# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_02_50a7400546e28c93f414g-1.jpg?height=309&width=526&top_left_y=233&top_left_x=794)

We can assume that the soup bowl (i.e., a hemisphere) has its flat base on top. The volume of the soup is equal to the volume of a spherical cap. Since the height of the hemisphere is equal to the radius of the sphere, the height of the cap is \( m = \frac{r}{2} \). The volume of the entire sphere is:

\[
V_{1} = \frac{4 \pi}{3} r^{3}
\]

If the volume of the hemisphere is 8 liters, then the volume of the sphere is 16 liters, so: \(\frac{4 \pi}{3} r^{3} = 16\), from which \(\pi r^{3} = 12\).

The volume of a spherical cap with height \( m \) is:

\[
V_{2} = \frac{\pi}{3} m^{2} (3 r - m)
\]

Substituting \( m = \frac{r}{2} \) into the formula:

\[
V_{2} = \frac{\pi}{3} \left( \frac{r}{2} \right)^{2} \left( 3 r - \frac{r}{2} \right) = \frac{5 \pi r^{3}}{24}
\]

Substituting \(\pi r^{3} = 12\) into the formula: \( V_{2} = \frac{12 \cdot 5}{24} = 2.5 \).

Therefore, 2.5 liters of soup fill the bowl to half its height.
A bowl-shaped pot with a hemispherical shape has a volume of 8 liters. How much soup fills the pot to half its height?