A cylindrical glass stands on a horizontal table, the water level inside is at a height of $m_{1}$ above the table. By tilting the glass, the plane of its base forms an angle with the table, the water does not spill out, and the new water level height is $m_{2}$. How much water is in the glass? How does the result change if we take into account that the thickness of the glass at the bottom is $d_{1}$ and on the side wall is $d_{2}$? Numerical data: $m_{1}=40 \mathrm{~mm}, a=25^{\circ}, m_{2}=48 \mathrm{~mm}, d_{1}=7 \mathrm{~mm}, d_{2}=3 \mathrm{~mm}$.
We can only answer with the assumption that the water will still cover the rim of the glass after tilting. The radius $r$ of the glass is calculated based on the fact that the center of the water surface does not change relative to the glass during the tilting, and the center of the elliptical surface after tilting coincides with the center of the original circular surface. To understand this, consider that the glass is initially half-filled with water and is sealed at the top like a lid. Thus, before tilting, the glass, water, and air form a centrally symmetric system where the water-air boundary passes through the center. Since the volumes do not change with tilting and the boundary remains horizontal, it must pass through the center of symmetry.

![](https://cdn.mathpix.com/cropped/2024_05_02_2cc5108db6c6679c0776g-1.jpg?height=702&width=696&top_left_y=454&top_left_x=690)

Now, according to the proven statement, $m_{2}=r \sin \alpha + m_{1} \cos \alpha$, hence $r = (m_{2} - m_{1} \cos \alpha) / \sin \alpha$, and thus the volume of the water is $V = \pi m_{1} (m_{2} - m_{1} \cos \alpha)^{2} / \sin^{2} \alpha$. This formula remains valid in the modified case if we interpret the heights from the deepest point of the water. The height of this point above the table is $M_{1} = d_{1}$, and $M_{2} = d_{1} \cos \alpha + d_{2} \sin \alpha$. Therefore, $m_{1}$ and $m_{2}$ are replaced by $m_{1}' = m_{1} - d_{1}$ and $m_{2}' = m_{2} - d_{1} \cos \alpha - d_{2} \sin \alpha$, respectively. The difference $m_{2} - m_{1} \cos \alpha$ in the expression for the radius becomes $m_{2}' - m_{1}' \cos \alpha = m_{2} - m_{1} \cos \alpha - d_{2} \sin \alpha$, so $r' = (m_{2} - m_{1} \cos \alpha - d_{2} \sin \alpha) / \sin \alpha = r - d_{2}$, which is also expected. Thus, the volume is $V' = \pi (m_{1} - d_{1}) (m_{2} - m_{1} \cos \alpha - d_{2} \sin \alpha)^{2} / \sin^{2} \alpha$.

In the numerical example, $r = 27.8 \, \text{mm}$ and $V = 97.1 \, \text{cm}^{3}$, while $r' = 24.8 \, \text{mm}$ and $V' = 63.7 \, \text{cm}^{3}$ (it is not worth writing more than three significant figures, and even this may be too many, as the data itself contains only two, and the modifying data only one significant figure).

Zenkl Edit (Budapest, Building Materials Technology III. o. t.)

Remarks. 1. Many solutions only calculated $V'$ with the numerical data, although general results should be applied wherever possible in numerical examples. Comparing $V$ and $V'$ is intended to show how a small approach to reality modifies the results obtained with simplifications (e.g., $V' \approx 2V / 3$), and also to demonstrate that (with appropriate simplifications) taking into account the modifying conditions is not difficult.
2. The expressions for $M_{2}$ and $m_{2}$ essentially represent the rotation of the coordinate system.