thus, $m_{0}=15.3-3.525=11.775$,

and therefore

$$
k=\frac{m_{0}}{m}=\frac{11.725}{15.3}=0.76971=769.7 \mathrm{~cm}^{3}
$$