Derive the one-third volume factor

Try

A square pyramid has a base 6 cm by 6 cm and perpendicular height 9 cm. A prism with that same base and height has volume 36×9=324 cm³. The pyramid occupies less space because its horizontal sections shrink toward the apex. But “it gets narrower” does not establish the fraction one third.

The purpose of this lesson is to justify that factor without requiring integration. We will compare the pyramid with stacks of thin prisms whose volumes are already understood. The same argument will work for a cone once circle area is available. Keep perpendicular height separate from a sloping edge or the cone’s slant height.

Predict

Measure t downward from the apex, so t=0 at the tip and t=9 at the base. A section parallel to the base is a similar square. Its length scale is t/9, and its area scale is the square of that fraction. At t=3 the side is 2 cm and the area is 4 cm²; at t=6 the side is 4 cm and the area is 16 cm².

If lengths shrink by one half, predict whether area shrinks by one half or one quarter. This square relationship is the source of the volume fraction.

Experiment

Build a numerical bound using four equal-height layers. Each layer has height 9/4. For the centered square pyramid, every section is contained in the next larger section. In a layer, the smaller endpoint square gives an inscribed prism and the larger gives a containing prism. Add the four prism volumes to bracket the true volume.

Equal layers n	Total inner-prism volume, cm³	Total outer-prism volume, cm³
2	40.5	202.5
4	70.875	151.875
10	92.34	124.74

Reproduce the four-layer row: the lower section areas are 36 times 0², (1/4)², (2/4)² and (3/4)². Multiply their sum by 9/4. The upper sum instead uses (1/4)² through 1². These finite stacks bound the answer; neither stack is claimed to equal it.

Understand

Let B be base area and h perpendicular height. Similarity gives section area B(t/h)². Start with a right pyramid whose convex base contains the projection of its apex, so the sections are nested; a right circular cone also has this property. With n layers, the lower bound is Bh times the sum of k² for k=0 through n−1, divided by n³. The upper bound uses k=1 through n.

The needed sum is not a calculus result. Expand (k+1)³−k³=3k²+3k+1 and add for k=0 through n−1. The left side telescopes to n³. The integers sum to n(n−1)/2, which follows by pairing first with last. Solving gives the square sum n(n−1)(2n−1)/6. The lower and upper volume bounds are therefore

Bh(1/3−1/(2n)+1/(6n²)) and Bh(1/3+1/(2n)+1/(6n²)).

Both approach Bh/3 as n increases, and their gap is Bh/n, which can be made smaller than any positive requested tolerance. We use the ordinary volume principle that containing solids have at least as much volume as contained solids. Since the pyramid lies between these stacks for every n, its volume must be Bh/3. This is a limiting argument, not an assertion that a finite staircase fits perfectly.

For a cone, B=πr², so the result is πr²h/3. For an oblique pyramid or cone, parallel sections at the same fractional height have the same areas as those of the corresponding right solid. Cavalieri’s principle then transfers the volume. This extension depends on that principle; the nested-stack construction is not silently assumed to fit an arbitrarily shifted oblique solid.

Apply

The worked square pyramid has volume 36×9/3=108 cm³. A separate right cone with radius 3 cm and perpendicular height 8 cm has volume π×9×8/3=24π cm³. Retaining π preserves the exact value; a decimal needs an indicated rounding rule.

Explain what would go wrong if someone used a cone’s slant height in place of h. The stack thickness is measured perpendicular to the base, and the similarity ratio is based on that same direction. A longer slant length would describe a different height and overestimate the volume. In practice, label the height and base area before substituting numbers.

Continue

A sphere can now be compared with a cylinder after removing two cones. That comparison needs the cone formula established here; using a sphere formula to derive the cone formula first would make the later argument circular.

For optional depth, use the bounds to choose n so that the gap is below 1 cm³ for the worked pyramid. Since Bh=324 cm³, any integer n greater than 324 suffices. This controls the uncertainty without a derivative, antiderivative or indefinite appeal to “very thin slices.”