HSG.GMD.A.2: Cavalieri's Principle and the Volume of a Sphere
In plain English: HSG.GMD.A.2 is an advanced (+) Common Core geometry standard that asks students to use Cavalieri's principle to give an informal argument for the volume of a sphere and of other solids. Students compare a hemisphere, slice by slice, with a cylinder that has a cone removed, which gives V = (4/3)πr³. It is usually taught in honors Geometry or a later course.
(+) Give an informal argument using Cavalieri's principle for the formulas for the volume of a sphere and other solid figures.
Common Core State Standards for Mathematics · Domain: Geometric Measurement and Dimension (GMD) · Cluster: Explain volume formulas and use them to solve problems Also written as HSG-GMD.A.2 or G-GMD.2 · Official standard
Students already accept Cavalieri's principle and the volume formulas for cylinders and cones. Here they use them to explain where V = (4/3)πr³ comes from. The key move, due to Archimedes and later to Cavalieri, is to find a solid with a known volume whose slices match the slices of a hemisphere at every height. A cylinder of radius r and height r with a cone carved out of it (apex at the center of the bottom, base at the top) does exactly that: at height y, both slices have area πr² - πy².
After the sphere, students use the same strategy on other solids named "other solid figures" in the standard: a napkin ring left when a cylinder is drilled through a sphere, a spherical cap such as the water in a bowl, and the solid common to two crossing cylinders. In every case the work is the same: describe a slice at height y, find its area, and match it with a slice of a solid whose volume is known.
Learning Objectives
By the end of this lesson, students will be able to:
Find the area of the slice of a hemisphere at height y using the Pythagorean Theorem
Show that a cylinder with a cone removed has slices of the same area as a hemisphere at every height
Use Cavalieri's principle to argue that a sphere has volume (4/3)πr³
Use the same slicing strategy to find the volumes of other solids, such as napkin rings, spherical caps and bicylinders
Prior Knowledge Required
Students should already be comfortable with:
Cavalieri's principle and informal arguments for cylinder and cone volumes HSG.GMD.A.1
Volume formulas for cylinders, cones and spheres 8.G.C.9
The Pythagorean Theorem 8.G.B.7
Cross-sections of three-dimensional solids HSG.GMD.B.4
Area of a ring as the difference of two circle areas
Draw a semicircle of radius 5 on the board, standing on its diameter, with a horizontal line 3 units above the diameter. Students find half the length of the chord on that line.
Warm-Up Prompt
"The line is 3 units above the center, and the radius is 5. How long is half the chord? Now spin the picture around its vertical axis to make a hemisphere. What shape is the slice at that height, and what is its area?"
The radius to the end of the chord is the hypotenuse, so half the chord is √(25 - 9) = 4, and the slice of the hemisphere is a disk of area 16π. Ask: "Which solid we already know has a slice of area 16π at height 3?" Keep the question open; the lesson answers it.
Direct Instruction25 minutes
Reminder. Cavalieri's principle: two solids of equal height have equal volumes if, at every level, their cross-sections have equal areas. To use it, we need a partner solid whose volume we already know. Use Diagram 1 and build the argument in five steps:
Slice the hemisphere: a plane at height y above the flat face cuts a disk. Its radius s satisfies s² + y² = r² (the radius of the sphere is the hypotenuse), so the slice has area π(r² - y²).
Build the partner solid: take a cylinder with radius r and height r, and remove a cone whose apex is at the center of the bottom face and whose base is the top face. The partner has the same height r as the hemisphere.
Slice the partner: the cone widens from radius 0 at the bottom to radius r at the top, so by similar triangles its radius at height y is y. The slice is a ring with outer radius r and inner radius y, with area πr² - πy².
Apply Cavalieri's principle: π(r² - y²) = πr² - πy² for every y from 0 to r, so the hemisphere and the partner have equal volumes: πr² · r - (1/3)πr² · r = (2/3)πr³.
Double it: a sphere is two hemispheres, so V = (4/3)πr³. For the same r and height r, the cone, the hemisphere and the cylinder have volumes in the ratio 1 : 2 : 3.
Slice areas for r = 10
Height y
Hemisphere slice π(r² - y²)
Partner slice πr² - πy²
0
100π
100π
2
96π
96π
5
75π
75π
8
36π
36π
10
0
0
Other solids. The same plan works whenever we can write the slice area at height y and recognize it as the slice of a known solid. Work through the examples; the last two are solids that have no simple formula in the textbook.
Matching one slice
A hemisphere has a radius of 13 cm. Find the area of its slice 5 cm above the flat face, and the area of the partner solid's slice at the same height.
Equation: slice radius √(13² - 5²) = 12, so the area is 144π cm²; partner ring: 169π - 25π = 144π cm²
Volume of a sphere
Use the partner solid to find the volume of a sphere with a radius of 6 cm.
Equation: hemisphere: cylinder minus cone, 216π - 72π = 144π cm³, so the sphere is 288π ≈ 904.8 cm³
Archimedes' ratio
A cone, a hemisphere and a cylinder all have a radius of 3 in and a height of 3 in. Find the three volumes and their ratio.
Equation: cone 9π, hemisphere 18π, cylinder 27π in³, so the ratio is 1 : 2 : 3
Napkin ring (other solid)
A cylindrical hole is drilled straight through the center of a wooden ball with a radius of 5 cm, leaving a ring 8 cm tall. Find the volume of the ring.
Equation: hole radius √(25 - 16) = 3; slice at height y: π(25 - y²) - 9π = π(16 - y²), the slice of a sphere of radius 4, so V = (4/3)π(4³) = 256π/3 ≈ 268.1 cm³
Bicylinder (other solid)
Two cylinders with a radius of 2 cm cross at right angles through the same center. Find the volume of the solid common to both.
Equation: each slice is a square of side 2√(4 - y²), area 16 - 4y² = (cube slice 16) - (double pyramid slice 4y²), so V = 64 - 2 · (1/3)(16)(2) = 128/3 ≈ 42.7 cm³
Point out the pattern: in every example the argument never needed the shape of the slice, only its area. That is why a disk can be matched with a ring, and a square with a square minus a smaller square.
Guided Practice10-15 minutes
Pairs work on whiteboards. (a) A hemisphere has a radius of 17. Find the slice area 8 units above the flat face, both ways (225π). (b) Use the partner solid to find the volume of a sphere with a radius of 3 (36π). (c) A hemispherical mixing bowl has an inner radius of 9 cm. How much does it hold? (486π ≈ 1,526.8 cm³, about 1.5 liters.) (d) A cylinder with a radius of 4 and a height of 4 has a cone removed as in Diagram 1. Find its volume (128π/3) and name the solid with the same volume (a hemisphere of radius 4). Ask each pair to state the two conditions of Cavalieri's principle before they compute.
Independent Practice10 minutes
Students work alone. (a) Write the full Cavalieri argument for the volume of a sphere, with a labeled sketch of both solids and one slice. (b) Spherical cap: find the volume of the top 3 cm of a hemisphere with a radius of 6 cm, the part from height 3 to height 6. Use the partner solid between the same two heights: the cylinder part is π · 36 · 3 = 108π, the cone part is (1/3)π(6³ - 3³) = 63π, so the cap is 45π ≈ 141.4 cm³.
Closure5 minutes
Exit ticket: "In two sentences, explain why a hemisphere and a cylinder with a cone removed have the same volume. Then explain why the cone must have its apex at the bottom, not at the top." Look for the slice areas π(r² - y²) and πr² - πy², and for the idea that the hemisphere is widest at the bottom, where the hole must be smallest.
Differentiation Strategies
For Struggling Students
Give a slice table with the heights already filled in, so students only compute the two areas and compare them
Use a clay hemisphere cut with a plastic knife to show that each slice is a disk
Keep one worked slice, such as r = 5 and y = 3, visible on the board as a model
For Advanced Students
Derive the spherical cap formula V = πh²(3r - h)/3 from the partner solid, for a cap of height h
Explain why a napkin ring's volume depends only on its height, and find the volume of a ring 10 cm tall
Write the slice area A(y) = π(r² - y²) and describe how adding up thin slabs of this area leads to a calculus volume formula
Assessment Guidance
What to Look For
A complete argument has four parts: the slice of the hemisphere at height y and its area from the Pythagorean Theorem, the partner solid with the cone's apex at the bottom, the ring slice with inner radius y, and a statement of Cavalieri's principle with both conditions (equal heights, equal slice areas at every level). Watch for students who write the hemisphere slice radius as r - y, and for students who stop at (2/3)πr³ without doubling. For other solids, check that students name the partner solid, not only a formula.
02
Classroom Activities
3 Activities
1
Slice Tables
20 minPairs
Pairs fill a table of slice areas for a hemisphere and for its partner solid, then turn the numbers into a written argument. The table makes "equal slices at every level" concrete before students write it with variables.
Procedure
For a hemisphere with a radius of 5 cm, compute the slice radius and slice area at heights y = 0, 1, 2, 3, 4 and 5 cm. Expected areas: 25π, 24π, 21π, 16π, 9π and 0 cm²
For the partner solid (cylinder with radius 5 and height 5, cone removed), compute the outer and inner radius of the ring at the same heights and its area
Compare the two columns and write the general rule with r and y
Use Cavalieri's principle to find the volume of the hemisphere (250π/3 ≈ 261.8 cm³) and of the whole sphere
Discussion Questions
At which height are the slices largest? Where are they smallest?
Why does the table alone not prove the volumes are equal, and what does the rule with r and y add?
What would change if the cone had its apex at the top?
Modification for Distance Learning
Share a spreadsheet with columns for y, the hemisphere slice area and the partner slice area. Students enter formulas instead of numbers, change r, and post a sentence explaining why the two columns always agree.
2
Cone, Hemisphere, Cylinder Pour
15 minGroups of 3-4
Groups use a set of hollow solids with the same radius and a height equal to the radius to test Archimedes' ratio with rice. The experiment supports the argument; it does not replace it.
Procedure
Fill the cone with rice and pour it into the hemisphere. Count how many cones fill the hemisphere (2)
Pour one full cone and one full hemisphere into the cylinder. The cylinder should be full
Write the volumes with the formulas: (1/3)πr³, (2/3)πr³ and πr³, and check the ratio 1 : 2 : 3
Explain how the pour matches the partner solid: cylinder minus cone equals hemisphere
Discussion Questions
Why must the height of the cone and the cylinder equal the radius for this ratio?
Rice leaves small gaps. Why does the experiment still give good evidence?
How could you use the ratio to predict the volume of a sphere with a radius of 7 cm?
Modification for Classes Without Hollow Solids
Use clay: shape a hemisphere and a cone with the same radius and a height equal to the radius, and compare their masses on a kitchen scale. The hemisphere should weigh about twice as much as the cone.
3
Other Solids Stations
20 minGroups of 3-4
Groups rotate through four stations. At each one they describe a slice at height y, find its area, and name a partner solid with the same slice areas. This is the "other solid figures" part of the standard.
Stations
Station 1, napkin rings: a ring 6 cm tall is cut from a sphere with a radius of 5 cm (hole radius 4 cm). Show that each slice has area π(9 - y²) and that the ring has the volume of a sphere with a radius of 3 cm: 36π ≈ 113.1 cm³
Station 2, same height, different sphere: a ring 6 cm tall is cut from a sphere with a radius of 3.5 cm (hole radius about 1.8 cm). Explain why its volume is also 36π cm³
Station 3, bicylinder: sketch two crossing pipes and explain why each slice of the common solid is a square
Station 4, oblique cone: a stack of paper disks of shrinking size is pushed into a slant. Explain why it keeps the volume (1/3)πr²h of the straight cone
Challenge Variation
Going further: use the partner solid between heights r - h and r to derive the cap formula V = πh²(3r - h)/3, then check it with the cap from Independent Practice.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: A Hemisphere and Its Partner Solid
Left: the slice of the hemisphere at height y has radius √(r² - y²). Right: a cylinder with radius r and height r with a cone removed (apex at the bottom center). At height y the cone's radius is y, so the slice is a ring from y to r. Both slices have area πr² - πy², shown in the top views at half scale. Drawn with r = 110 and y = 66 units, so the disk radius is 88.
Diagram 2: Napkin Rings of the Same Height
Two rings of the same height h are left when holes are drilled through spheres of different radii. The dark segments show the slices at the same level y. Each ring slice is a ring from the hole to the sphere with area π((h/2)² - y²), the same as the slice of a sphere of diameter h, so all three solids have volume πh³/6. Drawn to scale with h = 120 and sphere radii 100, 75 and 60 units.
04
Homework Assignment
~30 min
HSG.GMD.A.2 Homework: Cavalieri's Principle for Spheres and Other Solids
Directions: For each solid, sketch one slice at height y, find its area, and name the partner solid you compare it with. State both conditions of Cavalieri's principle when you use it. Give exact answers in terms of π and decimals to the nearest tenth.
Part 1: The Sphere (Problems 1-3)
A hemisphere has a radius of 25 cm. Find the radius and area of its slice 7 cm above the flat face. Then find the area of the partner solid's slice (cylinder minus cone) at the same height and compare.
A globe is a sphere with a radius of 12 cm. Find the volumes of the cylinder and the cone that make up the partner solid of one hemisphere, then use Cavalieri's principle to find the volume of the globe.
A classmate builds the partner solid with the cone upside down: its apex is at the center of the top face. For a radius of 8, compare the slice areas of the hemisphere and of this solid at height y = 2. Explain why the argument fails with the cone this way up.
Part 2: Other Solids (Problems 4-6)
A cylindrical hole is drilled through the center of a sphere with a radius of 10 cm, leaving a napkin ring 16 cm tall. Find the radius of the hole, write the area of the ring's slice at height y, and find the volume of the ring.
A hemispherical bowl has an inner radius of 15 cm and holds water 6 cm deep at the center. The water fills the part of the hemisphere between heights 9 and 15 cm, measured from the flat face at the rim. Use the partner solid between the same heights to find the volume of the water.
Two pipes with a radius of 3 cm cross at right angles through the same center. Each slice of the solid common to both is a square. Find the side and area of the slice at height y, compare it with a cube with an edge of 6 cm that has two square pyramids removed, and find the volume of the common solid.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Slice Areas
Slice at height y described and its area correct
Area set up with one error
Slice missing or wrong
Partner Solid
Partner solid named and its slice matched at every level
Partner named, matching not shown
No partner solid
Cavalieri Argument
Both conditions stated and used
Only one condition stated
Principle not used
Volumes
All volumes correct, exact and rounded
One or two calculation errors
Many errors
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
A hemisphere with radius r is cut by a plane at height y above its flat face. What is the radius of the slice?
Answer: A
The sphere's radius r runs from the center to the edge of the slice, so it is the hypotenuse of a right triangle with legs y and the slice radius s: s = √(r² - y²). Choice B subtracts lengths instead of squares, which gives the wrong value at every height between the base and the top. Choice C adds the squares, which would be larger than r.
Question 2 of 20 · Multiple Choice
A hemisphere has a radius of 29 cm. What is the area of its slice 20 cm above the flat face?
Answer: B
The slice radius is √(29² - 20²) = √441 = 21, so the area is 441π cm². Choice A uses r - y = 9 as the radius. Choice C adds the squares: 841 + 400 = 1,241. Choice D uses y as the radius, which is the cone in the partner solid.
Question 3 of 20 · Multiple Choice
In the partner solid (a cylinder of radius r and height r with a cone removed, apex at the bottom center), what is the radius of the cone at height y?
Answer: C
The cone grows from radius 0 at the bottom to radius r at height r, so by similar triangles its radius at height y is y. Choice A describes a cone with its apex at the top. Choice D is the slice radius of the hemisphere, not of the cone.
Question 4 of 20 · Multiple Choice
What is the area of the partner solid's slice at height y?
Answer: D
The slice is a ring: the cylinder's disk of radius r minus the cone's disk of radius y, so its area is πr² - πy². Choice A ignores the hole. Choice C treats the ring as a disk of radius r - y, which has a smaller area.
Question 5 of 20 · Multiple Choice
Which two facts let you conclude that the hemisphere and the partner solid have equal volumes?
Answer: B
Cavalieri's principle needs equal heights and equal slice areas at every level. The slices do not have the same shape, a disk and a ring, so choice C is false and also not needed. Choice A mentions surface area, which does not control volume.
Question 6 of 20 · Multiple Choice
What is the volume of the partner solid for radius r?
Answer: C
The cylinder has volume πr² · r = πr³ and the cone has (1/3)πr³, so the partner has πr³ - (1/3)πr³ = (2/3)πr³, which is the volume of the hemisphere. Choice D is the whole sphere, two hemispheres. Choice B is only the cone that was removed.
Question 7 of 20 · Multiple Choice
A ball is a sphere with a radius of 5 in. What is its volume?
Answer: D
V = (4/3)π(5³) = (500/3)π ≈ 523.6 in³. Choice A is 4πr², the surface area, not the volume. Choice C is the volume of one hemisphere, before doubling.
Question 8 of 20 · Multiple Choice
A cone, a hemisphere and a cylinder all have radius r and height r. The cone holds 30 cm³ of rice. How much do the hemisphere and the cylinder hold?
Answer: A
The volumes are (1/3)πr³, (2/3)πr³ and πr³, a ratio of 1 : 2 : 3, so the hemisphere holds 2 · 30 = 60 cm³ and the cylinder 3 · 30 = 90 cm³. This matches the partner solid: cylinder minus cone equals hemisphere, 90 - 30 = 60. Choice C would make the cylinder four cones, but πr³ is three times (1/3)πr³. Choice B uses the ratio 1 : 3 : 4, and choice D uses 2 : 3 : 4, treating the cone as two parts.
Question 9 of 20 · Multiple Choice
Why must the removed cone have its apex at the bottom of the cylinder, where the hemisphere's flat face is?
Answer: C
With the apex at the bottom, the ring at height y has area πr² - πy², equal to the hemisphere slice. With the apex at the top, the ring would be πr² - π(r - y)², which is different, for example at y = 0 it is 0 instead of πr². Choice B is false: the height is r either way.
Question 10 of 20 · Multiple Choice
Two napkin rings are both 10 cm tall. One is cut from a sphere with a radius of 13 cm and the other from a sphere with a radius of 50 cm. Which ring has the larger volume?
Answer: B
At height y, each ring slice has area π(R² - y²) - π(R² - 5²) = π(25 - y²), which does not depend on R. Equal heights and equal slices give equal volumes, both (4/3)π(5³). Choice A is the intuitive guess, but the larger sphere also has a much wider hole. Choice D misses that the hole radius is fixed by R and the height.
Question 11 of 20 · Multiple Choice
A napkin ring is 18 cm tall. What is its volume?
Answer: A
Its slices match those of a sphere of diameter 18 cm, so V = (4/3)π(9³) = 972π ≈ 3,053.6 cm³. Choice B is only a hemisphere of radius 9. Choice C forgets the factor 1/3 in 4/3. Choice D is the common intuition that the slices disprove.
Question 12 of 20 · Multiple Choice
Two cylinders of radius r cross at right angles through the same center. What is the slice of the solid common to both at height y?
Answer: D
Each cylinder cuts the plane in a strip of width 2√(r² - y²), and the two strips cross at right angles, so the common slice is a square with that side. Choice A is the slice of the sphere of radius r, which fits inside the square. Choice C is the slice of the cube around the solid.
Question 13 of 20 · Multiple Choice
Solid A is a square prism with base area 50 cm² and height 12 cm. Solid B is also 12 cm tall, and every horizontal slice of B is a disk with an area of 50 cm². What is the volume of B?
Answer: A
The heights match and every slice has the same area, so by Cavalieri's principle B has the volume of A: 50 · 12 = 600 cm³. Choice D is the error of thinking the slices must have the same shape. Choice C multiplies by π again, but 50 cm² is already the slice area.
Question 14 of 20 · Multiple Choice
An oblique cone has a base area of 30 cm² and a height of 8 cm. What is its volume?
Answer: C
At every level its slices have the same areas as a right cone with the same base and height, so V = (1/3)(30)(8) = 80 cm³. Choice A leaves out the one-third. Choice B uses one half instead of one third.
Question 15 of 20 · Short Answer
Write a complete informal argument, using Cavalieri's principle, for the formula V = (4/3)πr³.
Slice a hemisphere of radius r at height y: the slice is a disk with radius √(r² - y²) (Pythagorean Theorem), so its area is π(r² - y²). Take a cylinder with radius r and height r and remove a cone with apex at the bottom center and base at the top; at height y the cone's radius is y, so the slice is a ring with area πr² - πy². The solids have the same height and equal slice areas at every level, so by Cavalieri's principle they have equal volumes: πr³ - (1/3)πr³ = (2/3)πr³. A sphere is two hemispheres, so V = (4/3)πr³.
Question 16 of 20 · Short Answer
A hemisphere has a radius of 20 cm. Find the area of its slice 12 cm above the flat face, and show that the partner solid's slice at the same height has the same area.
Slice radius: √(20² - 12²) = √256 = 16, so the area is 256π cm². Partner solid: outer radius 20, inner radius 12, so the ring has area 400π - 144π = 256π cm². The areas agree, as they must at every height.
Question 17 of 20 · Short Answer
A spherical water tank has a radius of 1.5 m. Use the partner solid to find its volume to the nearest tenth of a cubic meter.
For one hemisphere: the cylinder has π(1.5²)(1.5) = 3.375π m³ and the cone has one third of that, 1.125π m³, so the hemisphere has 2.25π m³. The tank is two hemispheres: 4.5π ≈ 14.1 m³, which agrees with (4/3)π(1.5³).
Question 18 of 20 · Short Answer
Explain why the volume of a napkin ring depends only on its height h and not on the radius of the sphere it was cut from.
Let the sphere have radius R. The hole reaches the sphere's surface at heights ±h/2, so its radius is √(R² - (h/2)²). At height y the ring slice is a ring from the hole to the sphere, with area π(R² - y²) - π(R² - (h/2)²) = π((h/2)² - y²). R cancels. This is the slice of a sphere with radius h/2, and the heights match, so by Cavalieri's principle every ring of height h has volume (4/3)π(h/2)³ = πh³/6.
Question 19 of 20 · Short Answer
A hemisphere has a radius of 5 cm. Find the volume of the cap between heights 3 cm and 5 cm, using the partner solid between the same heights.
The slices match at every height, so the cap equals the part of the partner solid between heights 3 and 5. Cylinder part: π(5²)(2) = 50π. Cone part: the cone up to height y has volume (1/3)πy³, so between 3 and 5 it is (1/3)π(125 - 27) = 98π/3. Cap: 50π - 98π/3 = 52π/3 ≈ 54.5 cm³.
Question 20 of 20 · Short Answer
Two cylinders with a radius of 5 cm cross at right angles through the same center. Use a cube with two square pyramids removed to find the volume of the solid common to both.
At height y the common slice is a square with side 2√(25 - y²), area 100 - 4y². A cube with an edge of 10 cm has slices of area 100, and two square pyramids with apex at its center (base 100 cm², height 5 cm) have slices of area 4y² at height y. So the slices match, and V = 1,000 - 2 · (1/3)(100)(5) = 1,000 - 1,000/3 = 2,000/3 ≈ 666.7 cm³.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.GMD.A.2 mean?
HSG.GMD.A.2 asks students to explain, using Cavalieri's principle, why a sphere has volume (4/3)πr³, and to use the same principle for other solids. The explanation is informal: it compares slices instead of using calculus.
What does the (+) in HSG.GMD.A.2 mean?
The (+) marks an advanced standard. Common Core describes these as additional mathematics for students who take advanced courses, such as calculus. HSG.GMD.A.2 is usually taught in honors Geometry, and some Geometry courses show the sphere argument as enrichment.
How does Cavalieri's principle give the volume of a sphere?
A hemisphere of radius r and a cylinder of radius r and height r with a cone removed have equal slice areas, πr² - πy², at every height y. So they have equal volumes, (2/3)πr³, and the sphere has twice that: (4/3)πr³.
Why do we remove a cone from the cylinder?
Because the hemisphere's slices shrink as y grows, and a cylinder's slices do not. Removing a cone with its apex at the bottom takes away a disk of area πy² at height y, which is exactly the amount the hemisphere's slice is missing compared with πr².
Is the Cavalieri argument a real proof?
It is a valid argument if Cavalieri's principle is accepted, and Common Core calls it informal because the principle itself is not proved in high school. Calculus proves it: the volume is the integral of the slice area, and equal slice areas give equal integrals.
Which other solids can you find with Cavalieri's principle?
Oblique prisms, cylinders, pyramids and cones; spherical caps, such as water in a round bowl; napkin rings left after drilling through a sphere; and the solid common to two crossing cylinders. In each case students match the slices with a solid whose volume they know.
What is the napkin ring problem?
If a cylindrical hole is drilled through the center of a sphere so that the remaining ring has height h, the ring's volume is πh³/6, whatever the size of the sphere. Its slices at every level have the same area as the slices of a sphere of diameter h, so Cavalieri's principle gives the result.
What mistakes do students make with HSG.GMD.A.2?
Common errors include:
writing the slice radius of the hemisphere as r - y instead of √(r² - y²)
putting the cone's apex at the top of the cylinder
stopping at (2/3)πr³ and forgetting to double for the whole sphere
mixing up the volume (4/3)πr³ with the surface area 4πr²
How is HSG.GMD.A.2 different from HSG.GMD.A.1?
HSG.GMD.A.1 covers informal arguments for the circle, the cylinder, the pyramid and the cone, using dissection, Cavalieri's principle and limits. HSG.GMD.A.2 builds on those results: it uses the cylinder and cone formulas inside a Cavalieri argument for the sphere and for other solids.
How does HSG.GMD.A.2 connect to calculus?
The slice area A(y) = π(r² - y²) is exactly what a calculus student integrates from 0 to r to get the volume of a hemisphere, (2/3)πr³. Comparing solids slice by slice is the idea behind the disk and washer methods, where the partner solid's ring is a washer.
07
Related Standards
5 standards
These standards connect to HSG.GMD.A.2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSG.GMD.A.1Prerequisite
Give informal arguments for circle, cylinder, pyramid and cone formulas
