Solve real-world and mathematical problems involving volumes of cylinders, cones and spheres

Learning goal: State the volume formulas for a cylinder, a cone and a sphere with their conditions, use them to solve real-world and mathematical problems, give answers in exact form with π and as decimal approximations with correct cubic units, and work backwards from a known volume to a missing dimension.

Volume and its units. Volume measures the space inside a solid, in cubic units: cubic centimeters (cm³), cubic meters (m³), cubic inches. When every length in a formula is in centimeters, the volume is in cubic centimeters, because the formula multiplies three lengths together. Report the unit every time.

The three formulas and their conditions.

Cylinder: V = πr²h, where r is the radius of the circular base and h is the perpendicular height between the two bases. The reasoning is the same as for a prism: the base area, πr², multiplied by the height, because the cylinder is a stack of identical circles.

Cone: V = ⅓πr²h, where r is the radius of the base and h is the perpendicular height from the base to the apex. The cones here are right cones, with the apex directly above the center of the base; the volume formula holds for any cone with the perpendicular height, but the slant-height link below needs the right cone. A cone with the same base and the same height as a cylinder holds exactly one third as much; filling a cone with water three times exactly fills the matching cylinder.

Sphere: V = (4/3)πr³, where r is the radius. Only one measurement is needed, because every point on the surface is the same distance from the center. A sphere of radius r fits exactly inside a cylinder of radius r and height 2r, and its volume is two thirds of that cylinder's volume.

The conditions matter. The height in the cone formula is the perpendicular height, not the slant height along the side. The radius is half the diameter; a problem that gives a diameter must be halved first. Using a diameter as a radius makes a cylinder or cone volume four times too large and a sphere volume eight times too large.

Exact and approximate forms. π is irrational, so a volume such as 90π cm³ is exact, and any decimal is an approximation. Keep π in the answer until the last step, then multiply once by 3.14, or by a calculator's value of π, and round to a stated precision. Rounding π to 3 early, or rounding a middle step, changes the answer.

A fully worked model with all three solids. Take radius 3 cm and, for the cylinder and cone, height 10 cm.

Cylinder: V = π × 3² × 10 = π × 9 × 10 = 90π cm³ ≈ 282.7 cm³.

Cone: V = ⅓ × π × 9 × 10 = 30π cm³ ≈ 94.2 cm³, which is one third of 90π.

Sphere: V = (4/3) × π × 3³ = (4/3) × π × 27 = 36π cm³ ≈ 113.1 cm³. Check against the two-thirds rule: the cylinder that just holds this sphere has radius 3 and height 6, volume π × 9 × 6 = 54π, and two thirds of 54π is 36π. The diagram compares the three values.

The step students skip is 3³ = 27, not 9: the sphere formula cubes the radius, while the cylinder and cone formulas square it and multiply by a separate height.

A real-world problem. A cylindrical water tank has diameter 4 m and height 3 m. The radius is 2 m. V = π × 2² × 3 = 12π m³ ≈ 37.70 m³. Since 1 m³ holds 1,000 liters, the tank holds about 37,700 liters. The reasoning step to write down is the halving of the diameter; the wrong calculation π × 4² × 3 = 48π gives four times too much.

When the height is not given directly. A cone might be described by its radius and slant height. Then, for a right cone, the Pythagorean Theorem supplies the height: radius, height and slant height form a right triangle with the slant height as hypotenuse, because the height meets the base at its center. A cone of radius 5 cm and slant height 13 cm has height √(169 − 25) = √144 = 12 cm, and volume ⅓ × π × 25 × 12 = 100π cm³ ≈ 314.16 cm³.

Working backwards. If a cylinder has volume 150π cm³ and radius 5 cm, then 150π = π × 25 × h, so h = 150π ÷ 25π = 6 cm. Dividing both sides by 25π leaves the unknown alone. The same reversal works for a cone (divide by ⅓πr²) and, with a cube root, for a sphere.

Composite solids. A grain silo shaped like a cylinder of radius 2 m and height 7 m with a hemisphere (half a sphere) of radius 2 m on top has volume equal to the cylinder plus half a sphere: π × 4 × 7 = 28π, plus ½ × (4/3) × π × 8 = 16π/3. Total 28π + 16π/3 = 84π/3 + 16π/3 = 100π/3 m³ ≈ 104.72 m³. Add volumes only when the solids do not overlap.

Misconceptions, and why they are wrong. Using the slant height of a cone in the formula gives too large a volume, because the slant is longer than the perpendicular height. Forgetting the ⅓ for a cone reports the volume of the surrounding cylinder. Writing cm² for a volume confuses area with volume; three lengths were multiplied. Adding volumes of solids that overlap counts the shared region twice.

Worked cylinder and cone from a diameter: A can has diameter 10 cm and height 4 cm. Radius 5 cm. Cylinder volume: π × 25 × 4 = 100π cm³ ≈ 314.16 cm³. A separate cone with radius 4 cm and height 9 cm: V = ⅓ × π × 16 × 9 = ⅓ × 144π = 48π cm³ ≈ 150.80 cm³. In each case the exact form keeps π; the decimal uses a calculator value of π and is rounded to two places.

Worked sphere from a diameter: A ball is 24 cm across, so its radius is 12 cm. V = (4/3) × π × 12³ = (4/3) × π × 1728 = 2304π cm³ ≈ 7,238.2 cm³. The cube is 12 × 12 × 12 = 1,728, and (4/3) × 1,728 = 2,304. A student who used r = 24 would get (4/3) × π × 13,824 = 18,432π, eight times too large, because doubling the radius multiplies r³ by 2³ = 8.

Worked cone with a slant height, then a missing height: A cone has radius 6 cm and slant height 10 cm. Height: √(100 − 36) = √64 = 8 cm. Volume: ⅓ × π × 36 × 8 = 96π cm³ ≈ 301.59 cm³. Using the slant height 10 instead would give 120π, too large. Separately, a cylinder with volume 200π cm³ and radius 5 cm has height h with 200π = 25πh, so h = 8 cm.

Formula and comparison record: Make a table with one row for each of a cylinder (r = 3 cm, h = 10 cm), a cone (r = 3 cm, h = 10 cm) and a sphere (r = 3 cm). Columns: formula with its conditions, substitution, exact volume with π, decimal volume to one place, and unit. Confirm that the cone entry is one third of the cylinder entry (30π and 90π) and that the sphere entry (36π) is two thirds of a cylinder of radius 3 and height 6 (54π). Then repeat the table for a diameter of 6 cm instead of a radius of 3 cm and explain why the numbers do not change.

Independent performance: Design a container problem: choose a real object shaped like a cylinder, a cone, a sphere or a combination of two, state its measurements with units, and find its volume. Success criteria: the formula is written with r and h identified on a labeled sketch, including the perpendicular height if a slant height is given; any diameter is halved before use; the exact volume is given with π and the decimal is rounded to a stated precision; the unit is cubic; and one reverse question is included, such as the height a cylinder of the same radius would need to hold the same volume, solved by dividing both sides.