Composite solids and volume-preserving conversions

Learning outcome

Separate composite solids, calculate exposed area, and conserve volume when counting recast objects and leftover material.

Concepts and assumptions

Composite volumes add for parts without overlap; cavities subtract volume. Joining faces have no thickness or volume.

Adding separate total areas counts every contact patch twice, although both copies become hidden. Subtract both. Count remaining outside faces and requested cavity walls; drilling can reduce volume yet increase area.

Recasting conserves volume, not area, assuming no material loss or volume change. Divide original volume by one new object’s volume. Objects must be complete: never round a nonintegral count up. Take the greatest whole count and retain leftover material; conversion into only complete objects is then impossible.

Use exact centimetre dimensions and π = 22/7 as the prescribed approximation, without further rounding. Joins are flush, without overlaps or glue thickness; all other outside faces remain exposed.

Worked examples

Example 1 — Joined cubes. Two cubes of edge 4 cm join along a complete face. Volume = 2 × 4³ = 128 cm³. Separate areas total 2 × 6 × 4² = 192 cm². Subtract both hidden faces: exposed area = 192 - 2 × 16 = 160 cm².

Example 2 — Cone on cylinder. A cylinder has radius 7 cm and height 10 cm. A right cone of equal radius and perpendicular height 24 cm covers its top; the bottom stays exposed. Base area = π × 7² = 154 cm²; cone slant height = √(49 + 576) = 25 cm. Volume = 154 × 10 + 154 × 24 ÷ 3 = 2772 cm³. Curved areas are 2 × π × 7 × 10 = 440 cm² and π × 7 × 25 = 550 cm². Adding the bottom gives 440 + 550 + 154 = 1144 cm².

Example 3 — Recasting. A sphere of radius 6 cm becomes cones of radius 3 cm and height 4 cm without loss. Sphere volume = (4/3) × π × 6³ = 288 × π cm³. Each cone needs π × 3² × 4 ÷ 3 = 12 × π cm³. Count = 288 ÷ 12 = 24 complete cones; π cancels.

Common mistakes

Do not conserve area during melting, subtract only one hidden contact face, or discard unaccounted material.

Practice questions

1. Two cubes with 6 cm edges join at one full face. Find volume and exposed area.
2. A cylinder (radius 7 cm, height 10 cm) has an equal-radius hemisphere covering its top. Find exposed area including the bottom.
3. A 20 cm × 15 cm × 10 cm cuboid has a central cylindrical hole of radius 3.5 cm drilled perpendicularly through its 10 cm height. Find remaining volume and exposed area, including the hole wall.
4. Recast a 17 cm × 8 cm × 5 cm cuboid into cubes of edge 4 cm without loss. Find maximum complete count and leftover volume.

Worked solutions

1. Volume = 2 × 6³ = 432 cm³. Area = 2 × 6 × 6² - 2 × 6² = 360 cm².
2. Cylinder wall = 2 × (22/7) × 7 × 10 = 440 cm². Hemisphere curved area = 2 × (22/7) × 7² = 308 cm²; bottom = 154 cm². Exposed area = 440 + 308 + 154 = 902 cm².
3. Hole volume = (22/7) × 3.5² × 10 = 385 cm³. Remaining volume = 20 × 15 × 10 - 385 = 2615 cm³. Outside area = 2 × (300 + 150 + 200) = 1300 cm². Remove two disks of 38.5 cm² each and add hole wall 2 × (22/7) × 3.5 × 10 = 220 cm². Area = 1300 - 77 + 220 = 1443 cm².
4. Original volume = 17 × 8 × 5 = 680 cm³; each cube needs 4³ = 64 cm³. Ten cubes use 640 cm³, leaving 40 cm³. All material cannot form only complete cubes of this size.