Give informal arguments for the circumference and area of a circle and the volumes of a cylinder, a pyramid and a cone using dissection, Cavalieri's principle and informal limits

Learning goal: Give an informal argument for the formulas for the circumference of a circle, the area of a circle, and the volume of a cylinder, a pyramid and a cone, using dissection arguments, Cavalieri's principle and informal limit arguments, and say at each step what is assumed and what is being shown.

What an informal argument is. A formal proof that a circle of radius r has circumference 2πr needs a definition of the length of a curved path, which belongs to calculus. An informal argument uses shapes whose measures you already trust, such as polygons, parallelograms, prisms and cubes, shows that the circle or the solid is trapped between them, and explains why the gap can be made as small as you like. Three tools do all the work in this lesson. A dissection argument cuts a figure into pieces and rearranges them; because rearranging does not change total area or volume, whatever the rearranged figure measures, the original measured too. Cavalieri's principle says that 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 solids have equal volume. It needs equal areas, not equal shapes. An informal limit argument produces a sequence of approximations that get as close as you like to the true value while staying on one side of it (always below, or always above), and concludes that the true value is the number the approximations approach.

Circumference: first a definition, then a limit. Every circle is similar to every other one, because a translation carrying one center to the other, followed by a dilation about that center with scale factor equal to the ratio of the radii, carries one circle onto the other. Similar figures have all lengths in the same ratio, so circumference ÷ diameter is the same number for every circle. That constant is what π means. So C = πd = 2πr is not a discovery; it is the definition of π applied to a circle of radius r. What needs an argument is the value of π, and inscribed and circumscribed regular polygons supply it. A regular hexagon inscribed in a circle of radius r is six equilateral triangles of side r, so its perimeter is 6r exactly. Doubling the number of sides, the inscribed 12-gon has side 2r sin 15° ≈ 0.5176r and perimeter about 6.2117r; the 24-gon has perimeter about 6.2653r; the 96-gon about 6.2821r. Each inscribed perimeter is less than the circumference, because a straight segment is the shortest path between two points and so each polygon side is shorter than the arc it cuts off. The circumscribed polygons close in from above: 6.9282r for the hexagon, 6.4308r for the 12-gon, 6.2854r for the 96-gon. The circumference is trapped between 6.2821r and 6.2854r, so 2π lies between 6.2821 and 6.2854, and it does: 2π ≈ 6.2832. Nothing stops us continuing; that is the informal limit argument.

Area of a circle by dissection. Cut a circle of radius r into 8 congruent sectors and lay them side by side, alternately point up and point down. The result is close to a parallelogram. Its lower edge is made of 4 arcs whose total length is half the circumference, πr, and its upper edge is the other 4 arcs, also πr. Its height is close to r, the radius, because each sector has straight sides of length r. A parallelogram has area base × height, so the arrangement has area close to πr × r = πr². With 16 sectors the scalloped edges are flatter and the slanted sides are steeper; with 100 sectors the figure is almost exactly a rectangle of base πr and height r. Since rearranging never changes area, the area of the circle is the value these arrangements approach, πr². For r = 6 cm the half-circumference is 6π ≈ 18.85 cm, the height is 6 cm, and the area is 36π ≈ 113.10 cm². A second route uses inscribed polygons again: a regular n-gon inscribed in the circle has area ½ × perimeter × apothem, where the apothem is the distance from the center to a side. As n grows the perimeter approaches 2πr and the apothem approaches r, so the area approaches ½ × 2πr × r = πr². In units of r², the inscribed hexagon has area about 2.5981, the 12-gon exactly 3, the 24-gon about 3.1058, the 96-gon about 3.1394, heading to π ≈ 3.1416.

Volume of a cylinder: stacking and Cavalieri. A cylinder of base radius r and height h can be thought of as a stack of very thin circular discs. Each disc has area πr², and the discs together are h tall, so the volume is πr²h, exactly as a prism has volume base area × height. The prism connection can also be run as a limit: a prism whose base is a regular n-gon inscribed in the base circle has volume (area of the n-gon) × h, and for r = 5 the base areas 64.95, 75.00, 77.65 and 78.48 square units (for 6, 12, 24 and 96 sides) approach 25π ≈ 78.54. Now push the stack of discs sideways so that it leans: the same discs, the same total volume. That is Cavalieri's principle in action. An oblique cylinder, with its axis tilted, has the same volume as the right cylinder with the same base and the same perpendicular height, πr²h. The height h is the perpendicular distance between the two base planes, never the length of the tilted side. For r = 4 cm and h = 10 cm, V = π × 16 × 10 = 160π ≈ 502.65 cm³, right or oblique.

Volume of a pyramid: dissect a cube, then generalize with Cavalieri. A pyramid has a polygon base, an apex, and a height that is the perpendicular distance from the apex to the plane of the base. Take a cube of side s and one vertex V of it. The three faces of the cube that do not contain V are the bases of three pyramids with apex V, and these three pyramids fill the cube without overlapping. They are congruent, because a rotation of the cube about the diagonal through V carries each onto the next. So each has volume s³ ÷ 3, which is ⅓ × s² × s, one third of base area times height. For s = 6 cm the cube is 216 cm³ and each pyramid is 72 cm³ = ⅓ × 36 × 6. That settles one very special pyramid. To reach every pyramid, look at cross-sections. Cut a pyramid of base area B and height h by a plane parallel to the base at distance d from the apex. The cross-section is the base dilated from the apex by factor d/h, so its area is B(d/h)². For a square pyramid with base 6 cm and height 9 cm, the section 3 cm below the apex has factor 1/3, side 2 cm and area 4 cm², which is 36 × (1/9). Any two pyramids with the same base area B and the same height h therefore have cross-sections of equal area B(d/h)² at every level, whatever the shape of the bases and wherever the apex sits. By Cavalieri's principle they have equal volumes. A stretch argument fills in the last gap: stretching a solid in one direction by a factor t makes every thin slab t times thicker without changing its area, so the volume is multiplied by t; stretching the cube pyramid this way produces a pyramid with any rectangular base and any height, still with volume ⅓ × base × height. Together these give V = ⅓Bh for every pyramid: the square pyramid above has volume ⅓ × 36 × 9 = 108 cm³.

Volume of a cone: a limit of pyramids. Inscribe a regular n-gon in the base circle of a cone of radius r and height h and build the pyramid on that polygon with the cone's apex. Its volume is ⅓ × (area of the n-gon) × h, and as n grows the n-gon area approaches πr², so the pyramid volumes approach ⅓πr²h. For r = 5 cm and h = 12 cm the pyramid volumes are 259.81, 300.00, 310.58 and 313.94 cm³ for 6, 12, 24 and 96 sides, approaching 100π ≈ 314.16 cm³. The same conclusion comes straight from Cavalieri: a cone and a pyramid with base area πr² and the same height have equal cross-section areas πr²(d/h)² at every level.

Misconceptions, and why they are wrong. "π equals 3.14" confuses a rounded value with a definition; the limit argument pins π down to as many decimal places as you like, and that its decimals never end or repeat (π is irrational) is a separate theorem not proved here. "The eight sectors make an exact parallelogram" is false; the edges are arcs, and only the limit gives equality. "Cavalieri needs the cross-sections to be the same shape" is false; a triangle and a square of equal area count as equal, and that is exactly what lets a square pyramid stand in for every pyramid. "The ⅓ comes from the three sides of a triangle" is a coincidence of numbers; it comes from three congruent pyramids filling a cube. Finally, a tilted cylinder or cone with a slanted edge of length 9 and perpendicular height 8 has volume computed with 8, not 9.

Worked trap for the circumference: Trap the circumference of a circle of radius 10 cm between the inscribed and circumscribed regular 12-gons. Inscribed side: 2 × 10 × sin 15° = 20 × 0.25882 = 5.1764 cm, so the perimeter is 12 × 5.1764 = 62.117 cm. Circumscribed side: 2 × 10 × tan 15° = 20 × 0.26795 = 5.3590 cm, perimeter 64.308 cm. So 62.12 cm < C < 64.31 cm. The formula gives C = 2π × 10 ≈ 62.83 cm, inside the trap. With 96 sides the trap narrows to 62.821 cm < C < 62.854 cm, and the true value is still inside.

Worked dissection for the area: A circle of radius 6 cm is cut into 8 sectors of 45° each. Each arc has length 2π × 6 ÷ 8 = 1.5π ≈ 4.71 cm, so the 4 arcs along the bottom of the rearranged figure total 6π ≈ 18.85 cm, which is half the circumference. The straight sides are radii of 6 cm, so the height is close to 6 cm. Multiplying 6π by 6 uses the limiting dimensions, base exactly πr and height exactly r, so 6π × 6 = 36π ≈ 113.10 cm² is already the exact area: the parallelogram picture shows where πr² comes from, not an estimate that improves. To watch an estimate improve, replace each sector by the triangle formed by its two radii and its chord. Eight triangles have total area ½ × 8 × 6² × sin 45° ≈ 101.82 cm², 16 give ½ × 16 × 36 × sin 22.5° ≈ 110.21 cm², and 96 give about 113.02 cm², closing in on 36π ≈ 113.10 cm².

Worked Cavalieri for a cylinder: A stack of 40 coins, each of radius 1.2 cm and thickness 0.2 cm, stands 40 × 0.2 = 8 cm tall. Its volume is π × 1.44 × 8 = 11.52π ≈ 36.19 cm³. Push the stack so that it leans: every horizontal plane still cuts it in a circle of area 1.44π, the stack is still between planes 8 cm apart, so by Cavalieri's principle the leaning stack also has volume 11.52π cm³. The tilted edge is longer than 8 cm, and that length is irrelevant.

Worked pyramid and cone: A square pyramid has base 8 cm by 8 cm and height 9 cm; a triangular pyramid has a right-triangle base with legs 16 cm and 8 cm (area ½ × 16 × 8 = 64 cm²) and the same height 9 cm. Both have base area 64 cm². At 3 cm above the base, 6 cm below the apex, both cross-sections are scaled by 6/9 = 2/3 and have area 64 × 4/9 ≈ 28.44 cm². Equal areas at every level, so equal volumes: ⅓ × 64 × 9 = 192 cm³ each. For a cone of radius 5 cm and height 12 cm, V = ⅓ × π × 25 × 12 = 100π ≈ 314.16 cm³, one third of the cylinder with the same base and height, 300π ≈ 942.48 cm³.

Approximation record: Make a table with rows for n = 6, 12, 24 and 96 and columns for the inscribed perimeter ÷ r (6.0000, 6.2117, 6.2653, 6.2821), the circumscribed perimeter ÷ r (6.9282, 6.4308, 6.3193, 6.2854) and the inscribed area ÷ r² (2.5981, 3.0000, 3.1058, 3.1394). Below the table write the targets 2π ≈ 6.2832 and π ≈ 3.1416 and explain in one sentence each why the inscribed values can only rise toward the target and why the circumscribed values can only fall toward it. Then draw the 8-sector rearrangement of a circle of radius 4 cm, label its base 4π ≈ 12.57 cm and height about 4 cm, and record the estimate 16π ≈ 50.27 cm².

Independent performance: Write three informal arguments in full sentences: one for the area of a circle by dissection, one for the volume of an oblique cylinder using Cavalieri's principle, and one for the volume of a cone as a limit of pyramids. Success criteria: each argument names the principle it uses (dissection, Cavalieri or limit) and states the condition that principle requires (rearrangement preserves area; solids between the same parallel planes with equal cross-section areas; approximations that approach the value from one side); each argument includes a numerical instance with units, such as r = 5 and h = 12 giving pyramid volumes 259.81, 300.00, 310.58 and 313.94 cm³ approaching 100π cm³; and each argument says explicitly which step is exact and which step is an approximation that the limit makes exact.