Chapter 18 — Mensuration

A cuboid is a solid with six rectangular faces, such as a brick, a matchbox or a book. Its three dimensions are the length l, the breadth b and the height h. The six faces come in three pairs: two of area lb, two of area bh and two of area hl. So

Total surface area (TSA) of a cuboid = 2(lb + bh + hl).

The lateral surface area (LSA) is the area of the four vertical faces only, the walls of a room without the floor and the ceiling: LSA = 2(l + b)h = perimeter of base × height. The volume is the space the cuboid occupies: V = l × b × h, which is base area × height. The diagonal of a cuboid has length √(l² + b² + h²).

A cube is a cuboid whose length, breadth and height are equal, say a. Then TSA = 6a², LSA = 4a², V = a³, and the diagonal is a√3.

Units. Surface area is measured in square units (cm², m²) and volume in cubic units (cm³, m³). Capacity of a container is the volume of liquid it holds; 1 litre = 1000 cm³, 1 m³ = 1000 litres. Converting units of volume needs care: 1 m = 100 cm, so 1 m³ = 100³ = 1,000,000 cm³, not 100 cm³.

Worked example 1. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. Its capacity is 6 × 5 × 4.5 = 135 m³ = 135,000 litres. The area of the tin sheet needed to make it (open at the top) is the base plus the four walls: 6 × 5 + 2(6 + 5) × 4.5 = 30 + 99 = 129 m².

Worked example 2. The walls and ceiling of a room 5 m × 4 m × 3 m are to be painted at ₹20 per m². Area = LSA + ceiling = 2(5 + 4) × 3 + 5 × 4 = 54 + 20 = 74 m². Cost = 74 × 20 = ₹1480.

Worked example 3. A cube has TSA 216 cm². Then 6a² = 216, a² = 36, a = 6 cm, and its volume is 216 cm³. (The numbers coincide only because a = 6.)

Worked example 4. Two cubes of side 4 cm are joined face to face to make a cuboid. The cuboid is 8 cm × 4 cm × 4 cm, with TSA = 2(32 + 16 + 32) = 160 cm². Note that this is less than the TSA of the two separate cubes, 2 × 96 = 192 cm², because two faces of 16 cm² each are hidden at the join. This idea, that joined faces disappear, is the foundation of the whole chapter.

A right circular cylinder is the solid traced by a rectangle rotated about one of its sides, or equivalently a solid with two equal circular ends joined by a curved surface, with the axis perpendicular to the ends. Examples: a water pipe, a roller, a candle, a tin can. Its dimensions are the base radius r and the height h.

Curved surface area (CSA). If the curved surface is cut along a vertical line and opened out, it becomes a rectangle of length 2πr (the circumference) and breadth h. So CSA = 2πrh.

Total surface area. Add the two circular ends: TSA = 2πrh + 2πr² = 2πr(r + h). For an open cylinder such as a bucket or a pipe with no lids, only one end or no end is added.

Volume. Base area × height: V = πr²h.

Hollow cylinder. A pipe with outer radius R, inner radius r and length h has volume of material π(R² − r²)h, outer curved surface 2πRh, inner curved surface 2πrh, and two ring-shaped ends of total area 2π(R² − r²).

Worked example 1. A cylinder has radius 7 cm and height 10 cm (π = 22/7). CSA = 2 × (22/7) × 7 × 10 = 440 cm². TSA = 440 + 2 × (22/7) × 49 = 440 + 308 = 748 cm². Volume = (22/7) × 49 × 10 = 1540 cm³.

Worked example 2. A roller of diameter 84 cm and length 1 m takes 500 complete revolutions to level a playground. Area levelled = 500 × CSA = 500 × 2 × (22/7) × 0.42 × 1 = 500 × 2.64 = 1320 m².

Worked example 3. The CSA of a cylinder is 88 cm² and its height is 14 cm. Then 2 × (22/7) × r × 14 = 88, so 88r = 88 and r = 1 cm. Its volume is (22/7) × 1 × 14 = 44 cm³.

Worked example 4. A cylindrical pipe has inner diameter 4 cm, outer diameter 4.4 cm and length 20 cm. Volume of metal = (22/7)(2.2² − 2²) × 20 = (22/7)(4.84 − 4) × 20 = (22/7) × 0.84 × 20 = 52.8 cm³. Its mass at 8.9 g per cm³ is 469.92 g.

Worked example 5. Two cylinders have equal volumes and their heights are in the ratio 1 : 4. Then πr₁²h = πr₂² × 4h, so r₁² = 4r₂² and r₁ : r₂ = 2 : 1. Ratio problems of this kind are frequent one-mark questions.

A right circular cone is the solid traced by a right triangle rotated about one of its legs. The leg about which it turns is the height h, the other leg sweeps out the circular base of radius r, and the hypotenuse is the slant height l. The three are related by Pythagoras: l² = r² + h². Examples: an ice-cream cone, a conical tent, a funnel, a joker's cap, a heap of grain.

Curved surface area. If the curved surface is cut along a slant line and opened out, it becomes a sector of a circle of radius l whose arc length is 2πr. Its area is ½ × arc × radius = ½ × 2πr × l, so CSA = πrl.

Total surface area. Add the base: TSA = πrl + πr² = πr(l + r).

Volume. A cone holds exactly one third of a cylinder with the same base and height, a fact verified by filling: V = ⅓πr²h.

Worked example 1. A cone has radius 7 cm and height 24 cm (π = 22/7). l = √(49 + 576) = √625 = 25 cm. CSA = (22/7) × 7 × 25 = 550 cm². TSA = 550 + 154 = 704 cm². Volume = ⅓ × (22/7) × 49 × 24 = 1232 cm³.

Worked example 2. A conical tent is 10 m high with base radius 24 m. l = √(100 + 576) = 26 m. Canvas needed = CSA = 3.14 × 24 × 26 = 1959.36 m². At ₹70 per m² the canvas costs ₹1,37,155.20.

Worked example 3. The slant height of a cone is 21 m and the diameter of its base is 24 m. TSA = (22/7) × 12 × (21 + 12) = (22/7) × 12 × 33 = 1244.57 m².

Worked example 4. A heap of wheat is a cone of diameter 10.5 m and height 3 m. Volume = ⅓ × (22/7) × 5.25² × 3 = (22/7) × 27.5625 = 86.625 m³. To cover it with canvas, l = √(27.5625 + 9) = √36.5625 ≈ 6.05 m and CSA = (22/7) × 5.25 × 6.05 ≈ 99.83 m².

Worked example 5. A cone and a cylinder have the same base radius and the same height. The ratio of their volumes is 1 : 3. If the cone's volume is 100 cm³, the cylinder's is 300 cm³.

The student should remember that the slant height is never given by the height alone; whenever a surface area of a cone is required and l is not given, compute it first from r and h.

A sphere is the set of all points in space at a fixed distance r (the radius) from a fixed point (the centre): a ball, a marble, a globe. A plane through the centre divides the sphere into two hemispheres, each with a flat circular face of radius r and a curved surface.

Surface area of a sphere = 4πr². This is exactly four times the area of a great circle of the sphere, or the curved surface of the cylinder that just encloses the sphere (radius r, height 2r).

Volume of a sphere = (4/3)πr³.

For a hemisphere: curved surface area = ½ × 4πr² = 2πr²; total surface area = 2πr² + πr² = 3πr² (curved surface plus the flat circular face); volume = (2/3)πr³.

A spherical shell of outer radius R and inner radius r has volume of material (4/3)π(R³ − r³).

Worked example 1. A sphere has radius 7 cm (π = 22/7). Surface area = 4 × (22/7) × 49 = 616 cm². Volume = (4/3) × (22/7) × 343 = 1437.33 cm³. A hemisphere of the same radius has CSA 308 cm², TSA 462 cm² and volume 718.67 cm³.

Worked example 2. The surface area of a sphere is 154 cm². Then 4 × (22/7) × r² = 154, so r² = 154 × 7/88 = 12.25 and r = 3.5 cm. Its volume is (4/3) × (22/7) × 42.875 = 179.67 cm³.

Worked example 3. A hemispherical bowl has inner diameter 10.5 cm. Its capacity is (2/3) × (22/7) × 5.25³ = (2/3) × (22/7) × 144.703 = 303.19 cm³, about 0.303 litre. The inner curved surface to be tin-plated is 2 × (22/7) × 27.5625 = 173.25 cm².

Worked example 4. The radius of a balloon increases from 7 cm to 14 cm as air is pumped in. The ratio of its surface areas is 7² : 14² = 1 : 4, and the ratio of its volumes is 7³ : 14³ = 1 : 8. Surface area scales as the square of the radius and volume as the cube.

Worked example 5. A hollow spherical shell has outer radius 5 cm and inner radius 3 cm. Volume of metal = (4/3) × 3.14 × (125 − 27) = (4/3) × 3.14 × 98 = 410.29 cm³.

Remember the distinction between the CSA and the TSA of a hemisphere: a hemispherical dome to be painted on the outside has area 2πr², but a solid hemisphere to be painted all over has 3πr².

Most objects around us are not a single standard solid but a combination of two or more. The first skill of this chapter is to look at an object and name its parts, together with the dimensions that the parts share.

Common combinations. A toy top or a lattu: a cone standing on a hemisphere, the base of the cone and the flat face of the hemisphere coinciding, so they have the same radius. A medicine capsule: a cylinder with a hemisphere at each end, all of the same radius. A circus tent: a cylinder with a cone on top, sharing the radius. A pencil: a cylinder ending in a cone. A funnel: a cone joined to a narrow cylinder. A rocket model: a cone on a cylinder. A gulab jamun: a cylinder with hemispherical ends. A bucket-shaped pen stand: a cuboid with conical or hemispherical depressions. A dome: a hemisphere on a cylinder. An ice-cream cone with a scoop: a cone with a hemisphere on its open end. A solid with a cavity: a cylinder with a cone removed, a cube with a hemisphere scooped out, a hemispherical cavity in a cylinder.

Shared dimensions. When two solids are joined along a circular face, the radius of that face is the same for both solids. When the total length or height of the object is given, the heights of the parts must be found by subtraction. For example, a capsule 14 mm long and 5 mm in diameter has hemispheres of radius 2.5 mm at each end, and the cylindrical part has length 14 − 2 × 2.5 = 9 mm. A toy of total height 15.5 cm whose hemisphere has radius 3.5 cm has a cone of height 15.5 − 3.5 = 12 cm.

The two rules of the chapter. (1) The volume of a combination is the sum of the volumes of the parts, and the volume of a solid with a cavity is the volume of the whole minus the volume of the cavity. Volumes add because the parts fill separate regions of space. (2) The surface area of a combination is not the sum of the surface areas of the parts. When two solids are joined the faces along which they meet are hidden, and when a cavity is scooped out the cut surface is new. The total surface area is the area of what is actually exposed.

Worked example. A toy is a cone of radius 3.5 cm and height 12 cm on a hemisphere of the same radius. Its exposed surface is the curved surface of the cone plus the curved surface of the hemisphere; the base of the cone and the flat face of the hemisphere are glued together and hidden. Its volume is the cone plus the hemisphere. The figures are worked out in the next topics.

Draw the figure, label each part with its own dimensions, and list which surfaces are exposed, before writing any formula. This planning step is where most of the marks are won or lost.

Worked example 1: a toy. A toy is in the shape of a cone of radius 3.5 cm mounted on a hemisphere of the same radius; the total height is 15.5 cm. Find its total surface area (π = 22/7).

Height of the cone h = 15.5 − 3.5 = 12 cm. Slant height l = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm. Exposed surfaces: the curved surface of the cone and the curved surface of the hemisphere. CSA of cone = πrl = (22/7) × 3.5 × 12.5 = 137.5 cm². CSA of hemisphere = 2πr² = 2 × (22/7) × 12.25 = 77 cm². TSA of the toy = 137.5 + 77 = 214.5 cm².

Worked example 2: a capsule. A medicine capsule is a cylinder with two hemispherical ends. Its total length is 14 mm and its diameter is 5 mm. Find its surface area.

Radius r = 2.5 mm, cylinder length h = 14 − 5 = 9 mm. Exposed surfaces: curved surface of the cylinder and the two hemispherical caps, which together make one sphere. Area = 2πrh + 4πr² = 2 × (22/7) × 2.5 × 9 + 4 × (22/7) × 6.25 = 141.43 + 78.57 = 220 mm².

Worked example 3: a solid with a hemisphere on a cylinder. A wooden article is a cylinder of height 10 cm and radius 3.5 cm with a hemisphere of the same radius on top. Its total surface area = curved surface of the cylinder + curved surface of the hemisphere + base circle = 2πrh + 2πr² + πr² = 2 × (22/7) × 3.5 × 10 + 3 × (22/7) × 12.25 = 220 + 115.5 = 335.5 cm².

Worked example 4: a gulab jamun. A sweet is shaped like a cylinder with two hemispherical ends, of total length 5 cm and diameter 2.8 cm. r = 1.4 cm, cylinder length = 5 − 2.8 = 2.2 cm. Surface area = 2 × (22/7) × 1.4 × 2.2 + 4 × (22/7) × 1.96 = 19.36 + 24.64 = 44 cm². Its volume, needed for a syrup problem, is (22/7) × 1.96 × 2.2 + (4/3) × (22/7) × 2.744 = 13.552 + 11.499 = 25.05 cm³.

The pattern. In every case, first list the exposed surfaces in words, then write the formula for each, then substitute. The surfaces along which the parts are joined never appear. When the two parts share a radius, a common factor πr can often be taken out to shorten the arithmetic: for the toy, TSA = πr(l + 2r) = (22/7) × 3.5 × (12.5 + 7) = 11 × 19.5 = 214.5 cm², the same answer.

Worked example 1: a tent. A tent is in the shape of a cylinder surmounted by a cone. The cylindrical part has height 2.1 m and diameter 4 m, and the conical part has slant height 2.8 m. Find the area of canvas used and its cost at ₹500 per m² (π = 22/7).

r = 2 m. Exposed surfaces: curved surface of the cylinder and curved surface of the cone (the ground is not covered, and the circle where cone meets cylinder is internal). Canvas = 2πrh + πrl = 2 × (22/7) × 2 × 2.1 + (22/7) × 2 × 2.8 = 26.4 + 17.6 = 44 m². Cost = 44 × 500 = ₹22,000.

Worked example 2: a cubical block with a hemisphere on top. A hemisphere is placed on a cubical block of side 7 cm; the hemisphere has the largest possible diameter, which is 7 cm. Find the surface area of the solid.

r = 3.5 cm. The top face of the cube loses a circle of area πr² where the hemisphere sits, and gains the curved surface 2πr² of the hemisphere. TSA = 6 × 49 − πr² + 2πr² = 294 + πr² = 294 + (22/7) × 12.25 = 294 + 38.5 = 332.5 cm².

Worked example 3: a hemisphere scooped out of a cube. A hemispherical depression of the largest possible diameter is cut out of one face of a cubical block of side l. The surface area = 6l² − πr² + 2πr² = 6l² + πr² with r = l/2, that is 6l² + πl²/4 = l²(24 + π)/4. For l = 7 cm this is again 332.5 cm²: scooping out and adding on a hemisphere of the same radius change the surface area by the same amount, since in both cases a flat circle is replaced by a curved hemispherical surface.

Worked example 4: a cylinder with hemispherical ends scooped out. From a solid cylinder of height 20 cm and radius 7 cm, a hemisphere of radius 7 cm is scooped out from each end. TSA = curved surface of cylinder + 2 × curved surface of hemisphere = 2 × (22/7) × 7 × 20 + 2 × 2 × (22/7) × 49 = 880 + 616 = 1496 cm². The two flat ends are entirely removed, being exactly the size of the scoops.

Worked example 5: a conical cavity in a cylinder. A solid cylinder of radius 6 cm and height 8 cm has a cone of the same radius and height hollowed out from one end. TSA of the remaining solid = curved surface of the cylinder + the other flat end + the curved surface of the cavity = 2πrh + πr² + πrl, with l = √(36 + 64) = 10 cm: 2 × 3.14 × 6 × 8 + 3.14 × 36 + 3.14 × 6 × 10 = 301.44 + 113.04 + 188.4 = 602.88 cm².

Worked example 6: a pen stand. A cuboidal wooden pen stand 15 cm × 10 cm × 3.5 cm has four conical depressions of radius 0.5 cm and depth 1.4 cm to hold pens. Surface area exposed = TSA of cuboid − 4 × base circles + 4 × curved surfaces of the cones; l = √(0.25 + 1.96) = √2.21 ≈ 1.487 cm; area ≈ 2(150 + 35 + 52.5) − 4 × (22/7) × 0.25 + 4 × (22/7) × 0.5 × 1.487 = 475 − 3.14 + 9.35 = 481.21 cm².

The volume of a composite solid is the sum of the volumes of its parts. There are no hidden surfaces to worry about; the only care needed is to find the correct dimensions of each part.

Worked example 1: the toy. A cone of radius 3.5 cm and height 12 cm on a hemisphere of radius 3.5 cm (π = 22/7). Volume of cone = ⅓ × (22/7) × 12.25 × 12 = 154 cm³. Volume of hemisphere = (2/3) × (22/7) × 42.875 = 89.83 cm³. Total = 243.83 cm³.

Worked example 2: a rocket model. A rocket is a cylinder of radius 2.5 cm and height 21 cm with a cone of the same radius and height 6 cm on top. Volume = (22/7) × 6.25 × 21 + ⅓ × (22/7) × 6.25 × 6 = 412.5 + 39.29 = 451.79 cm³.

Worked example 3: a shed. A shed is a cuboid 7 m × 15 m × 8 m with a half-cylinder of diameter 7 m and length 15 m as its roof. Volume of air inside = 7 × 15 × 8 + ½ × (22/7) × 3.5² × 15 = 840 + 288.75 = 1128.75 m³. If 20 workers each occupy 0.08 m³ of space and machinery occupies 300 m³, the air remaining is 1128.75 − 1.6 − 300 = 827.15 m³.

Worked example 4: an ice-cream cone with a scoop. A cone of radius 3 cm and height 12 cm is filled with ice cream and a hemispherical scoop of radius 3 cm sits on top. Volume of ice cream = ⅓ × 3.14 × 9 × 12 + (2/3) × 3.14 × 27 = 113.04 + 56.52 = 169.56 cm³. If a container holds 8 litres = 8000 cm³, it fills 8000/169.56 ≈ 47 such cones.

Worked example 5: a gulab jamun in syrup. Each gulab jamun (cylinder of length 2.2 cm with hemispherical ends, radius 1.4 cm) has volume 25.05 cm³ as computed earlier. If it contains syrup up to 30% of its volume, the syrup in 45 pieces is 45 × 0.3 × 25.05 = 338.2 cm³, roughly 338 mL.

Worked example 6: a building. A building is a cylinder of radius 7 m and height 10 m surmounted by a hemispherical dome of the same radius. Volume = (22/7) × 49 × 10 + (2/3) × (22/7) × 343 = 1540 + 718.67 = 2258.67 m³. Its outside surface (curved cylinder + dome) is 440 + 308 = 748 m².

Method. Write the volume of each part on its own line with its formula, keep fractions exact as long as possible (⅓ × 12 cancels to 4, (2/3) × 3.5³ is (2/3) × 42.875), and add at the end. Give the unit as cm³ or m³, or convert to litres when the question asks for capacity.

When a solid has a part cut away, its volume is the volume of the original solid minus the volume removed. The removed part is itself a standard solid, so the calculation is a subtraction of two familiar quantities.

Worked example 1: cylinder with hemispherical ends scooped out. From a solid cylinder of height 20 cm and radius 7 cm, a hemisphere of radius 7 cm is scooped out from each end (π = 22/7). Volume remaining = πr²h − 2 × (2/3)πr³ = (22/7) × 49 × 20 − (4/3) × (22/7) × 343 = 3080 − 1437.33 = 1642.67 cm³.

Worked example 2: conical cavity in a cylinder. A solid cylinder of radius 6 cm and height 8 cm has a cone of the same radius and height removed from it. Volume remaining = πr²h − ⅓πr²h = (2/3)πr²h = (2/3) × 3.14 × 36 × 8 = 602.88 cm³. In general, removing a cone of the same base and height leaves two thirds of the cylinder.

Worked example 3: a hemispherical bowl of given thickness. A hemispherical bowl has inner radius 5 cm and is 0.25 cm thick. Volume of steel = (2/3)π(5.25³ − 5³) = (2/3) × 3.14 × (144.70 − 125) = (2/3) × 3.14 × 19.70 = 41.25 cm³.

Worked example 4: a lead shot inside a cone of water. A conical vessel of radius 5 cm and height 8 cm is full of water; when lead shots of radius 0.5 cm are dropped in, one fourth of the water flows out. Volume of water displaced = ¼ × ⅓ × π × 25 × 8 = (50/3)π cm³. Volume of one shot = (4/3)π(0.5)³ = π/6 cm³. Number of shots = (50/3)/(1/6) = 100.

Worked example 5: a spherical ball in a cylinder of water. A cylinder of radius 6 cm contains water; a solid sphere of radius 3 cm is dropped in and fully submerged. Rise in water level h satisfies π × 36 × h = (4/3)π × 27 = 36π, so h = 1 cm.

Worked example 6: a hollow cylinder. An iron pipe has external diameter 8 cm, thickness 1 cm and length 21 cm; find the mass of the pipe at 7.5 g per cm³. R = 4, r = 3: volume = (22/7) × (16 − 9) × 21 = 462 cm³; mass = 3465 g = 3.465 kg.

Worked example 7: a wooden toy with a cavity. A solid wooden toy is a cylinder of radius 3.5 cm and height 10 cm with a hemispherical cavity of the same radius at one end and a hemispherical bump at the other. Its volume equals that of the plain cylinder, (22/7) × 12.25 × 10 = 385 cm³, because the scooped hemisphere and the added hemisphere cancel exactly.

In cavity problems the surface area often rises while the volume falls, so read the question for which is asked. The displacement of water by an immersed solid is the same idea from the other side: the water that rises equals the volume of the solid that enters.

When a solid is melted and recast into a different shape, or a lump of clay is reshaped, or a heap of earth is moved, the volume does not change. This is the whole principle of conversion problems: volume of the original solid = volume of the new solid (or the total volume of the new solids). The surface area, by contrast, does change, and problems sometimes ask by how much.

Worked example 1: sphere to cylinder. A metallic sphere of radius 4.2 cm is melted and recast into a cylinder of radius 6 cm. Find the height of the cylinder. (4/3)π(4.2)³ = π(6)²h, so h = (4 × 74.088)/(3 × 36) = 296.352/108 = 2.744 cm.

Worked example 2: cone to sphere. A cone of height 24 cm and radius 6 cm is melted into a sphere. ⅓π × 36 × 24 = (4/3)πr³, so 288 = (4/3)r³, r³ = 216, r = 6 cm.

Worked example 3: three spheres into one. Metallic spheres of radii 6 cm, 8 cm and 10 cm are melted into a single sphere. (4/3)πR³ = (4/3)π(216 + 512 + 1000), so R³ = 1728 and R = 12 cm.

Worked example 4: a big cone into small spheres. A cone of radius 12 cm and height 24 cm is melted into small spheres of diameter 1 cm, that is, radius 0.5 cm. Cone volume = ⅓π × 144 × 24 = 1152π cm³. Sphere volume = (4/3)π × 0.125 = π/6 cm³. Number = 1152 × 6 = 6912.

Worked example 5: a well into a platform. A well of diameter 7 m is dug 20 m deep and the earth is spread evenly to form a rectangular platform 22 m × 14 m. Find the height of the platform. Volume of earth = (22/7) × 3.5² × 20 = 770 m³. Height = 770/(22 × 14) = 770/308 = 2.5 m.

Worked example 6: a well into an embankment. A well of diameter 3 m is dug 14 m deep and the earth is spread evenly around it to form a ring-shaped embankment of width 4 m. Volume = (22/7) × 1.5² × 14 = 99 m³. The embankment is a ring with inner radius 1.5 m and outer radius 5.5 m: area = (22/7)(30.25 − 2.25) = (22/7) × 28 = 88 m². Height = 99/88 = 1.125 m.

Worked example 7: coins into a cuboid. Silver coins of diameter 1.75 cm and thickness 2 mm are melted into a cuboid 5.5 cm × 10 cm × 3.5 cm. Volume of cuboid = 192.5 cm³. Volume of a coin = (22/7) × 0.875² × 0.2 = 0.48125 cm³. Number of coins = 192.5/0.48125 = 400.

Worked example 8: a rod into a wire. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m. Volume = π × 0.25 × 8 = 2π cm³; wire: πr² × 1800 = 2π, r² = 1/900, r = 1/30 cm, so the thickness (diameter) is 1/15 cm ≈ 0.067 cm.

Write 'volume before = volume after' as the first line of every solution; the examiner looks for that statement.

Water flowing through a pipe or a canal at a steady speed is a moving solid. In time t the water that passes any cross-section is a cylinder (for a round pipe) or a cuboid (for a rectangular canal) whose length is the distance the water travels in that time, speed × t, and whose cross-section is that of the pipe or canal. So

Volume delivered in time t = (area of cross-section) × (speed × t).

This volume then fills a tank, or spreads over a field to a certain depth, and the conversion principle applies. The chief difficulty is units: speeds are often given in km per hour and dimensions in cm or m. Convert everything to metres and hours (or cm and seconds) before multiplying.

Worked example 1: pipe filling a tank. Water flows at 3 km per hour through a pipe of internal diameter 20 cm into a cylindrical tank of diameter 10 m and depth 2 m. How long does it take to fill the tank? Pipe radius = 0.1 m; in one hour the water column is 3000 m long, so the volume per hour = (22/7) × 0.01 × 3000 = 94.29 m³. Tank volume = (22/7) × 25 × 2 = 157.14 m³. Time = 157.14/94.29 = 5/3 hours = 1 hour 40 minutes = 100 minutes.

Worked example 2: canal irrigating a field. A canal 6 m wide and 1.5 m deep carries water at 10 km per hour. How much area can it irrigate in 30 minutes if 8 cm of standing water is needed? In 30 minutes the water travels 5 km = 5000 m; volume = 6 × 1.5 × 5000 = 45,000 m³. Area = 45,000/0.08 = 562,500 m² = 56.25 hectares.

Worked example 3: rain on a roof. Rainwater from a flat roof 22 m × 20 m drains into a cylindrical tank of radius 1 m. A rainfall of 2 cm fills the tank to what height? Volume of rain = 22 × 20 × 0.02 = 8.8 m³. Height = 8.8/((22/7) × 1) = 2.8 m.

Worked example 4: a pipe emptying a hemispherical tank. A hemispherical tank of radius 1.75 m is full of water and is emptied by a pipe at 7 litres per second. Volume = (2/3) × (22/7) × 1.75³ = (2/3) × (22/7) × 5.359 = 11.23 m³ = 11,229 litres. Time = 11,229/7 ≈ 1604 seconds ≈ 26.7 minutes.

Worked example 5: a pipe in cm and seconds. Water flows through a pipe of diameter 2 cm at 6 m per second into a cylindrical vessel of radius 30 cm. Rise per second = volume per second ÷ base area = (π × 1 × 600)/(π × 900) = 2/3 cm per second, so the level rises 40 cm in one minute.

Method. Draw the pipe, write its cross-section area, multiply by the distance travelled in the given time, and equate to the volume of the receiving container. Give the answer in the unit the question asks for: minutes, hectares, litres.

The formulas. Cuboid: TSA = 2(lb + bh + hl), V = lbh. Cube: TSA = 6a², V = a³. Cylinder: CSA = 2πrh, TSA = 2πr(r + h), V = πr²h. Cone: l² = r² + h², CSA = πrl, TSA = πr(l + r), V = ⅓πr²h. Sphere: SA = 4πr², V = (4/3)πr³. Hemisphere: CSA = 2πr², TSA = 3πr², V = (2/3)πr³.

The principles. For a combination, add volumes; for surface area, add only the exposed surfaces and leave out the faces along which parts are joined. For a solid with a cavity, subtract the cavity's volume and add the cavity's inner curved surface while removing the flat face it replaces. For conversion, volume before equals volume after. For flow, volume delivered equals cross-section times distance travelled.

How the board asks. One-mark questions ask for a formula, a ratio (volumes of a cone, hemisphere and cylinder of equal base and height are 1 : 2 : 3), or the number of parts in a named object. Two-mark questions give a single standard solid with two dimensions and ask for a third quantity, such as the slant height of a cone or the radius of a sphere of given surface area. Four-mark questions are the composite-solid problems: the toy, the tent, the capsule, the scooped cylinder, the well and the platform, the pipe and the tank. The marks are divided among the figure, the identification of parts, the formulas, the substitution, the arithmetic and the unit.

Common errors. Using the diameter where the radius is needed. Forgetting to compute the slant height before the curved surface of a cone. Adding the hidden faces of a combination. Using (2/3)πr³ for a sphere or (4/3)πr³ for a hemisphere. Mixing cm and m in one calculation, especially in flow problems. Dropping the ⅓ in the cone's volume. Forgetting to convert cm³ to litres when capacity is asked. Writing cm² for a volume.

Checking. The volume of a composite solid must be more than the volume of its largest part; the volume remaining after a cavity must be less than the whole; a height obtained by conversion must be plausible (a sphere of radius 4.2 cm recast into a wide cylinder of radius 6 cm must be short, and 2.744 cm is short). When in doubt estimate with π ≈ 3.

What lies ahead. In the Intermediate course the volumes of these solids are derived by integration, the frustum of a cone (a bucket shape) is treated with the formula V = ⅓πh(R² + Rr + r²), and the surface area of a sphere is proved rather than accepted. The habits of this chapter, naming the parts, writing the formula, keeping units straight and checking the size of the answer, serve in physics (density, pressure, flow), chemistry (molar volumes) and every engineering course.