Does the water surface rise proportionally to the volume of an object sunk? I have a big bucket which has the constant cross section, like a cylinder.
Now I fill the bucket with water, but not fully so that the surface level can still increase, and I put an object in it. The object may have no symmetrical shape. The surface level will rise up as I put more part of the object.
Here is my question: is the increment of the surface height proportional to the volume of the sunk part of the object? That is, if I raise the surface level by an amount h, is the sunk-volume increment independent to the shape of the object?
I guess the volume increment is equal to Sh, where S is the cross section of the bucket, but can't feel certain of it...
 A: The volume of the object immersed in water will have the same volume as the water that rises above the original height so that:
$$V_{object} = V_{displaced}$$
If the water container has constant cross section in height the volume of displaced water is equal to the product of displaced height and the cross section area. So the answer is that the displacement height depends proportionally on the volume of the sunken object (and the shape of the object is irrelevant) as:
$$h_{displaced} = \frac{V_{object}}{A_{cylider}}$$
A: Yes. And the object displaces a volume of water equal to his own volume once submerged (it's the space it needs), what means that you can also use your equation to find out the object's volume.
A: As others have answered the rise in height of water level is indeed proportional to submerged volume of the object. What follows is a rigorous justification.
Let $A$ be the cross-section area of the container. Take Z-axis perpendicular to water surface, and fix the origin at water surface before immersion of body in it. A body of arbitrary shape is now immersed. Let the body's lowermost point be at $z=-h$, and let the new water level be at $z=\delta h$. Let $A_b(z)$ be the cross-section area of the submerged body at given coordinate $z$. Now mass balance gives:
\begin{align}
\textrm{Volume of body submerged beneath the initial water level}& =\textrm{Volume of water that has risen above the initial water level}\\
\int_{- h}^0dz~A_b(z) &= A~\delta h-\int_0^{\delta h}A_b(z)\\
\int_{- h}^0dz~A_b(z)+\int_0^{\delta h}A_b(z) &= A~\delta h\\
\int_{- h}^{\delta h}dz~A_b(z)&= A~\delta h\\
\end{align}
The quantity on the LHS is nothing but the total submerged volume of the body viz. the volume of the body beneath the water level after it has risen. This is indeed proportional to $\delta h$.
