Give an informal argument using Cavalieri's principle for the volume of a sphere and of other solid figures

Learning goal: Give an informal argument using Cavalieri's principle for the formula for the volume of a sphere, and use the same principle to find the volumes of other solid figures whose cross-sections can be matched, level by level, with those of a solid you already know.

Scope note. G.GMD.2 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full as part of the Geometry course, as extension content for students on that path.

The principle, with its conditions. Cavalieri's principle: if two solids lie between the same pair of parallel planes, and every plane parallel to those two cuts the solids in cross-sections of equal area, then the two solids have equal volume. Three conditions are built in. The solids must occupy the same band between two parallel planes. The comparison must hold for every plane parallel to those, not just a few. And what must match is area, not shape; a circle and a square ring of equal area count as equal. The reason the principle is believable is the slab picture: slice both solids into very thin slabs with the same cutting planes; matching slabs have equal face areas and equal thickness, so equal volumes, and the totals agree. The principle by itself never produces a formula. It transfers a volume from a solid you can compute to one you cannot, so the art is choosing the comparison solid.

The comparison solid for a hemisphere. Take a hemisphere of radius r resting flat side down on a table. Beside it place a cylinder of radius r and height r, also standing on the table, and remove from it a cone whose apex is at the center of the cylinder's base and whose circular top, of radius r, is the top face of the cylinder. Both solids lie between the table and the plane r above it. Now cut both with a horizontal plane at height y above the table, where 0 ≤ y ≤ r. The hemisphere's cross-section is a circle. Its radius is the horizontal leg of a right triangle with hypotenuse r (from the center of the sphere to a point of the circle) and vertical leg y, so its radius is √(r² − y²) and its area is π(r² − y²). The cylinder-minus-cone's cross-section is a ring: the outer circle has radius r, and the inner circle is the cone's cross-section, which at height y has radius y because the cone widens from 0 at the table to r at height r, one unit of radius per unit of height. The ring's area is πr² − πy² = π(r² − y²). The two areas are equal at every height, so by Cavalieri the volumes are equal. The cylinder has volume πr² × r = πr³ and the removed cone has volume ⅓πr² × r = ⅓πr³, so the hemisphere has volume πr³ − ⅓πr³ = ⅔πr³, and the sphere, two hemispheres, has volume (4/3)πr³.

A numerical run of the argument. Let r = 10 cm and cut at y = 6 cm. The hemisphere's slice has radius √(100 − 36) = √64 = 8 cm and area 64π cm². The ring has outer radius 10 cm and inner radius 6 cm, area 100π − 36π = 64π cm². Cut at y = 0: the hemisphere gives a circle of radius 10 (area 100π) and the ring has inner radius 0, area 100π. Cut at y = 10: the hemisphere gives a single point (area 0) and the ring has inner radius 10, area 0. For r = 3 cm the slice areas at y = 0, 1, 2, 3 are 9π, 8π, 5π and 0 on both sides. The hemisphere of radius 3 cm then has volume 27π − 9π = 18π cm³ and the sphere 36π ≈ 113.10 cm³. The whole sphere can also be done at once: a cylinder of radius r and height 2r with a double cone removed (two cones apex to apex at the center) matches the sphere slice for slice; for r = 5 cm that is 250π − 2 × (⅓ × π × 25 × 5) = 250π − 250π/3 = 500π/3 ≈ 523.60 cm³, which is (4/3)π × 125.

Other solid figures. The same principle handles anything whose slices you can match. A leaning stack of coins is an oblique cylinder; each horizontal slice is a circle of the same area as in the straight stack, so the volumes agree, πr²h with h the perpendicular height. An oblique cone or pyramid, apex not above the center of its base, has slices that are dilations of the base with scale factor d/h, and so of area B(d/h)², at distance d from the apex, exactly as for the right solid, so its volume is ⅓Bh. A solid with a hole, such as a cylinder with a cylindrical hole through it, has ring cross-sections; if another solid has slices of the same area at every level, the volumes match. And a slightly stronger form of the principle, which follows from the same slab picture, says that if every cross-section of one solid has c times the area of the matching cross-section of another, the volume is c times as large. The intersection of two cylinders of radius r whose axes cross at right angles has square cross-sections, of side 2√(r² − y²) at height y, while the sphere of radius r has circular ones of radius √(r² − y²). Square area 4(r² − y²) is 4/π times circle area π(r² − y²) at every level, so the intersection has volume (4/π) × (4/3)πr³ = 16r³/3. For r = 3 that is 144 cubic units, against the sphere's 36π ≈ 113.10; the ratio 144 ÷ 113.10 ≈ 1.273 is 4/π.

A worked model inside the lesson. A bowl is a hemisphere of inner radius 6 cm. Its capacity is ⅔ × π × 216 = 144π ≈ 452.39 cm³, about 452 mL. The argument that justifies this is the one above with r = 6: at height y the bowl's slice has area π(36 − y²) and the comparison cylinder-minus-cone's ring has area 36π − πy², so the bowl holds π × 36 × 6 − ⅓ × π × 36 × 6 = 216π − 72π = 144π cm³.

Misconceptions, and why they are wrong. "The cross-sections must be the same shape" is false; the sphere argument compares circles with rings, and only areas need to agree. "The removed cone has its apex at the top" reverses the picture; the cone points downward, with its apex at the center of the base on the table (y = 0), and it widens to radius r at the top (y = r), which is what makes its slice radius equal to the height y. "Cavalieri gives the formula by itself" is false; it only equates the hemisphere with cylinder-minus-cone, and the formulas for cylinder and cone volume then finish the job. Mixing reference levels breaks the computation: the formulas √(r² − y²) and y both measure y up from the table, so substituting a distance measured down from the top into them gives the wrong slice radii, even though the planes being compared are the same. Finally, comparing solids of different heights, such as a cone of height r with a hemisphere of radius 2r, violates the same-two-planes condition, and no conclusion follows.

Worked matching at three heights: For a hemisphere of radius 5 cm and its comparison solid, compute both cross-section areas at y = 0, 3 and 4 cm. Hemisphere: π(25 − 0) = 25π, π(25 − 9) = 16π, π(25 − 16) = 9π. Ring: 25π − 0 = 25π, 25π − 9π = 16π, 25π − 16π = 9π. Equal at each level, and the same algebra gives equality at every level. So the hemisphere volume is π × 25 × 5 − ⅓ × π × 25 × 5 = 125π − 125π/3 = 250π/3 ≈ 261.80 cm³, and the sphere of radius 5 cm has volume 500π/3 ≈ 523.60 cm³, matching (4/3)π × 125.

Worked oblique cone: A cone has base radius 4 cm and its apex is 9 cm above the plane of the base but displaced sideways, so that the axis is tilted. At distance d below the apex the slice is a circle of radius 4d/9, area 16π(d/9)², exactly the slice area of the right cone with the same base and height. Between the same planes, with equal slices at every level, Cavalieri gives equal volumes: ⅓ × π × 16 × 9 = 48π ≈ 150.80 cm³. The slanted axis length, which is longer than 9 cm, does not appear.

Worked square-sectioned solid: A solid lies between the planes y = −3 and y = 3 and its horizontal cross-section at height y is a square of side 2√(9 − y²). The sphere of radius 3 has circular slices of radius √(9 − y²) between the same planes. At every height the square's area 4(9 − y²) is 4/π times the circle's area π(9 − y²). By the proportional form of Cavalieri's principle the solid's volume is (4/π) × 36π = 144 cubic units. Check with the sphere: 36π ≈ 113.10, and 113.10 × 4/π ≈ 144.0.

Worked cylinder with a hole: A solid is a cylinder of radius 6 cm and height 10 cm with a coaxial cylindrical hole of radius 2 cm. Every horizontal slice is a ring of area 36π − 4π = 32π cm². A plain cylinder of radius √32 cm and height 10 cm has the same slice area at every level, so both have volume 32π × 10 = 320π ≈ 1005.31 cm³; directly, 360π − 40π = 320π cm³ agrees.

Cross-section record: Draw a side view of a hemisphere of radius 4 cm next to a cylinder of radius 4 cm and height 4 cm with the cone removed, both standing on the same line. Draw the cuts at y = 0, 1, 2, 3 and 4 cm and make a table with columns y, hemisphere slice radius √(16 − y²), hemisphere slice area, ring inner radius (equal to y), ring outer radius 4, ring area. The areas should read 16π, 15π, 12π, 7π and 0 in both area columns. Under the table write the volume calculation 64π − 64π/3 = 128π/3 ≈ 134.04 cm³ for the hemisphere and 256π/3 ≈ 268.08 cm³ for the sphere.

Independent performance: Write a complete informal argument, in full sentences, for the volume of a sphere of radius r using Cavalieri's principle, then apply the same principle to one other solid figure of your choice (an oblique cylinder, an oblique pyramid, a solid with a hole, or a solid with square cross-sections matched to a sphere). Success criteria: the argument names the two solids being compared and the two parallel planes they lie between; it computes both cross-section areas at a general height y and shows they are equal (or in a fixed ratio); it includes one numerical check at a specific height, such as r = 10 and y = 6 giving 64π on both sides; it finishes with the formula (4/3)πr³ and a numerical volume with units; and the second application states which condition of the principle each step is using.