Brian Bi

## Hydrology: For a given column of water, is the water pressure at the center of the base greater in a vessel with a right conical bottom versus a flat bottom?

This problem is known as the "hydrostatic paradox". The problem is that if the pressure at a point inside a fluid depends only on its depth, then how can a conical flask filled with water weigh less than a beaker with the same base diameter, filled with water to the same height? Clearly the former must weigh less, since it contains less water, and yet both have the same pressure at the base and the same base area.

If you search for "hydrostatic paradox" on Google, a lot of websites will tell you that the pressure really does only depend on the height of the column of water above, but I couldn't find any site that actually explains the resolution of the paradox. So, here it is.

If you have a conical flask and a beaker with the same base surface area and filled with water to the same height, of of course the latter's weight will be greater. It's tempting to attribute the entire weight of the container to the force (pressure times area) exerted by the water on the inside base, which gives rise to the paradox. But that's wrong!

Why? Because consider the sides of the conical flask. Because of the way they are slanted, the net force exerted by the liquid on the side has an upward vertical component. There are two downward forces keeping the glass on the side in place: gravity, and internal forces within the glass. As we increase the slant of the conical flask, the internal forces within the glass have to provide more and more of the restraining force at any given depth, since the vertical component of the hydrostatic pressure increases while the gravitational force stays the same. Ultimately, the internal forces transmit the upward pressure of the water on the sides of the conical flask to the base, which consequently feels lighter than the base of the beaker.

Now, you might be skeptical. Everything I said should sound reasonable: the water pressure on the walls does indeed have an upward component when the walls are slanted as in a conical flask. But does the total upward force exerted by the water on the walls of the conical flask really exactly equal $$\rho g$$ times the difference in volume between the conical flask and the beaker? How can it be that by just integrating over the walls of the container, we can get a result that's exactly consistent with the container's volume? Is this really the correct resolution to the paradox? And that suggests that we need to apply the divergence theorem somehow, because the divergence theorem relates integrals over the surface of a region to integrals over the region itself. And I'll do this calculation now, to convince you that it really does work out.

Suppose we have a container filled with water (whose density I will denote by $$\rho$$). We'll use a coordinate system in which the top surface of the water (exposed to air) is at height $$z = 0$$, with $$z$$ increasing toward the bottom. Let the shape of the container be described by an oriented surface $$\partial V$$ which denotes the boundary of the region $$V$$ occupied by the water. Consider an infinitesimal element $$\mathrm{d}a$$ of the container walls located at depth $$z$$, where the water pressure is $$\rho gz$$. Then the downward component of the force exerted by the water is $\mathrm{d}F_z = \rho gz n_z \, \mathrm{d}a$ where $$\hat{\mathbf{n}}$$ is the outward unit normal, and $$n_z = \hat{\mathbf{n}} \cdot \hat{\mathbf{z}}$$. (You can sanity-check this by considering the flat base of the container: here $$\hat{\mathbf{n}} = (0, 0, 1)$$ (remember, $$z$$ increases downward) and so the force on an infinitesimal element of the base is simply $$\mathrm{d}F_z = \rho g z \, \mathrm{d}a$$ as you would expect.) Now let's integrate over the entire surface, \begin{align*} F_z &= \int_{\partial V} \rho g z n_z \, \mathrm{d}a \\ &= \int_{\partial V} \rho g z \hat{\mathbf{z}} \cdot \mathrm{d}\mathbf{a} \end{align*} Applying the divergence theorem, we then obtain: \begin{align*} F_z &= \int_V \rho g \nabla \cdot (z\hat{\mathbf{z}}) \, \mathrm{d}^3x \\ &= \int_V \rho g \, \mathrm{d}^3 x \end{align*} which, of course, is precisely the total weight of the fluid. So it is indeed the case that the contact force of the pressure of the fluid upon the walls and base of the container precisely communicates the total weight of the fluid, which resolves the paradox.