Why volumes do add even though surface areas do not

What to assume they know

• Decomposing an everyday object into the basic solids — decomposing a composite object into basic solids and converting overall dimensions into part dimensions
• Which faces vanish at the join, and why you cannot simply add — why a joined solid's surface area falls short of what the two pieces' surfaces would give, the claim this topic is set against
• The Class IX volume formulas: cuboid, cylinder πr²h, cone (1/3)πr²h, sphere (4/3)πr³, and hemisphere (2/3)πr³
• That a cone of the same base and height as a cylinder has one third of its volume
• Cubic units, and converting between cubic centimetres and cubic metres
• That a plane region has area but occupies no space

What they should be able to do

• State the additivity rule for the volume of a joined solid and give the reason, not just the rule
• Explain why the same reason does not apply to surface area, in terms of what each quantity measures
• Compute the volume of a solid built from a cuboid and a fraction of a cylinder
• Subtract occupied space from a total volume to find the free space remaining
• Compute the volume of a hemisphere carrying a cone, and of the cylinder that just encloses it
• Show that when a cone's height equals its radius, the cone-on-hemisphere toy fills exactly half its circumscribing cylinder, and that this does not depend on the radius
• Give a volume as an exact multiple of π when the question asks for one, and as a decimal when it does not
• Recognise that the additive rule needs the pieces not to overlap, and say what would go wrong if they did

Where it usually goes wrong

• "Volumes add, areas do not — two rules to memorise." One rule, read on two measurements. The shared face is what changes hands, and it has area but not volume.
• "Then surface areas must add too, if you are careful enough." They cannot: the contact region belongs to both pieces' boundaries and to neither's afterwards. The chapter's own examples put a number on the discrepancy.
• "A half cylinder needs a formula of its own." It needs half the cylinder's. Any fraction of a solid contributes that fraction of its volume, because volume scales with the region.
• "The air in the shed is the shed's volume." Only until you put anything in it. The question deliberately asks twice — once for the enclosure, once for what is left after 300 m³ of machinery and 20 people.
• "The workers' 0.08 m³ is negligible, so ignore it." It is small — 1.6 m³ against 1128.75 — but the question asks for it and the answer changes. Deciding something is negligible is a decision to be stated, not assumed.
• "Example 7's difference came out equal to the toy by luck." It could not have come out otherwise. A cone whose height equals its radius, on a hemisphere of that radius, always fills exactly half its enclosing cylinder.
• "The enclosing cylinder is 2 cm high, like the cone." It has to clear the whole toy: the hemisphere contributes its radius below the join and the cone its height above, so 2 + 2 = 4 cm.
• "Give the answer in terms of π only when the arithmetic is ugly." Exercise 12.2 q. 1 asks for it explicitly, and the exact form π is what shows the structure. Read what the question wants.
• "A pole's second cylinder has diameter 8 cm, matching the first." The first gives a diameter of 24 cm and the second gives a radius of 8 cm. Mixed conventions inside one question are the most reliable source of wrong answers in this exercise.

Questions to check understanding

• Find the volume of a solid made of a cylinder with a cone or hemisphere at one or both ends, given an overall length
• Find the volume of an enclosure made of a cuboid with a half-cylindrical roof
• Subtract the space taken by contents from an enclosure's volume to find the air remaining
• Give a volume in terms of π when asked, and as a decimal otherwise
• Find the volume of the smallest cylinder that encloses a given composite solid, and the volume left over
• Convert a volume to a mass using a given density in grams per cubic centimetre
• Find a stated percentage of a computed volume, and multiply by a count of items
• Explain in words why volumes of joined solids add although surface areas do not

Examples worth working on the board

Inputs only. Values marked verified are worked out here on data printed inside pp. 161–170.

• The chapter's own turn (§12.3, opening paragraph, p. 167). Having spent §12.2 refusing to add surface areas, the chapter opens the volume section by recalling why it refused — part of the surface was lost when the pieces were joined — and then states that volume behaves differently and simply adds. The lost thing was a face, a face has area but no volume, so there is nothing for the volume sum to lose.
• The condition the rule needs. Volumes add when the pieces meet along a boundary face and do not overlap. Interiors that overlapped would be counted twice; that is the only way additivity can fail, and the chapter's objects never do it. Worth one slide, because it is what makes the rule a theorem rather than a slogan.
• Example 5, Shanta's shed (pp. 167–168, Fig. 12.12). Inputs: the base measures 7 m by 15 m; the cuboidal part is 8 m high; a half cylinder sits on top of it as a roof; π taken as 22/7. Then: machinery inside occupies 300 m³, and 20 workers occupy about 0.08 m³ each. Verified: the cuboid holds 15 × 7 × 8 = 840 m³; the half cylinder has diameter 7 m, so radius 3.5 m, and length 15 m, giving (1/2)(22/7)(3.5)²(15) = 288.75 m³; together 1128.75 m³. The workers occupy 20 × 0.08 = 1.6 m³, so the air left is 1128.75 − 301.6 = 827.15 m³. Read from the printed page, Fig. 12.12 marks 7 m across the front, 15 m along the side and 8 m up the wall, and the roof is drawn as a smooth barrel with no ridge.
• Why the half cylinder is legitimate. The rule was stated for whole solids, and the shed's roof is half of one. Nothing breaks: the half cylinder is itself a solid occupying space, and the flat rectangle where it meets the cuboid has no volume either. The half cylinder appears once in the whole chapter, here in Example 5, and in neither exercise. The only sub-whole solid the exercises use is the hemisphere, which the p. 170 summary lists among the five basic solids the chapter works with, so that one is not left unjustified in the same way.
• Example 7, the toy in its cylinder (p. 169, Fig. 12.14). Inputs: a hemisphere with a right circular cone standing on it; cone height 2 cm; base diameter 4 cm; π taken as 3.14. Then a right circular cylinder is drawn just enclosing the whole toy, and the difference of the two volumes is wanted. Verified: the shared radius is 2 cm; the toy measures (2/3)π(8) + (1/3)π(4)(2) = (16/3 + 8/3)π = 8π, which is 25.12 cm³ at π = 3.14; the enclosing cylinder has radius 2 cm and height 2 + 2 = 4 cm, so 16π = 50.24 cm³; the difference is 8π = 25.12 cm³ again. Read from the printed page: Fig. 12.14 letters the figure inside the artwork — A at the cone's apex, E and F at the top corners of the cylinder, B and C where the hemisphere's flat face meets the cylinder's wall, O at the centre, H and G at the bottom corners, P at the lowest point of the hemisphere.
• The identity behind that coincidence — this is the section the topic exists for. Verified as algebra: take any radius r and let the cone's height equal r, as it does here. Hemisphere (2/3)πr³ plus cone (1/3)πr³ is exactly πr³. The enclosing cylinder has radius r and height r + r = 2r, so its volume is 2πr³. The toy is therefore precisely half of it, whatever r is, and the leftover must equal the toy. The answer was never going to be anything else. Note also what makes it work: the hemisphere is two thirds of its own enclosing cylinder and the cone is one third of its own, and two thirds plus one third is one — the two Class IX fractions were built to fit together.
• Exercise 12.2 q. 1 (p. 169). Inputs: a hemisphere carrying a cone, both of radius 1 cm, the cone's height matching that radius; the answer is asked for in terms of π. Verified: (1/3)π + (2/3)π = π cm³ exactly. This is the identity of the previous item at r = 1, which is why the answer is a bare π.
• Exercise 12.2 q. 2 (p. 169). Inputs: a model shaped as a cylinder with a cone attached at each end, made of thin aluminium sheet; overall length 12 cm; diameter 3 cm; each cone 2 cm high; inner and outer dimensions to be treated as the same. Verified: radius 1.5 cm, cylinder length 12 − 4 = 8 cm, volume π(2.25)(8) + 2 × (1/3)π(2.25)(2) = 18π + 3π = 21π = 66 cm³ at π = 22/7. Three pieces, one rule.
• Exercise 12.2 q. 3, the gulab jamun (p. 170, Fig. 12.15). Inputs: each sweet is a cylinder with a hemisphere at each end, overall length 5 cm, diameter 2.8 cm; syrup fills about 30% of the volume; 45 of them. Verified: radius 1.4 cm, cylinder length 5 − 2.8 = 2.2 cm, one sweet measures π(1.96)(2.2) + (4/3)π(1.4)³ = π(4.312 + 3.65867) ≈ 25.05 cm³, so 45 of them come to about 1127.3 cm³ and the syrup to roughly 338 cm³. Read from the printed page, Fig. 12.15 shows a glass jar of stippled oblong sweets in liquid, and carries no measurements — the dimensions are in the question only.
• Exercise 12.2 q. 6, the iron pole (p. 170). Inputs: a cylinder 220 cm high of base diameter 24 cm carrying a second cylinder 60 cm high of radius 8 cm; iron masses about 8 g per cm³; π taken as 3.14. Verified: 3.14 × 144 × 220 = 99 475.2 cm³ and 3.14 × 64 × 60 = 12 057.6 cm³, so 111 532.8 cm³ in all, and the mass is about 892 262 g, near 892 kg. Note the trap the question sets: the first cylinder is given by diameter and the second by radius.
• Exercise 12.2 q. 4, q. 5 and q. 7 (p. 170). These three remove material or displace water rather than assemble pieces, and belong to When a piece has been scooped out: apparent capacity against actual. Name them here only to show that the additive rule and the subtractive one cover the whole exercise between them.
• Exercise 12.2 q. 8 (p. 170). Worth naming separately, because it is not a removal or a displacement: a spherical vessel with a cylindrical neck is a sum, a cylinder plus a sphere, exactly the rule this topic teaches. It is routed to When a piece has been scooped out: apparent capacity against actual for a different reason — it asks the student to test a reported measurement rather than to produce one — and the sibling brief computes it as a sum accordingly.

Figures to have open

• A single face shown as a plane region with a thickness slider running to zero: the area stays fixed, the volume goes to nothing. Standard schematic, not in the book, and it carries the whole thesis.
• Fig. 12.12's shed redrawn with the cuboid and the half cylinder separable, so the rectangle where they meet can be lifted out and shown to be flat. Chapter figure (p. 167).
• Fig. 12.14 redrawn: the toy and its circumscribing cylinder, with the lettering kept, since Example 7's solution refers to the points by name and the labels sit inside the artwork on the printed page (p. 169).
• A general-r version of the same picture with the halving shown as two equal stacked regions rather than as numbers. Standard schematic; not in the book, because the chapter works only the r = 2 case.
• A ledger graphic that can be reused across sections 6, 7 and 11: pieces down the left, volume of each, running total at the foot. Standard schematic.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 12 "Surface Areas and Volumes", §12.3 "Volume of a Combination of Solids", pp. 167–169, and in particular its opening paragraph on p. 167
• Example 5 with Fig. 12.12, pp. 167–168; Example 7 with Fig. 12.14, p. 169
• Exercise 12.2 questions 1 and 2, p. 169, and questions 3 and 6, p. 170, with Fig. 12.15 (p. 170)
• §12.2, pp. 162–166, for the surface-area behaviour this topic is contrasted with
• The chapter's summary, §12.4, p. 170, second point