Mensuration: Cylinder, Cone and Sphere notes: ICSE Class 10 Mathematics

1. The cylinder

A right circular cylinder of base radius $r$ and height $h$ has two circular ends and a curved surface; the curved surface unrolls into a rectangle $2\pi r$ by $h$ .
Curved surface area $= 2\pi r h$ ; total surface area $= 2\pi r(r + h)$ ; volume $= \pi r^2 h$ .
An open tank or can has no top: curved surface $+$ one base. A closed tin has both ends.
Volumes are in cubic units: $1$ m $^3 = 1000$ litres and $1$ litre $= 1000$ cm $^3$ .

Example: a tin with $r = 10.5$ cm and $h = 20$ cm: $S = 1320$ cm $^2$ , total $= 2013$ cm $^2$ , $V = 6930$ cm $^3$ (using $\pi = \dfrac{22}{7}$ ).

A cylinder: radius $r$, height $h$ (not to scale).

The curved surface unrolls into a rectangle $2\pi r$ by $h$ (not to scale).

The tin (not to scale).

The tank is open at the top (not to scale).

Formulas

Cylinder: curved
$\text{curved surface} = 2\pi r h$
Cylinder: total
$\text{total surface} = 2\pi r(r + h)$
Cylinder: volume
$V = \pi r^2 h$

Key terms

Curved surface area: The area of the curved face of a solid, without its flat ends.
Total surface area: The curved surface area plus the areas of the flat faces.
Capacity: The volume of liquid a hollow container holds; $1$ litre $= 1000$ cm $^3$ .

2. The hollow cylinder

A pipe or tube has outer radius $R$ , inner radius $r$ and length $h$ ; the thickness is $R - r$ .
Metal $= \pi R^2 h - \pi r^2 h = \pi h(R^2 - r^2)$ .
Outside curved surface $2\pi R h$ , inside curved surface $2\pi r h$ , and two ring-shaped ends of area $\pi(R^2 - r^2)$ each.
Always find $r$ from the thickness first when it is given instead of $r$ .

Example: $R = 6$ cm, $r = 5$ cm, $h = 14$ cm: $V = 484$ cm $^3$ ; the two curved surfaces measure $528 + 440 = 968$ cm $^2$ .

Hollow cylinder: outer radius $R$, inner radius $r$, length $h$ (not to scale).

The pipe (not to scale).

Formulas

Material in a hollow cylinder
$V = \pi h(R^2 - r^2)$

Key terms

Hollow cylinder: A cylinder with a cylindrical bore through it, such as a pipe; it has an outer and an inner radius.
Thickness: The difference $R - r$ between the outer and inner radii.

3. The cone

A right circular cone has base radius $r$ , vertical height $h$ and slant height $l$ with $l^2 = r^2 + h^2$ .
Curved surface $= \pi r l$ ; total surface $= \pi r(r + l)$ ; volume $= \dfrac{1}{3}\pi r^2 h$ (one third of the cylinder with the same base and height).
Find $l$ first when only $r$ and $h$ are given; when the curved area is given, $l = \dfrac{\text{curved area}}{\pi r}$ .
Not in the ICSE 2027 syllabus: the frustum of a cone.

Example: $r = 10.5$ cm, $h = 14$ cm: $l = 17.5$ cm, curved $= 577.5$ cm $^2$ , total $= 924$ cm $^2$ , $V = 1617$ cm $^3$ .

A cone: radius $r$, height $h$, slant height $l$ (not to scale).

The cone (not to scale).

The cone: find $h$ (not to scale).

Formulas

Slant height
$l = \sqrt{r^2 + h^2}$
Cone: curved
$\text{curved surface} = \pi r l$
Cone: volume
$V = \dfrac{1}{3}\pi r^2 h$

Key terms

Slant height: The distance along the sloping side of a cone from its apex to the rim of its base: $l = \sqrt{r^2 + h^2}$ .
Apex: The pointed top of a cone.

4. The sphere and the hemisphere

Sphere: surface area $4\pi r^2$ (four times a great circle), volume $\dfrac{4}{3}\pi r^3$ . Halve a diameter to get $r$ .
Hemisphere: curved surface $2\pi r^2$ , total surface of a solid hemisphere $3\pi r^2$ (the flat face adds $\pi r^2$ ), volume $\dfrac{2}{3}\pi r^3$ .
A hollow bowl is open at the top: only the curved surface is counted.

Example: a sphere of radius $21$ cm has $S = 5544$ cm $^2$ and $V = 38808$ cm $^3$ ; a solid hemisphere of the same radius has curved surface $2772$ cm $^2$ , total $4158$ cm $^2$ and volume $19404$ cm $^3$ .

A sphere of radius $r$.

A solid hemisphere: curved surface $2\pi r^2$, flat face $\pi r^2$ (not to scale).

The sphere (not to scale).

The solid hemisphere (not to scale).

Formulas

Sphere
$S = 4\pi r^2, \; V = \dfrac{4}{3}\pi r^3$
Hemisphere: curved, total, volume
$2\pi r^2, \; 3\pi r^2, \; \dfrac{2}{3}\pi r^3$

Key terms

Hemisphere: Half of a sphere, cut by a plane through its centre.
Great circle: The circle in which a plane through the centre cuts a sphere; its area is $\pi r^2$ .

5. Spherical shells; inner and outer volume

A spherical shell (hollow sphere) of outer radius $R$ and inner radius $r$ has metal of volume $\dfrac{4}{3}\pi (R^3 - r^3)$ ; a hemispherical bowl has $\dfrac{2}{3}\pi (R^3 - r^3)$ .
Material $=$ outer volume $-$ inner volume; capacity $=$ inner volume.
Mass $=$ volume $\times$ mass of $1$ cm $^3$ (density).

Example: $R = 5.5$ cm, $r = 4$ cm: $V = 429$ cm $^3$ of metal; a hemispherical bowl with $R = 6.5$ cm and $r = 5$ cm holds 313.5 cm $^3$ of steel, and its mass is $313.5 \times 8 = 2508$ g.

A spherical shell: outer radius $R$, inner radius $r$ (not to scale).

The hollow sphere in section (not to scale).

Metal $=$ outer volume $-$ inner volume (not to scale).

The bowl: outer radius $6.5$ cm, inner radius $5$ cm (not to scale).

Formulas

Spherical shell
$V = \dfrac{4}{3}\pi (R^3 - r^3)$
Hollow solid
$V_{\text{material}} = V_{\text{outer}} - V_{\text{inner}}$

Key terms

Spherical shell: A hollow sphere: the material between two concentric spheres.
Density: The mass of one unit of volume of a material, for example $8$ g per cm $^3$ .

6. Cost problems

A rate per square unit (painting, polishing, plating, canvas) needs a surface area; a rate per litre (water, oil, fuel) needs a volume in litres.
Convert first: $1$ m $^3 = 1000$ litres, $1$ m $= 100$ cm, $1$ m $^2 = 10000$ cm $^2$ .
Count only the surfaces that are really painted: an open tank has no top, and the ground under a dome is not painted.

Example: a dome of radius $14$ m has curved surface $1232$ m $^2$ ; at ₹ $15$ per m $^2$ painting costs ₹ 18480. A drum holding $0.462$ m $^3$ holds $462$ litres, which cost ₹ 39270 at ₹ $85$ per litre.

Convert before you multiply by a rate.
Conversion	Value
1 cubic metre	1000 litres
1 litre	1000 cubic centimetres
1 metre	100 centimetres
1 square metre	10000 square centimetres
1 kilolitre	1 cubic metre

Match the rate to the quantity.

The dome (not to scale).

The oil drum (not to scale).

Formulas

Volume to litres
$1 \text{ m}^3 = 1000 \text{ litres}$

Key terms

Rate: The price of one unit of area or volume, such as ₹ $15$ per m $^2$ or ₹ $85$ per litre.

7. Melting and recasting

When a solid is melted and recast, volume before $=$ volume after; surface area and dimensions change.
One solid into $n$ equal solids: $V = n \times v$ ; the common factor $\pi$ (and often $\dfrac{4}{3}$ ) cancels, so $n$ is a ratio of cubes for spheres.
A rod drawn into a wire: $\pi R^2 h = \pi r^2 l$ , so $l = \dfrac{R^2 h}{r^2}$ .
A solid submerged in an upright cylinder raises the water by $x$ with $\pi R^2 x = V_{\text{solid}}$ .

Example: a sphere of radius $7.5$ cm gives $125$ spheres of radius $1.5$ cm; a rod of radius $1.4$ cm and length $25$ cm becomes a wire of radius $0.1$ cm and length $49$ m; a hemisphere of radius $6$ cm becomes a cone of base radius $6$ cm and height $12$ cm.

Melting and recasting changes the shape, not the volume.

Volume of 1 big sphere $= n \times$ volume of 1 small sphere.

The copper rod (not to scale).

A submerged solid raises the water by $x$ (not to scale).

The new cone: find $h$ (not to scale).

The cone dropped into the vessel (not to scale).

Formulas

Recasting
$V_{\text{before}} = V_{\text{after}}$
Rise of water
$\pi R^2 x = V_{\text{solid}}$

Key terms

Recasting: Melting a solid and forming a new solid of a different shape from the same material.
Displaced volume: The volume of water pushed aside by a submerged solid; it equals the volume of the solid.

8. Combinations of solids

Add the volumes of the parts. For the surface, count only the faces on the outside: a joined face is hidden.
A hemisphere, a cylinder and a cone of the same radius: outside area $=$ hemisphere $2\pi r^2$ $+$ cylinder $2\pi r h$ $+$ cone $\pi r l$ (find $l$ first).
A hemisphere or cone hollowed out of a cylinder removes volume and may remove or add surface: check which faces remain.

Example: a cylinder with $r = 4.2$ cm and $h = 6$ cm under a cone of height $5.6$ cm ( $l = 7$ cm): outside area $55.44 + 158.4 + 92.4 = 306.24$ cm $^2$ and volume $332.64 + 103.488 = 436.128$ cm $^3$ .

Hemisphere + cylinder + cone, all of radius $r$ (not to scale).

Cylinder and cone (not to scale).

Hemisphere, cylinder and cone (not to scale).

Key terms

Combination of solids: A solid made by joining simple solids along faces; its volume is the sum of the volumes of the parts.

Exam tips

How marks are awarded

Surface area or volume of one solid (2 marks): formula 1, substitution and result with unit 1.
Cone (3 marks): slant height 1, formula with substitution 1, result with unit 1.
Cost problem (3–4 marks): area or volume 1–2, conversion of units 1, cost 1.
Melting and recasting (3–4 marks): statement "volume before $=$ volume after" 1, the two volumes 1–2, answer 1.
10-mark structured question: parts (a), (b), (c), usually $3 + 3 + 4$ ; show the formula for every solid.

Common mistakes

Using the diameter where the radius is needed.
Using the vertical height instead of the slant height in $\pi r l$ .
Counting a top that an open tank or a bowl does not have, or counting joined faces in a combination.
Forgetting the factor $\dfrac{1}{3}$ for a cone, or the factor $\dfrac{4}{3}$ or $\dfrac{2}{3}$ for a sphere or hemisphere.
Forgetting to convert cubic metres to litres or centimetres to metres before multiplying by a rate.
Equating surface areas instead of volumes when a solid is melted and recast.

1. The cylinder

Formulas

Key terms

2. The hollow cylinder

Formulas

Key terms

3. The cone

Formulas

Key terms

4. The sphere and the hemisphere

Formulas

Key terms

5. Spherical shells; inner and outer volume

Formulas

Key terms

6. Cost problems

Formulas

Key terms

7. Melting and recasting

Formulas

Key terms

8. Combinations of solids

Key terms

Exam tips

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