math-2017:G.GMD.3
math-2017:G.GMD.3 Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.
Learning goal: Use the volume formulas for cylinders, pyramids, cones and spheres to solve problems, including composite solids, reverse problems that ask for a missing dimension, and problems that convert between cubic units and capacity, always with the perpendicular height, the radius rather than the diameter, and a stated unit.
The formulas and their conditions. A cylinder with base radius r and perpendicular height h has volume V = πr²h. A pyramid with base area B and perpendicular height h has V = ⅓Bh. A cone with base radius r and perpendicular height h has V = ⅓πr²h. A sphere of radius r has V = (4/3)πr³, and a hemisphere has half that, ⅔πr³. The conditions are not decoration. Height means the perpendicular distance from the apex or top face to the plane of the base; a slant height, the distance along the surface of a cone or a pyramid's face, is longer and must be converted with the Pythagorean Theorem before use. Radius means half the diameter; a problem that gives "a can 10 cm across" is giving a diameter. The formulas hold for oblique solids as well as right ones, by Cavalieri's principle, provided the perpendicular height is used.
Exact and approximate answers, and units. Keep π in the answer until the last step. A cylinder of radius 4 cm and height 10 cm has volume π × 16 × 10 = 160π cm³ exactly, and about 502.65 cm³ when π is replaced by a calculator value. State the rounding: "≈ 502.65 cm³ to two decimal places" or "about 503 cm³". Volume units are cubic because three lengths were multiplied: cm³, m³, in³, ft³. Capacity units are linked exactly to some of these: 1 mL = 1 cm³, 1 L = 1,000 cm³, and 1 m³ = 1,000 L. So the cylinder above holds 160π mL ≈ 0.503 L. A common error is to convert the lengths after computing the volume as if volume converted like length; 1 m = 100 cm but 1 m³ = 1,000,000 cm³, because each of the three factors is converted.
A workflow that avoids most mistakes. First name the shape or shapes that model the object. Second, write down every measurement with its unit and decide whether it is a radius, a diameter, a perpendicular height or a slant height. Third, write the formula with the conditions checked. Fourth, substitute and compute an exact answer with π. Fifth, approximate, round as the situation demands, and attach the unit. Sixth, sanity-check the size: a coffee mug holds a few hundred milliliters, a bathtub holds a couple of hundred liters, a room holds tens of cubic meters.
Composite solids. When an object is made of solids that touch but do not overlap, add the volumes. A grain silo that is a cylinder of radius 3 m and height 10 m topped by a hemisphere of radius 3 m has volume 90π + 18π = 108π ≈ 339.29 m³: the cylinder contributes π × 9 × 10 = 90π and the hemisphere ⅔ × π × 27 = 18π. When a solid has a hole or a piece removed, subtract. A frustum, the shape of a paper cup, is a large cone with a smaller similar cone cut from its top; the missing cone's height comes from similar triangles. A cup has top radius 4 cm, bottom radius 2 cm and height 9 cm. If the full cone (extended below the cup) had height H, then by similarity H/4 = (H − 9)/2, so 2H = 4H − 36 and H = 18 cm. The large cone has volume ⅓ × π × 16 × 18 = 96π cm³, the removed cone has height 9 cm and radius 2 cm, volume ⅓ × π × 4 × 9 = 12π cm³, and the cup holds 84π ≈ 263.89 cm³, about 264 mL.
Reverse problems. When the volume is known and a dimension is not, substitute what is known and solve. A cone holds 100π cm³ and is 12 cm tall; then ⅓ × π × r² × 12 = 100π, so 4πr² = 100π, r² = 25 and r = 5 cm (the negative root is not a length). A cylinder of radius 5 cm holding 200π cm³ has 25πh = 200π, so h = 8 cm. A sphere with volume 36π cm³ has (4/3)πr³ = 36π, so r³ = 27 and r = 3 cm.
A worked model inside the lesson. A cylindrical rain tank has inside radius 1.2 m and height 2.5 m. Volume: π × 1.44 × 2.5 = 3.6π m³ ≈ 11.31 m³. Since 1 m³ = 1,000 L, the tank holds about 11,310 L. If a pump removes 60 L per minute, emptying a full tank takes 11,310 ÷ 60 ≈ 188.5 minutes, a little over three hours. The height is perpendicular because the tank stands upright on flat ground, and the radius was given, not the diameter, so no halving was needed; both checks belong in a written solution.
Misconceptions, and why they are wrong. Using the diameter in place of the radius multiplies a cylinder's or cone's volume by 4 and a sphere's by 8, since r² and r³ are involved. Using the slant height instead of the height makes a cone's volume too large. Leaving out the ⅓ reports the surrounding cylinder or prism. Writing cm² for a volume confuses area with volume. Adding the volumes of overlapping parts counts the overlap twice; for example, a cone sitting inside a cylinder is not a cylinder plus a cone. And converting 3.6π m³ to liters by multiplying by 100, as if it were a length in meters, is wrong by a factor of 10; the correct factor is 1,000.
Worked composite capsule: A medicine capsule is a cylinder of radius 2 mm and length 6 mm with a hemisphere of radius 2 mm on each end. The two hemispheres make one sphere of radius 2 mm, volume (4/3) × π × 8 = 32π/3 mm³. The cylinder has volume π × 4 × 6 = 24π mm³. Total 24π + 32π/3 = 72π/3 + 32π/3 = 104π/3 ≈ 108.91 mm³, which is about 0.109 mL since 1 mm³ = 0.001 cm³ and 1 cm³ = 1 mL. The parts touch along circles and do not overlap, so adding is correct.
Worked cone with a slant height: A funnel shaped as a right circular cone has diameter 6 cm and slant height 5 cm. Radius 3 cm. Height from the right triangle with hypotenuse 5 and leg 3: √(25 − 9) = √16 = 4 cm. Volume ⅓ × π × 9 × 4 = 12π ≈ 37.70 cm³, about 37.7 mL. Using the slant height 5 would give 15π ≈ 47.12 cm³, too large, and using the diameter 6 as the radius would give ⅓ × π × 36 × 4 = 48π cm³, four times the correct 12π; both errors together give 60π, five times the correct value.
Worked square pyramid and a comparison: A hypothetical monument is a right square pyramid with base 30 m by 30 m and height 20 m. V = ⅓ × 900 × 20 = 6,000 m³. A cone with the same height whose base has the same area 900 m² would also have volume 6,000 m³, whatever its radius (here r = √(900/π) ≈ 16.93 m), because the volume of a pyramid or cone depends only on base area and height.
Worked filling-rate problem: A cylindrical drum of radius 20 cm and height 50 cm is filled by a hose delivering 4 L per minute. Volume π × 400 × 50 = 20,000π cm³ ≈ 62,832 cm³ = 62.83 L (dividing by 1,000). Time: 62.83 ÷ 4 ≈ 15.7 minutes. A solid ball of radius 11 cm that sinks (it is denser than water) dropped into the full drum would displace (4/3) × π × 1,331 ≈ 5,575 cm³ ≈ 5.58 L of water, which overflows.
Formula and comparison record: Four containers all have the same key measurement of 10 cm: a cylinder of radius 10 cm and height 30 cm, a cone of radius 10 cm and height 30 cm, a sphere of radius 10 cm and a square pyramid with base 20 cm by 20 cm and height 30 cm. Make a table with columns shape, formula with its conditions, substitution, exact volume in cm³, volume in liters to two decimal places. The entries should be 3,000π cm³ = 9.42 L, 1,000π cm³ = 3.14 L, (4,000/3)π cm³ = 4.19 L and 4,000 cm³ = 4.00 L. Below the table explain why the cone is exactly one third of the cylinder and why the pyramid's entry has no π.
Independent performance: Choose a real object that is a composite of two of the four solids or a solid with a piece removed (a silo, a capsule, a funnel with a cylindrical spout, a pencil with a conical tip, a bowl). Measure or estimate its dimensions with units, draw a labeled sketch, and compute its volume. Success criteria: each part's formula is written with its conditions; any diameter is halved and any slant height is converted to a perpendicular height with a shown Pythagorean calculation; the exact volume with π is given before the decimal; the decimal is rounded to a stated precision and converted to milliliters or liters correctly (1 cm³ = 1 mL, 1,000 cm³ = 1 L); one reverse question is posed and solved, such as the height a cylinder of the same radius would need to hold the same volume; and one sentence judges whether the answer is a sensible size for the object.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
A cylindrical water tank has inside diameter 2 m and height 1.5 m. Find its capacity in liters.
The radius is half the diameter, 1 m. V = π × 1² × 1.5 = 1.5π m³ ≈ 4.712 m³. Since 1 m³ = 1,000 L, the tank holds about 4,712 L. Using the diameter as the radius would have given 6π m³, four times too large.
A conical pile of sand has diameter 8 m and height 3 m. Find its volume.
Radius 4 m. V = ⅓ × π × 16 × 3 = 16π ≈ 50.27 m³. The height 3 m is the perpendicular height from the ground to the top of the pile, which is what the formula needs.
A cone has volume 100π cm³ and height 12 cm. Find its radius and explain why only one value is acceptable.
⅓ × π × r² × 12 = 100π gives 4πr² = 100π, so r² = 25. The equation has solutions 5 and −5, but a radius is a length and must be positive, so r = 5 cm. Check: ⅓ × π × 25 × 12 = 100π.
A paper cup is a frustum with top radius 4 cm, bottom radius 2 cm and height 9 cm. Find how much it holds.
Extend the cup to a full cone of height H. Similar triangles give H/4 = (H − 9)/2, so 2H = 4H − 36 and H = 18 cm. Large cone: ⅓ × π × 16 × 18 = 96π cm³. Removed cone (height 9 cm, radius 2 cm): ⅓ × π × 4 × 9 = 12π cm³. Cup: 84π ≈ 263.89 cm³, about 264 mL since 1 cm³ = 1 mL.
A silo is a cylinder of radius 3 m and height 10 m with a hemispherical roof of radius 3 m. Find its total volume.
Cylinder: π × 9 × 10 = 90π m³. Hemisphere: ⅔ × π × 27 = 18π m³. The parts meet along a circle and do not overlap, so the total is 108π ≈ 339.29 m³, which is about 339,000 L.
A student computes the volume of a right circular cone with diameter 6 cm and slant height 5 cm as ⅓ × π × 6² × 5 = 60π cm³. Identify both errors and give the correct volume.
The radius is 3 cm, not 6 cm, and 5 cm is the slant height, not the height. The height is √(5² − 3²) = 4 cm. Correct volume ⅓ × π × 9 × 4 = 12π ≈ 37.70 cm³. The student's answer is five times too large: four times from squaring the diameter and 5/4 from the slant height.
A glass paperweight is a right square pyramid with base edges 10 cm. The slant height of each triangular face, measured from the midpoint of a base edge up the face to the apex, is 13 cm. Find the volume of glass, and say what answer using the slant height as the height would give.
The base area is B = 10 × 10 = 100 cm². The perpendicular height runs from the apex to the center of the base, which is 5 cm from the midpoint of each base edge, so the height, the 5 cm half-edge and the 13 cm slant height form a right triangle: h = √(13² − 5²) = √144 = 12 cm. V = ⅓Bh = ⅓ × 100 × 12 = 400 cm³. Using the slant height gives ⅓ × 100 × 13 ≈ 433.33 cm³, too large, because the slant height is longer than the perpendicular height. There is no π because the base is a square.
Which holds more: a sphere of radius 6 cm or a cone of radius 6 cm and height 24 cm? Justify with exact volumes.
Sphere: (4/3) × π × 216 = 288π cm³. Cone: ⅓ × π × 36 × 24 = 288π cm³. They hold exactly the same, 288π ≈ 904.78 cm³; a cone with the same radius as a sphere needs height 4r to match it, since ⅓πr²(4r) = (4/3)πr³.