Emptying a tank from the top takes work, and the water at the bottom costs more than the water at the top. Slice the water into slabs, add up what each one costs, and the sum turns into an integral. Scroll down: at each stop a short note explains the next idea, then the picture is yours to play with before the next one arrives.
Everything on this page follows from the assumptions and definitions below. They are granted, not proved, so judge each one for yourself. A step that leans on one is tagged with its number.
Lift something with a steady force through a distance and the work you do is force × distance. A newton of force through a meter is one joule.
The picture is a side view of a tank full of water: a cylinder 3 m tall and 1.5 m in radius. A pump has to lift all of it up to the outlet at the top. The trouble is that the water does not all travel the same distance. Water near the top barely moves. Water at the bottom has to climb the whole 3 m.
So cut the water into horizontal slabs. Every drop in one thin slab rises about the same distance, and for one slab force × distance works. One slab is green, and the arrow shows how far it rises.
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Try itDrag the green slab, or tap another one. Which slab costs the most to lift out? Which costs the least?
The panel on the right works out one slab in four lines.
In this tank every slab has the same radius, so every slab weighs the same: about 34.6 kilonewtons. Only the distance changes.
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Try itThe bottom slab is green now. Read its work, then drag to the top slab and read that. The bottom one costs exactly 11 times as much. Where does the 11 come from?
The work to empty the tank is the work for each slab, added together. The number at the top right is now that total.
A chart has appeared below it, with one bar for each slab. The height of a bar is the slab's work for each meter of its thickness, and the width is its thickness, so the area of a bar is the work for that slab. The total work is the area of all the bars together.
A new slider changes how many slabs the water is cut into.
Try itCut the water into more slabs, then fewer, down to one. The total does not move. Look at the tops of the bars: what shape do they follow, and why might one fat slab be enough here?
Let the slabs get thinner without end and the sum becomes an integral. A slab at height y weighs 9800 · π(1.5)² newtons for each meter of thickness, and rises 3 − y:
W = ∫03 9800 · π(1.5)² · (3 − y) dy
The limits are where the water is: from the bottom, y = 0, to the surface, y = 3. It works out to 99,225π joules, about 311.7 kilojoules.
The chart now shows the curve the integral adds up. Here it is a straight line, which is why the bars fit it exactly: a slab measured at its middle gets a straight line right, however thick the slab is.
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Try itFollow the six lines of the integral on the right. Then compare its answer with the slab total in the box above it, for any number of slabs.
A new slider lowers the water. The outlet stays at the top of the tank, so the only thing that changes in the integral is its upper limit: the slabs now run from the bottom up to the new surface.
Take out the top half of the water and you have removed the cheap half. What is left is the water with the farthest to go.
Try itSet the depth to 1.5 m. What fraction of the full-tank work is left? Then find the depth where the work is about half of the full tank's.
A menu has appeared, and the tank is now a cone with its point down: 3 m deep and 1.5 m in radius at the top. The slabs are no longer the same size. By similar triangles a slab at height y has radius
r = y/2
The picture draws each slab as the disk the sum treats it as, with the radius the tank has at the slab's middle. With only two slabs, those disks are a poor copy of a cone, and the total is off by an eighth.
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Try itRaise the number of slabs from 2 to 3, 6, 12 and on to 60. Watch the disks settle into the cone, the bars settle onto the curve, and the percentage in the top box settle at 100.
The recipe is the same. Only the radius changed, and it now depends on y:
W = ∫03 9800 · π(y/2)² · (3 − y) dy ≈ 52.0 kJ
Compare that with the cylinder. The cone holds one third as much water, and it takes only one sixth of the work. A cone with its point down keeps most of its water near the top, where the lift is short.
Try itDrag the green slab up the cone and watch the chart. The widest slab is at the top and the longest lift is at the bottom. Where is the slab that costs the most?
A new slider lifts the outlet above the top of the tank, as if the water had to reach a higher pipe. If the outlet is s meters above the top, every slab rises 3 + s − y, which is s more than before.
Every newton of water travels that extra distance, so the extra work is simply the weight of all the water times s.
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Try itWith the cone full, raise the outlet by 1 m. The work more than doubles. Do the same with the cylinder. Which tank's work grows by the bigger factor, and why?
The last tank on the menu is a sphere of radius 1.5 m. A slab at height y is a horizontal slice of a circle, and the Pythagorean theorem gives its radius:
r² = 1.5² − (y − 1.5)² = 3y − y²
The integrand is a polynomial again, so the integral is no harder than before. Pumping the full sphere out at the top takes about 207.8 kJ.
Try itThe widest slab is in the middle, at y = 1.5 m. Drag the green slab and find the one that costs the most. It is not the widest one. Where is it, and why is it below the middle?
A good habit after solving a problem is to get the answer again by a different route. Here is one. The total work equals the weight of all the water times the distance its center of mass rises.
For the full cylinder the center of mass is halfway up, at 1.5 m, so it rises 1.5 m. That is why one fat slab measured at its middle got the cylinder exactly right back at stop 3.
A blue box on the right now shows this second calculation beside the integral. For a full cone the center of mass is three quarters of the way up from the point, and for a full sphere it is at the center.
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Try itCheck that the two answers agree for each tank. Then lower the water in the cone. Is the center of mass still three quarters of the way up the water?
Every tank went the same way:
The number you get is the work done on the water against gravity, and nothing else. The story leaves out friction in the pipe, the speed the water leaves with, and whatever the pump itself wastes. Those come on top.
Everything is unlocked now. Play as long as you like, then scroll on to finish.
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Try itPredict before you look: rank the three full tanks by the work it takes for each cubic meter of water. Two of them tie. Use the blue box to see why.
| Claim | Status | Rests on | Detail |
|---|---|---|---|
| One 0.5 m slab of the cylinder weighs about 34.6 kN. The bottom slab costs exactly 11 times the top one (stops 1 and 2). | ✓ Proved | A1, A2, A3, A4, D1 | A 0.5 m slab of the cylinder holds 3.534 m^3 and weighs 34,636 N; the bottom slab takes 95.2 kJ and the top one 8.7 kJ, a ratio of 2.75/0.25 = 11. |
| Full cylinder: W = 99,225π ≈ 311.7 kJ. Half full leaves three quarters of that, and the work is half at a depth of about 0.88 m (stops 4 and 5). | ✓ Proved | A1, A2, A3, A4, D2 | Full cylinder: W = 99,225 pi = 311,725 J. Half full leaves 3/4 of that (233.8 kJ); the work is half at a depth of 3 - 3 sqrt(2)/2 = 0.88 m. |
| Slabs measured at their middles give the cylinder’s integral exactly, however many there are. Two slabs overshoot the cone by an eighth, and the sum closes in as the slabs thin (stops 3, 4 and 6). | ✓ Proved | A3, D2 | For the cylinder the midpoint slabs equal the integral for every number of slabs and every depth (the integrand is a straight line). For the cone, 2 slabs give 9/8 of the integral, and 60 slabs are within 0.014%. |
| Cone: r = y/2 and W ≈ 52.0 kJ. It holds a third of the cylinder’s water and takes a sixth of its work (stops 6 and 7). | ✓ Proved | A1, A2, A3, A4, D2 | Cone: r = y/2 and W = 9800 pi (27/16) = 51,954 J. It holds 1/3 of the cylinder's water and takes 1/6 of its work; the costliest thin slab is at y = 2 m. |
| Raising the outlet by s adds the weight of all the water times s. One meter more than doubles the cone’s work (stop 8). | ✓ Proved | A4, D2 | Raising the outlet s meters adds exactly (weight of the water) x s, for every tank and depth. With s = 1 the full cone goes from 52.0 to 121.2 kJ (x2.33) and the full cylinder from 311.7 to 519.5 kJ (x1.67). |
| Sphere: r² = 3y − y² and W ≈ 207.8 kJ. The costliest slab is at y = 1 m, below the widest one (stop 9). | ✓ Proved | A1, A2, A3, A4, D2 | Sphere: r^2 = 3y - y^2 and W = 66,150 pi = 207,816 J. The slab is widest at y = 1.5 m, but r^2 (3 - y) is largest at y = 1 m. |
| The integral equals the water’s whole weight times the rise of its center of mass, for every tank, depth and outlet. A full cone’s center is three quarters of the way up (stop 10). | ✓ Proved | D2, D3 | For every tank, depth and outlet the integral equals (weight of the water) x (rise of its center of mass). Full tanks: center at 1.5 m (cylinder), 1.5 m (sphere), 2.25 m (cone); in the cone it is 3/4 of the way up the water at every depth. Work per cubic meter, full: cyl 14.7 kJ, cone 7.3 kJ, sphere 14.7 kJ. |
| The numbers the tool shows as you drag. | ✓ Checked independently | — | The page's own functions, run at 180 settings across the three tanks, match SymPy's integrals, slab sums, weights and center-of-mass totals to 5.0e-16 (relative); the on-screen 2000-slab check is within 1.3e-07. |
| Units: newtons for weight, joules for work, shown as kN and kJ. | Reviewed | A1, D1 | N/m³ times m³ is N, and N times m is J. |
| The sentences on the cards. | Reviewed | — | Each card was read against the formulas, looking for anything assumed but not listed. No objection stands. |
| A1, A2, A3, A4, D1, D2, D3 | Assumed | — | Granted, not proved. Whether they suit a real tank and pump is yours to judge. |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: the page’s own functions were run against SymPy’s exact answers. Checked numerically: evidence, not proof. Reviewed: a model or a person read it. Assumed: an assumption or a definition, granted and not proved. Checks run 2026-11-11 with SymPy 1.14.0; rerun them with interactives/checks/pumping_work.py. Cards read by Claude Fable 5.2, 2026-11-11. How it’s checked
Part of Calc Practice: free Calculus 2 practice. Try the work: springs, pumping and cables problems or see more interactives.