Task B-3.5. A barrel has the shape of a right cylinder with its axis in a horizontal position, and the radius of the base is $2 \mathrm{dm}$, and the height is $6 \mathrm{dm}$. The barrel is submerged in water to half the radius of the base. Calculate the volume of the submerged part of the barrel.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_cf2c4989ba380bc8684dg-14.jpg?height=664&width=680&top_left_y=1804&top_left_x=668)

The angle $\angle A S C=60^{\circ}$, so the angle $\alpha=\angle A S B=120^{\circ}$.

We calculate the volume of the part of the cylinder whose base is a circular segment with a central angle $\alpha=120^{\circ}$ and height $v$, using the formula

$$
V=B \cdot v=\left(\frac{r^{2} \pi \alpha}{360^{\circ}}-\frac{1}{2} r^{2} \sin \alpha\right) v=\left(\frac{r^{2} \pi}{3}-\frac{1}{2} r^{2} \frac{\sqrt{3}}{2}\right) v=r^{2} v\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)=8 \pi-6 \sqrt{3} \mathrm{~cm}^{3}
$$

Note: The area of triangle $A B S$ can also be calculated as the sum of the areas of two triangles $A C S$ using trigonometry or without it.