Water Level in a Spherical Tank

Why the Water Level Does Not Rise at a Constant Speed

A spherical tank has a changing horizontal cross-section: the tank is narrow near its bottom and top, and widest at its middle. As a result, the same volume of incoming water can produce different changes in water depth, depending on the current level.

To find the level’s rising speed, describe the water’s volume as a function of its depth. Differentiating that relationship gives the volume change per unit depth. The inflow rate can then be related to the rate of change of depth using the chain rule.

Depth and Spherical-Cap Volume

Let \(R\) be the fixed radius of the spherical tank, and let \(h\) be the water depth measured vertically upward from the tank’s bottom. The water occupies a spherical cap. Its depth satisfies \(0\le h\le 2R\), because the sphere’s diameter is \(2R\).

Here is a derivation. Place the sphere so that its bottom is at vertical coordinate \(z=0\) and its center is at \(z=R\). At height \(z\), let \(r\) be the radius of the horizontal cross-section. The right triangle formed by \(r\), \(z-R\), and \(R\) gives

Therefore, the cross-section has area

The cap is made up of these circular cross-sections from the bottom, \(z=0\), to the water level, \(z=h\). Integrating their areas gives

This calculation also explains why using the total sphere volume alone does not describe the water at a general depth: the water occupies only the cap, not the whole sphere.

Relating Inflow to the Rising Speed

Differentiate the cap volume with respect to depth. Since \(R\) is constant,

At a particular height, this derivative is the area of the water’s horizontal surface. That makes geometric sense: raising the surface by a small amount adds a thin layer whose volume is approximately surface area times thickness.

Now suppose water flows into the tank at rate \(Q=\frac{dV}{dt}\). Because the volume depends on time through the changing depth \(h(t)\), the chain rule gives \(\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}\). Thus

Units provide a useful check. If \(R\) and \(h\) are measured in metres, then \(\pi h(2R-h)\) has units of square metres. Dividing an inflow rate in cubic metres per minute by this area gives metres per minute, the units of \(\frac{dh}{dt}\).

Worked Examples

Why the Rate Changes with Depth

The factor \(\pi h(2R-h)\) is largest at the tank’s middle, where \(h=R\). Indeed, its derivative with respect to \(h\) is \(2\pi(R-h)\), which is positive for \(h<R\) and negative for \(h>R\). So the surface area grows as the water approaches the middle, then shrinks as the level approaches the top.

For a fixed positive inflow \(Q\), a larger surface area means a smaller rise in depth per unit time, because the incoming water spreads across more area. A smaller surface area means the same volume input produces a faster rise in level.

Common Mistake: Differentiating the Whole-Sphere Volume

A tempting but incorrect calculation is to use the volume of the entire sphere, \(\frac{4}{3}\pi R^3\), and then differentiate it with respect to the water depth. For a fixed tank, \(R\) is constant, so this gives \(\frac{d}{dh}\left(\frac{4}{3}\pi R^3\right)=0\). That cannot describe the change in water volume as the level rises.

The whole-sphere formula applies when calculating the volume of the entire tank. At a partial fill, the relevant volume is the cap volume, which depends on \(h\). Using that formula gives \(\frac{dV}{dh}=\pi h(2R-h)\), the nonzero area of the water’s surface at an interior depth. The correction is not to vary the tank’s radius; it is to model the portion of the tank occupied by water.