Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.
Official wording from the Common Core State Standards for Mathematics (© 2010 National Governors Association Center for Best Practices and Council of Chief State School Officers). View on thecorestandards.org
Rounded solids complete the volume work begun with prisms. A cylinder is a prism with a circular base, so its volume is base area times height, V = πr²h. A cone with the same base and height holds exactly one third as much, V = (1/3)πr²h, something students can verify by pouring rice or water from a cone into a matching cylinder three times. A sphere of radius r has volume V = (4/3)πr³.
Students use these formulas to solve practical problems: how much a can holds, how many scoops of ice cream fit in a cone, or how much air is in a ball. Success depends on careful reading of the dimensions. Many problems give a diameter instead of a radius, mix units, or ask for a missing dimension when the volume is known. Answers can be left in terms of π (such as 36π cubic units) for exactness or approximated using 3.14. The relationships between the three shapes, such as a sphere filling two thirds of the cylinder that just encloses it, give students a way to check that answers are sensible.
A can 8 cm across has a radius of 4 cm. Using 8 makes the volume four times too large, because the radius is squared.
A cone's volume is not πr²h. Without the 1/3 the answer is the volume of the matching cylinder.
The sphere formula uses r³. Writing (4/3)πr² gives a number with the wrong units and far too small a value.
The height in the cone formula is the perpendicular height from the base to the tip, not the length along the sloping side.
A cylinder and a cone each have a radius of 3 cm and a height of 10 cm. Find each volume in terms of π, then approximate the cone's volume using 3.14.
Answer: Cylinder 90π cm³; cone 30π cm³, about 94.2 cm³.
Pouring activities with hollow plastic solids make the one-third and two-thirds relationships memorable. Follow up with real objects students can measure, like cans, cups and balls, and compare their calculated volume with the label. Building a formula card that shows how each formula connects to base area times height helps students reconstruct formulas rather than memorize three unrelated ones.
Assessments often give a diameter, ask for an answer in terms of π or rounded to the nearest tenth, and include a missing-dimension question such as finding a cylinder's height from its volume. Reinforce reading every given measurement twice before substituting.
Original questions written for this standard. Choose an option or type your answer, then press Check. Every question has a worked explanation.
Answer: B) 20π cm³
V = πr²h = π × 4 × 5 = 20π cm³.
Answer: C) 36π in³
V = (4/3)πr³ = (4/3) × π × 27 = 36π in³.
Answer: 263.76
V = (1/3) × 3.14 × 36 × 7 = (1/3) × 791.28 = 263.76 cm³.
Answer: 942
Radius = 10 ÷ 2 = 5 cm. V = 3.14 × 25 × 12 = 942 cm³.
Answer: 8
72π = π × 9 × h, so h = 72 ÷ 9 = 8 units.
Answer: C) 200 mL
A cone holds one third of the matching cylinder: 600 ÷ 3 = 200 mL.
Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.
A full lesson with slides, activities and an exit ticket on volume of cylinders, cones and spheres, pitched to grade 8 and editable in PowerPoint or Google Slides.
Make a lesson →A printable, differentiated worksheet on 8.G.C.9 with an answer key, ready in about a minute.
Make a worksheet →Turn volume of cylinders, cones and spheres into a quiz students answer online that marks itself, with a class summary for you.
Build a test →The standard says students should know the formulas. Many state tests also supply a formula sheet, but understanding where each comes from makes them easier to remember.
Both are used. In terms of π is exact; decimals using 3.14 or a calculator's π key are approximations. Follow the instructions in the question.