Case Study 1 — Draining a Spherical Water Tower

Field: Civil and environmental engineering (municipal water systems) Calculus used: Work integral (§18.7), with a sanity check via center of mass (§18.9)

The problem an engineer actually faces

Drive into almost any American town and you will see it: a fat steel sphere on stilts, the town's name painted across it, glinting above the rooftops. That is a spheroidal water tower, and it is not decoration. The sphere sits high so that gravity, not a continuously running pump, supplies household water pressure. Every morning, while demand is low and electricity is cheap, pumps fill the sphere; through the day, the stored water drains down to faucets and fire hydrants under its own weight.

Once a year, though, the tank must be emptied for inspection and cleaning — and here the calculus arrives. The maintenance crew cannot simply open a valve, because the inspection port and overflow spout sit at the top of the sphere; the residual water at the bottom has to be pumped up and out over that rim. The utility's engineer is asked a concrete question: how much energy will that drain-and-clean cycle cost in electricity? Underneath the dollars-and-cents framing is a pure §18.7 work integral, and getting the setup right is the entire job.

We will work it from the picture, exactly as the chapter insists: draw the slice, write its contribution, integrate. Then we will check the answer two independent ways, because an engineer who trusts a single unverified integral does not stay employed.

Setting up the slice

Idealize the tank as a sphere of radius $R$. Let the bottom of the sphere rest at height $0$, so its center is at height $R$ and its top — where the water exits — is at height $2R$. Let the variable $y$ measure height upward from the bottom, so $0 \le y \le 2R$.

The water does not all travel the same distance to the exit, so the constant-force law $W = F \cdot d$ fails outright. That failure is the signal to slice. Cut a thin horizontal slab of water at height $y$, thickness $dy$. Two facts about that slab decide everything:

Its size. The slab is a flat circular disk. Its radius $x$ is the half-width of the sphere at height $y$. The sphere's equation, with center at height $R$, is $x^2 + (y - R)^2 = R^2$, so

$$x^2 = R^2 - (y - R)^2 = 2Ry - y^2.$$

The slab's volume is therefore $dV = \pi x^2\,dy = \pi(2Ry - y^2)\,dy$, and — multiplying by water's weight density $\rho g$ — its weight is $\rho g\,\pi(2Ry - y^2)\,dy$.

Its lift distance. This is where students go wrong, and the §18.7 pitfall warns of it explicitly. The slab starts at height $y$ but must rise to the exit at height $2R$. It therefore travels a distance $2R - y$, not $y$. (Define the zero point, then write "exit height minus slab height.") A slab near the bottom, $y \approx 0$, must climb nearly the full $2R$; a slab near the top, $y \approx 2R$, barely moves. That asymmetry is exactly what an integral is for.

The work to hoist this one slab is its weight times its lift distance:

$$dW = \underbrace{\rho g\,\pi(2Ry - y^2)\,dy}_{\text{weight of slab}} \cdot \underbrace{(2R - y)}_{\text{lift distance}}.$$

Carrying out the integral

Sum over all slabs from $y = 0$ to $y = 2R$:

$$W = \rho g\,\pi \int_0^{2R} (2Ry - y^2)(2R - y)\,dy.$$

Expand the integrand carefully — this is the algebra step, mechanical once the setup is right:

$$(2Ry - y^2)(2R - y) = 4R^2 y - 2Ry^2 - 2Ry^2 + y^3 = 4R^2 y - 4R y^2 + y^3.$$

Integrate term by term:

$$\int_0^{2R}\!\big(4R^2 y - 4R y^2 + y^3\big)\,dy = \left[2R^2 y^2 - \tfrac{4R}{3}y^3 + \tfrac{1}{4}y^4\right]_0^{2R}.$$

Evaluate at $y = 2R$:

$$2R^2(2R)^2 - \tfrac{4R}{3}(2R)^3 + \tfrac14(2R)^4 = 8R^4 - \tfrac{32R^4}{3} + 4R^4 = 12R^4 - \tfrac{32R^4}{3} = \tfrac{4R^4}{3}.$$

Therefore

$$\boxed{\,W = \tfrac{4}{3}\pi\rho g R^4.\,}$$

A clean closed form — and notice it scales as $R^4$, not $R^3$. Doubling a tank's radius octuples its volume of water but multiplies the pumping work by sixteen, because the extra water is both more numerous and lifted farther. That fourth-power scaling is a genuine engineering insight, and it argues against very large single spheres.

Putting numbers on it

A typical small-municipality spheroid holds water in a sphere of radius $R = 5\ \text{m}$ (roughly $524\ \text{m}^3$, about $138{,}000$ gallons). With water's $\rho = 1000\ \text{kg/m}^3$ and $g = 9.81\ \text{m/s}^2$:

$$W = \tfrac{4}{3}\pi (1000)(9.81)(5)^4 = \tfrac{4}{3}\pi (1000)(9.81)(625) \approx 2.57\times 10^7\ \text{J} \approx 25.7\ \text{MJ}.$$

Converting to the units a utility bills in, $25.7\ \text{MJ} \div (3.6\times10^6\ \text{J/kWh}) \approx 7.1\ \text{kWh}$. At a commercial rate near $\$0.12$/kWh, the *theoretical minimum* electricity to lift the water is under a dollar. Real pumps are perhaps $60\%$ efficient, so call it $12\ \text{kWh}$ and a couple of dollars — trivial against the labor and downtime of the inspection itself. The integral's real payoff is not the dollar figure; it is confirming that pumping energy is a rounding error, which tells the utility to optimize scheduling and labor, not pump power. Good modeling answers the question behind the question.

Verifying the answer two ways

Check 1 — the center-of-mass shortcut (§18.9). The full sphere of water has total weight $\rho g \cdot \tfrac{4}{3}\pi R^3$. By symmetry its center of mass sits at the sphere's center, height $R$, which must rise to the exit at $2R$ — a lift of exactly $R$. If all the mass moved together that distance, the work would be

$$W = (\text{total weight})\times(\text{centroid lift}) = \rho g\,\tfrac{4}{3}\pi R^3 \cdot R = \tfrac{4}{3}\pi\rho g R^4.$$

It matches the slab integral exactly. This is the same shortcut that validated the hanging-cable computation in Worked Example 18.7.3, and it is no coincidence: lifting any rigid body's worth of mass costs (total weight) × (distance its center of mass rises). The integral derives that fact; the centroid argument confirms it.

Check 2 — numerical integration (theme 4). Hand calculus gives the exact answer; the machine confirms it. The code below evaluates the slab integral numerically and compares it to the closed form, never touching the antiderivative — the same theme-4 handshake from §18.10. (Values shown are computed by hand; the listing is for the reader to run, not for us to execute here.)

import numpy as np
from scipy.integrate import quad

rho, g, R = 1000.0, 9.81, 5.0
# Slab at height y: volume pi*(2Ry - y^2) dy, lift distance (2R - y)
integrand = lambda y: rho * g * np.pi * (2*R*y - y**2) * (2*R - y)
W_num, _ = quad(integrand, 0, 2*R)
W_exact = (4/3) * np.pi * rho * g * R**4
print(f"numerical : {W_num:.3e} J")   # 2.568e+07
print(f"closed form: {W_exact:.3e} J")  # 2.568e+07  -> agree

Three independent routes — the slice integral, the centroid shortcut, and the numerical quadrature — all land on $\tfrac{4}{3}\pi\rho g R^4 \approx 25.7\ \text{MJ}$. When three methods agree, you can sign the report.

Why the setup, not the integral, was the work

Step back and notice where the difficulty actually lived. The integration $\int(4R^2 y - 4Ry^2 + y^3)\,dy$ is freshman algebra. The hard, error-prone steps were geometric: writing the slab radius from the sphere's shifted-center equation, and — above all — recognizing that the lift distance is $2R - y$ and not $y$. Reverse those two and you compute a confident, precise, wrong number. That is the chapter's thesis in one example: master the picture and the calculus follows.

Discussion Questions

  1. We measured $y$ from the bottom of the sphere. Re-derive $W$ with $y$ measured downward from the top (the exit). The slab radius and lift distance both change form — show the final answer is identical, and explain why it must be.

  2. The pumping work scales as $R^4$ while the stored water scales as $R^3$. A town doubling its population could build one tank of radius $2R$ or eight tanks of radius $R$. Compare the total pumping energy of the two designs. Which is cheaper to drain, and why?

  3. Suppose the tank is only half full (water filling the bottom hemisphere, $0 \le y \le R$). Set up the new work integral. Is the answer more or less than half of $W$? Reason from the lift distances before computing.

  4. Real water towers are spheroidal (slightly flattened) or sit atop a narrow standpipe that also holds water. How would you modify the slab radius $x(y)$ to include the standpipe? Which §18.3 idea lets you handle a piecewise profile?

  5. The centroid shortcut (Check 1) made the whole integral unnecessary. When is the shortcut available, and when is it not? (Hint: it gives total work but hides the per-slab structure — when would an engineer need the structure anyway?)

A short annotated reading

  • Stewart, Calculus: Early Transcendentals (9th ed.), §6.4 "Work." The pumping-problem template, worked for cylindrical and conical tanks. Our spherical case is the natural next step; the lift-distance pitfall is Stewart's central caution too.
  • Çengel & Cimbala, Fluid Mechanics: Fundamentals and Applications, ch. 3. Treats hydrostatics and pumping energy from the engineering side, including pump efficiency — the factor that turns our $7.1\ \text{kWh}$ ideal into a real $12\ \text{kWh}$ bill.
  • AWWA (American Water Works Association), Steel Water-Storage Tanks (Manual M42). The industry standard for tank geometry, inspection, and drainage practice — the real-world document behind this case study's framing.

A water tower is a definite integral made of steel. The slab, the lift distance, the sum — set them up correctly and the energy to drain a town's water falls out in one line.