Compound solids and hidden join faces
Add or subtract volumes while counting only the surfaces exposed after assembly.
Core and Extended. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Use cylinder, cone and hemisphere formulas.
- Read the required boundary in a context.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many copies of a shared join circle should be excluded when adding two closed component areas?
What you will learn
- Calculate joined-solid and hollow-solid volumes.
- Exclude internal contact faces from outside surface area.
Volumes and exposed areas need different lists
For components joined without overlap, add their volumes. A cylinder of radius 3 cm and height 10 cm topped by a radius-3 hemisphere has volume 90π + 18π = 108π cm³. The shared circular face has no thickness, so it contributes no extra volume and is not subtracted from these volumes.
Outside area excludes both copies of the joined circular face. This capsule has cylinder wall 60π cm², hemisphere curve 18π cm² and one bottom circle 9π cm², totalling 87π cm². Adding the closed surface areas of both components would incorrectly include two internal circles.
Holes remove material and introduce walls
A cylindrical hole drilled through a cuboid removes a volume πr²h. It also removes the two circular opening areas from the cuboid's surfaces and creates the hole's curved wall 2πrh. If all exposed surfaces are requested, that inner wall counts; 'outside only' or a coating instruction may specify a different boundary.
For a 10 by 8 by 6 cm cuboid with a radius-2 hole through the 6 cm height, remaining volume is 480 − 24π cm³. Its original area is 376 cm²; removing 8π cm² of openings and adding 24π cm² of wall gives total exposed area 376 + 16π cm². The hole fits within the base, which is an essential physical check.
Pause and explain
A radius-3 cylinder of height 10 is topped by a matching hemisphere. Which exposed area includes the bottom?
Worked example
A radius-3 cm cylinder of height 10 cm is topped by a matching hemisphere. Find volume and exposed area including the bottom.
Show the worked solution
- Volumes add: 90π + 18π = 108π cm³.
- The joined circular faces are hidden and excluded from exposed area.
- Wall, hemisphere curve and bottom give 60π + 18π + 9π = 87π cm².
Answer Volume 108π cm³; exposed area 87π cm².
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A radius-3 cm cylinder of height 5 cm is topped by a matching right cone of height 4 cm. Find volume and exposed area including the bottom.
Give me a hint
The cone slant height is five; both joined circles are hidden.
Compare my reasoning
- Volumes add to 45π + 12π = 57π cm³.
- Exposed curved areas are 30π and 15π cm².
- Add one bottom circle 9π: exposed area 54π cm².
Volume 57π cm³; exposed area 54π cm².
Look for these in your work
- Used perpendicular height for cone volume.
- Excluded the contact circles.
A radius-2 cm hole passes through the 6 cm height of a 10 by 8 by 6 cm cuboid. Find remaining volume and all exposed surface area.
Give me a hint
Subtract hole volume; include the new inside wall.
Compare my reasoning
- Remaining volume is 480 − 24π cm³.
- Original cuboid area is 376 cm², and two circular openings remove 8π cm².
- The hole wall adds 24π cm², leaving area 376 + 16π cm².
Volume (480 − 24π) cm³; exposed area (376 + 16π) cm².
Look for these in your work
- Included the hole wall.
- Removed exactly two opening faces.
A tank has a radius-2 m cylindrical part of height 3 m and a matching hemispherical bottom. Give its ideal capacity, ignoring thickness.
Give me a hint
The bottom contributes volume even though its join face is hidden.
Compare my reasoning
- Cylinder capacity is 12π m³.
- Hemisphere capacity is 16π/3 m³.
- Total is 52π/3 m³, or 52 000π/3 litres.
52π/3 m³ = 52 000π/3 litres.
Look for these in your work
- Added both component volumes.
- Converted cubic metres to litres after summing.
Common mistakes
- Subtracting join-face areas from volume.
- Keeping hidden faces in exposed area.
- Removing hole volume but forgetting its new wall area.
What can you explain now?
Does a hidden join face contribute to outside surface area?
Compare with the explanation
No.
Both component contact faces become internal after assembly and are omitted from the exposed boundary.
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Make it stick
Tomorrow, list the volume pieces and exposed surfaces of one compound solid separately before calculating.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.