Recover sphere volume from a comparison solid

Try

Imagine a ball of radius 3 cm centered at height z=0. Its top is at z=3 and bottom at z=−3. At the equator, the horizontal section is a disk of radius 3. Halfway from the equator to the top, at z=1.5, the section is smaller, but its radius is not 1.5.

Draw a radius from the center to the circular edge of the section. Together with the vertical height z and horizontal section radius ρ, it forms a right triangle. Pythagoras gives ρ²+z²=3². Thus the disk area at z=1.5 is π(9−2.25)=6.75π cm².

Predict

Place the sphere beside a cylinder of radius 3 and height 6. Remove a centered double cone from the cylinder: both cone tips meet at the center height, and each cone’s base is an end disk of the cylinder. The removed radius grows linearly from 0 at the center to 3 at either end.

Predict the remaining section at z=1.5. Its outer radius is 3 and its missing inner radius is 1.5. Its annular area is 9π−2.25π=6.75π cm², exactly the sphere-section area. The sections have different shapes; Cavalieri compares their areas, not their outlines.

Experiment

Explore several heights and then change the radius. Predict which section areas will match before revealing the results. The comparison mode with one full-height cone is a deliberate nonexample: equal total volume does not force equal area at every height.

Match sections before comparing volumes

A sphere has radius R. Compare it with a cylinder of radius R and height 2R after removing cones. Height z is measured from the middle. Predict whether the two horizontal sections have equal area at the selected height.

Radius R (length units)12345 Height z / R-8/8-7/8-6/8-5/8-4/8-3/8-2/8-1/80/81/82/83/84/85/86/87/88/8 Removed solidTwo cones meeting at the middleOne cone across the full height Reset sections

R = 3; z = 3/2. Both solids occupy heights −3 through 3. The drawings keep a fixed scale of 20 SVG coordinate units per length unit; their size changes when R changes.

Reveal area comparison

Why matching every slice supports a volume argument

Split each hemisphere into n horizontal slabs, each of thickness R/n. A sphere’s disk areas decrease from its equator to its pole. The smaller endpoint disk gives an inscribed cylinder; the larger gives a containing cylinder. Reflect the same construction below the equator. Add their cylinder volumes to bound the sphere volume.

Slabs per hemisphere n248163264

With 8 slabs in all: lower bound (459/16)π ≤ sphere volume ≤ upper bound (675/16)π cubic units. Gap = (27/2)π.

The gap is 2πR³/n, which can be made arbitrarily small by taking n large. This uses the usual volume properties (additivity, containment and limits of such bounds). It explains the limiting step rather than claiming finitely many slices prove an exact volume. The available n values are bounded examples; the formula establishes the general limit.

All controls are temporary. Save your prediction, comparison and general argument in the lesson practice; moving a slider does not demonstrate mastery.

Record one positive, one zero and one negative height. For the centered double cone, compare the formulas as well as the picture. A handful of matching samples cannot prove the all-height premise. For the one-cone foil, locate a mismatch and explain why it prevents this particular Cavalieri argument, even if some heights agree.

Understand

Now let R be any positive radius and use −R≤z≤R. The sphere disk has radius squared R²−z², so its area is π(R²−z²). In the comparison cylinder the full disk area is πR². Similar triangles make the double-cone hole radius ∣z∣; subtracting its area gives πR²−πz². These expressions are identical for every allowed z, including the zero-area endpoint sections.

Both solids have height 2R. Cavalieri’s principle therefore equates their volumes. The cylinder volume is 2πR³. Each removed cone has base area πR² and height R, hence volume πR³/3. Their interiors do not overlap; sharing an apex adds no volume. Subtracting both gives

sphere volume = 2πR³−2πR³/3 = 4πR³/3.

The logical dependencies are circle area, Pythagoras, similarity, cylinder volume, the separately justified cone formula and Cavalieri. No sphere-volume formula was assumed to establish the section comparison.

Why is the foil useful? Remove instead one cone of radius R and height 2R, with its base at the bottom and tip at the top. It removes the same volume as the two small cones. Its hole radius at height z is (R−z)/2, however, so its remaining area is π[R²−(R−z)²/4]. This generally differs from the sphere’s. Equality of total volumes is a consequence of one calculation, not a converse form of Cavalieri’s sufficient condition.

Apply

For R=3 cm, the cylinder contributes 54π cm³ and the two cones together remove 18π cm³. The sphere volume is 36π cm³, approximately 113.1 cm³ to the nearest tenth. A diameter of 6 cm must first be halved; substituting 6 as the radius would multiply the answer by eight.

Your written argument should name the two solids, specify a common height coordinate, derive both section areas at an arbitrary height, and only then subtract volumes. A drawing of similar-looking profiles is insufficient. Also distinguish section area, measured in square centimeters, from volume, measured in cubic centimeters.

Continue

This argument suggests a general modeling habit: compare a difficult object with a combination of simpler ones using an invariant quantity. Here the invariant is area at every height. The actual objects need not be congruent.

Optional depth: the foil sections agree at z=−R and z=R/3, obtained by solving z²=(R−z)²/4 within the interval. Those two matching heights still cannot establish the required universal equality. The formulas, rather than a finite visual inspection, settle the question.