Compare volumes through complete section rules

Try

A rectangular box with integer side lengths can be tiled by unit cubes. A 2-by-3-by-4 box contains 24 cubes, so its volume is 24 cubic units. For rational side lengths, subdivide the unit cube into smaller equal cubes and count again. This explains the product of three lengths rather than presenting it as an unexplained rule.

For arbitrary positive real lengths, trap each length between increasingly close rational bounds. The products of the lower and upper bounds approach the same product. Containment traps the box volume between them, extending the rule without requiring calculus.

Predict

A right prism is formed by translating a planar base straight through perpendicular height h. Predict whether two differently shaped bases with equal area produce equal volumes when extruded through the same height. Now shear one prism sideways, preserving every parallel section area and the perpendicular height. Its slanted edge becomes longer, but should that alone create extra volume?

Experiment

Start with a right triangular prism whose base is a right triangle with legs 3 and 4 and whose height is 5. Two congruent copies tile a 3-by-4-by-5 box, so one has volume 30. For a general triangle, an altitude splits it into right triangles; if the altitude foot lies outside, use the difference of the two right triangles instead. Extruding this dissection proves the triangular-prism formula. For a polygonal base, triangulate it, extrude each triangle and add the nonoverlapping volumes. This gives V=Bh, where B is base area.

Solid	Base or section area	Perpendicular height	Volume
Right triangular prism	6	5	30
Right cylinder, radius 2	4π	5	20π
Straight tube, radii 2 and 1	3π	5	15π

For the cylinder, extrude the inner and outer circle polygons from the preceding lesson. Their prism volumes bound the cylinder and approach the same value πr²h. The tube subtracts a contained inner cylinder; it does not use the outer radius alone.

Understand

Cavalieri’s principle compares two bounded solids between parallel planes at the same distance h: if every plane parallel to them cuts sections of equal area in both solids, their volumes are equal. It does not require the sections to be congruent or in the same horizontal position. It does require every height, not a finite sample and not merely equal top and bottom areas.

The school-level justification is a limiting comparison by thin layers, with geometric control on the approximation error. It is not valid to say that a zero-thickness section itself has area times a positive thickness. The finite layer is a different set, and lower/upper enclosing prisms or boxes must bound it. For the ordinary polygonal and circular solids here, refining the layers makes the total discrepancy arbitrarily small. Equal section-area rules then give equal limiting layer totals. We use this regular-solid comparison principle; we are not asserting a theorem for arbitrary pathological sets.

One explicit control illustrates why sideways shifting does not defeat the argument. Suppose square sections have constant side s and their centers move at most LΔ in each coordinate within a slab of thickness Δ. A containing square prism can use side s+2LΔ; a contained one can use side s−2LΔ once this is positive. Across total height h their volume gap is h[(s+2LΔ)²−(s−2LΔ)²]=8sLhΔ, which tends to zero. The common limiting volume is s²h even though the centers shift. Polygon boundaries have finitely many sides; the corresponding thin boundary strips likewise have total area tending to zero. Circular boundaries can also be trapped by polygon bounds.

An oblique prism has translated copies of its base at every parallel height, so Cavalieri equates it with the right prism: V=Bh. Use perpendicular distance between base planes, not a lateral edge length. An oblique cylinder satisfies the same section comparison. Later cone and sphere comparisons will supply explicit variable area rules, rather than inferring them from the outline.

Apply

A sample table cannot establish the universal premise. A solid can bulge between sampled heights while agreeing with a cylinder at each recorded plane. The practice asks you to construct such a counterexample, not merely repeat the warning.

For a volume argument, state the height interval, derive the area rule for arbitrary height, and explain why the comparison solid’s volume is known. For a numerical application, check units: square centimeters multiplied by centimeters give cubic centimeters. Equal numerical labels with incompatible units cannot be added or compared.

Continue

Pyramids and cones have sections whose areas shrink quadratically with fractional height. The next lesson constructs genuine inner and outer stacks and proves that their gap tends to zero. Their one-third factor will then support the sphere comparison without circular reasoning.