Equal heights and matching slices mean equal volume — the rest follows.
A box is easy: stack length by width rows, pile them height high, and you have counted base area × height little cubes. But what about a cone, a pyramid, a tilted stack, a sphere? Instead of memorizing four unrelated formulas, we lean on one quietly powerful idea: slice every solid into paper-thin layers. If two solids have the same height and their slices match at every level, they must hold the same amount — this is Cavalieri's principle. From it, V = Sh, then V = ⅓Sh, then V = 4/3 πr³ all fall out, each one reasoned, none guessed.
Volume measures how much a solid holds — how many unit cubes it would take to fill it. Start where it is obvious. A rectangular box (a cuboid) of length a, width b, and height c is a tidy stack: one layer is an a × b grid of unit cubes — that is the base area S = a·b — and the box piles c such layers on top of each other. So
V = a · b · c = (base area) × (height) = S·h.
Read it as "area of one slice, times how many slices." That sentence — not the letters — is the whole lesson. A box's slices are all the same rectangle, stacked straight up. The moment we allow the slices to slide or shrink as we climb, we get tilted prisms, cones, and spheres, and we need one honest rule to keep the bookkeeping straight.
Volume = (area of a cross-section) × (height), when every cross-section is the same and stacked straight. The next four sections are all about what to do when the cross-sections change as you go up.
Picture a neat stack of identical coins — a little cylinder. Now nudge the stack so it leans, like a deck of cards pushed sideways: a sheared stack. Not one coin changed; not one was added or removed. The leaning stack and the upright stack hold exactly the same amount of metal. Why? Because at every height, a horizontal cut passes through one coin in each stack, and those two coin-cross-sections are congruent. Same height, matching slices all the way up ⇒ same volume. This is the principle, named for Bonaventura Cavalieri:
If two solids have the same height, and every horizontal plane cuts them in cross-sections of equal area, then the two solids have the same volume.
It is worth pausing on what is and isn't required. The slices must match at every single level — not just at the bottom. Equal bases alone is not enough: a cone and a cylinder share a base, but their slices disagree everywhere above it, and indeed their volumes differ. And the matching slices need not be the same shape as each other — only the same area. A pile of square cards and a pile of round coasters of equal area, same height, enclose the same volume. Let's make the shearing visible.
Cavalieri needs equal height AND matching cross-sections at every level — equal base area alone is not enough. Two solids that merely "look about the same size" can have very different volumes; you must check the slices all the way up.
Now Cavalieri pays its first dividend. Take any prism — straight or leaning (oblique) — or any cylinder. Stand a matching upright prism or cylinder of the same base S and the same height h right beside it. At every level, the oblique solid's slice is a copy of its base, slid sideways; the upright solid's slice is the same base, not slid. Same area, every level; same height. By Cavalieri their volumes are equal — and the upright one we already know is S·h. So both are:
V = S·h (any prism or cylinder; for a cylinder S = πr², so V = πr²h).
The leaning doesn't cost a thing. A tilted tower of bricks holds as much as the same bricks stacked plumb. For a cylinder of radius r the base is a circle of area πr², giving V = πr²h — a result you will recompute in the comparison widget below.
Here is the surprise the hero picture promised. Take a triangular prism. It can be cut into three pyramids, each with the same base area and the same height as the prism — three congruent-volume pieces filling one prism. (The classic proof slices the cube into three identical square pyramids; the same idea works for any prism.) Each pyramid is therefore one third of the prism. Run the slicing argument and the same factor appears for every pyramid and every cone, regardless of base shape:
V = ⅓ · S·h (any cone or pyramid; for a cone S = πr², so V = ⅓πr²h).
Why exactly one third — why not a half, which the eye might guess? Because a cone shrinks as it rises: at height t up a cone of total height h, the slice radius is r·(1 − t/h), so the slice area is πr²·(1 − t/h)² — it falls off like the square of the remaining height. Add up (integrate) those shrinking disks and the total is one third of the full cylinder πr²h, not one half. The widget below lets you stand a cone next to its cylinder and read the ratio.
A cone has radius 3 cm and height 4 cm. Then V = ⅓π·3²·4 = ⅓π·36 = 12π ≈ 37.7 cm³. The cylinder on the same base and height holds π·3²·4 = 36π cm³ — three times as much, as it must.
A cone (or pyramid) is ⅓ of the matching cylinder (or prism), not ½. The taper makes each slice shrink like the square of the remaining height, which is exactly why the factor is a third.
The boldest use of Cavalieri compares a hemisphere of radius r with a cylinder of radius r and height r that has a cone bored out of it (apex down, opening to radius r at the top). Slice both at height x above the base:
The two slice areas agree at every height x, and both solids have height r. By Cavalieri the hemisphere and the drilled cylinder have equal volume. But the drilled cylinder is easy: cylinder minus cone = πr²·r − ⅓πr²·r = ⅔πr³. So the hemisphere is ⅔πr³, and a whole sphere is twice that:
V = 4/3 · πr³.
A clean by-product drops out: a sphere of radius r sits snugly inside a cylinder of radius r and height 2r, whose volume is πr²·2r = 2πr³. The sphere is (4/3πr³)/(2πr³) = ⅔ of that bounding cylinder — Archimedes' favorite fact, the one he asked to have carved on his tomb. Try the sphere mode in the widget above and watch the ratio read ⅔.
Volume scales like the cube of the lengths. Double the radius and the volume multiplies by 2³ = 8, not 2: a beach ball twice the diameter of a basketball holds eight times as much air.
Every volume in this lesson came from one move — slice it and add up the slices — and Cavalieri's promise that matching slices mean matching volume.
| Solid | Volume | Where it comes from |
|---|---|---|
| Box / cuboid | V = abc = S·h | Stack S = ab of unit-cube layers, h high |
| Prism or cylinder | V = S·h | Cavalieri: shear it upright, slices unchanged |
| Cone or pyramid | V = ⅓ S·h | Three fill the prism; slices shrink like (1−t/h)² |
| Sphere | V = 4/3 πr³ | Hemisphere ≡ drilled cylinder by Cavalieri |
| The principle | matching slices ⇒ equal V | Equal height + equal cross-section at every level |
Cylinder V = πr²h · cone V = ⅓πr²h · sphere V = 4/3πr³. The cone is a third of its cylinder; the sphere is two-thirds of its bounding cylinder. And volumes scale by the cube of the lengths.
A cylinder has radius 5 cm and height 8 cm. Find its volume (leave the answer in terms of π, then as a decimal).
V = πr²h = π·5²·8 = 200π ≈ 628.3 cm³. (A radius-5 base has area 25π; eight such slices give 200π.)
A cone and a cylinder share the same base (radius 6) and the same height (10). The cylinder holds 360π cm³. Without recomputing from scratch, how much does the cone hold?
A cone is exactly ⅓ of the cylinder on the same base and height, so V = ⅓·360π = 120π ≈ 377.0 cm³. (Check: ⅓π·6²·10 = ⅓·360π = 120π. ✓)
An oblique (leaning) prism has a triangular base of area 12 cm² and a vertical height of 7 cm. Does the lean change its volume? Find it.
No — by Cavalieri, leaning the prism slides each slice sideways but keeps its area and the overall height, so the volume is unchanged: V = S·h = 12·7 = 84 cm³. Only the perpendicular height between the bases matters, not the slant.
Find the volume of a sphere of radius 3 cm, and state what fraction it is of the smallest cylinder that contains it.
V = 4/3·π·3³ = 4/3·π·27 = 36π ≈ 113.1 cm³. The bounding cylinder has radius 3 and height 2·3 = 6, so its volume is π·3²·6 = 54π. The ratio is 36π : 54π = ⅔ — Archimedes' result.
A spherical ball is melted and re-cast into a new ball of twice the radius. How many of the original balls' worth of material does the new ball need?
Volume scales like the cube of the radius. Doubling r multiplies the volume by 2³ = 8, so you would need eight of the original balls. (V₂/V₁ = (4/3π(2r)³)/(4/3πr³) = 8.)
A grain silo is a cylinder of radius 2 m and height 5 m, topped by a cone of the same radius and height 1.5 m. Find the total volume in terms of π.
Cylinder: πr²h = π·2²·5 = 20π. Cone: ⅓πr²h = ⅓·π·2²·1.5 = ⅓·6π = 2π. Total = 20π + 2π = 22π ≈ 69.1 m³. (Add the two slice-stacks; they sit on the same circle but have different heights.)
Six questions to lock it in. Tap the answer you think is right.
This lesson develops volume by dissection and Cavalieri's principle rather than rote formulas, aligning with CCSS G-GMD.1 (give an informal argument for the volume formulas of a cylinder, pyramid, and cone, using Cavalieri's principle) and G-GMD.3 (use volume formulas to solve problems). The sphere derivation and the cube-scaling caution touch G-MG.1 (modeling with geometric solids). The shearing widget makes the principle's hypothesis — equal height and matching cross-sections at every level — visible, and the comparison widget keeps every volume tied to a computed value, so the figures never outrun the arithmetic.