MathematicsClass 10ICSE

Why the Label on a Tin Can Unrolls Into a Perfect Rectangle

Find the curved surface area, total surface area and volume of cylinders, cones, spheres and hemispheres, use slant height correctly, and work out the cost of painting, plastering or covering a solid.

Where do the surface area formulas for round solids come from?

Peel the label off a tin can and it unrolls into a rectangle. Its width is the can's height

h

and its length is the distance round the can,

2 π r

. So the curved surface of a cylinder has area

2 π r h

.

Similar thinking gives the formulas for cones, spheres and hemispheres. Surface area measures covering; volume measures filling.

This part covers the four solids and the cost of covering them. Values below use

π = \frac{22}{7}

How do you find the curved surface area, total surface area and volume of a cylinder?

**For a cylinder of radius

r

and height

h

: curved surface area

= 2 π r h

, total surface area

= 2 π r (r + h)

, and volume

= π r^{2} h

.

Worked example.** A cylinder has

r = 7

cm and

h = 10

cm.

CSA = 2 \times \frac{22}{7} \times 7 \times 10 = 440 cm^{2}

TSA = 2 \times \frac{22}{7} \times 7 \times (7 + 10) = 748 cm^{2}

V = \frac{22}{7} \times 7^{2} \times 10 = 1540 cm^{3} = 1.54 litres

Reverse. If a cylinder of radius

7

cm holds

1540 cm^{3}

, then

h = \frac{1540}{154} = 10

cm.

An everyday example. A steel water drum on a rooftop is a cylinder; its capacity is the volume, and the paint needed for its side is the curved surface area.

The substance. Total surface area includes both circular ends, while an open pipe has only a curved surface.

How do you find the surface area and volume of a cone from its radius, height or slant height?

**For a cone of radius

r

, height

h

and slant height

l = r^{2} + h^{2}

: curved surface area

= π r l

, total surface area

= π r (l + r)

, and volume

= \frac{1}{3} π r^{2} h

.

Worked example.** A cone has

r = 7

cm and

h = 24

cm.

l = 7^{2} + 2 4^{2} = 625 = 25 cm

CSA = \frac{22}{7} \times 7 \times 25 = 550 cm^{2}

TSA = \frac{22}{7} \times 7 \times (25 + 7) = 704 cm^{2}

V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = 1232 cm^{3}

An everyday example. A conical birthday cap uses paper equal to its curved surface area, while an ice-cream cone's capacity is its volume.

The substance. A cone holds exactly one-third as much as a cylinder with the same base and height:

3 \times 1232 = 3696 = \frac{22}{7} \times 49 \times 24

.

The trap. **Curved surface area uses the slant height

l

, while volume uses the vertical height

h

.**

How do you find the surface area and volume of a sphere and a hemisphere?

**A sphere has surface area

4 π r^{2}

and volume

\frac{4}{3} π r^{3}

; a hemisphere has curved surface area

2 π r^{2}

, total surface area

3 π r^{2}

and volume

\frac{2}{3} π r^{3}

.

Worked example 1 — sphere.**

r = 21

cm.

SA = 4 \times \frac{22}{7} \times 441 = 5544 cm^{2} V = \frac{4}{3} \times \frac{22}{7} \times 9261 = 38808 cm^{3}

Worked example 2 — hemisphere.

r = 7

cm.

CSA = 2 \times \frac{22}{7} \times 49 = 308 cm^{2} TSA = 3 \times \frac{22}{7} \times 49 = 462 cm^{2}

V = \frac{2}{3} \times \frac{22}{7} \times 343 = 718.67 cm^{3}

Worked example 3 — diameter given. A ball has diameter

14

cm, so

r = 7

cm and

SA = 4 \times 154 = 616 cm^{2}

.

An everyday example. A steel katori shaped like a hemisphere holds

\frac{2}{3} π r^{3}

of dal, and polishing its outside covers

2 π r^{2}

.

The substance. A solid hemisphere's total surface adds the flat circle, giving

3 π r^{2}

, not

2 π r^{2}

How do you calculate the cost of painting, plastering or covering a solid?

Decide exactly which surface is being covered, calculate its area in the same unit as the rate, and multiply by the rate per square unit.

Worked example 1 — pillar. A cylindrical pillar has radius

35

= 0.35

m and height

4

m. Its curved surface is painted at ₹12 per

m^{2}

CSA = 2 \times \frac{22}{7} \times 0.35 \times 4 = 8.8 m^{2} cost = 8.8 \times 12 = ₹105.60

Worked example 2 — tent. A conical tent has radius

7

m and height

24

m, so

l = 25

m. Canvas costs ₹80 per

m^{2}

canvas = \frac{22}{7} \times 7 \times 25 = 550 m^{2} cost = 550 \times 80 = ₹44000

Worked example 3 — dome. A hemispherical dome of radius

7

m is whitewashed on the outside at ₹15 per

m^{2}

area = 2 \times \frac{22}{7} \times 49 = 308 m^{2} cost = 308 \times 15 = ₹4620

An everyday example. A painter quoting for the pillars of a school verandah measures only the curved sides, since the ends touch the floor and ceiling.

The substance. Convert all lengths to metres before using a rate per square metre.

Exam tip

What earns full marks on surface area and volume of solids?

**Write the formula, list

r

h

and

l

with units, substitute, and state the answer with square or cubic units.

- Cylinder**:

2 π r h

2 π r (r + h)

π r^{2} h

- Cone:

l = r^{2} + h^{2}

;

π r l

π r (l + r)

\frac{1}{3} π r^{2} h

- Sphere:

4 π r^{2}

\frac{4}{3} π r^{3}

- Hemisphere:

2 π r^{2}

3 π r^{2}

\frac{2}{3} π r^{3}

- Halve the diameter before substituting
- For costs, match units with the rate

The trap. Using the diameter as the radius. **A ball of diameter

14

cm has

r = 7

cm**, and using

14

multiplies the area by

4

Did you know

Why is a sphere's surface area the same as the curved side of the cylinder around it?

Fit a sphere of radius

r

snugly inside a cylinder: the cylinder has radius

r

and height

2 r

cylinder CSA = 2 π r \times 2 r = 4 π r^{2} = sphere’s surface area

The two are exactly equal. The volumes have a neat link too:

\frac{\frac{4}{3} π r ^{3}}{π r ^{2} \times 2 r} = \frac{2}{3}

The sphere fills exactly two-thirds of the cylinder that just holds it.

Exam relevance

How do volumes of solids lead into JEE Main calculus?

This is foundation work for Class 12 Application of Derivatives, a JEE Main chapter.

What gets built on. JEE Main uses these formulas in maxima and minima problems, such as finding the cylinder of greatest volume that fits inside a cone or sphere, and in rates of change, such as how fast the radius of a balloon grows as air is pumped in. Every such question starts by writing the correct volume or surface area formula.

Question types. Multiple-choice and numerical-value questions where a formula from this lesson is differentiated.

The trap that costs marks. Confusing slant height with vertical height in a cone, which gives the wrong function before any calculus begins.

Key takeaways

What must you be able to do from this part?

- **Cylinder

r = 7

h = 10

**: CSA

440

, TSA

748 cm^{2}

, volume

1540 cm^{3}

- **Cone

r = 7

h = 24

**:

l = 25

, CSA

550

, TSA

704 cm^{2}

, volume

1232 cm^{3}

- Cone volume is one-third of the matching cylinder
- **Sphere

r = 21

**: surface

5544 cm^{2}

, volume

38808 cm^{3}

- **Hemisphere

r = 7

**: CSA

308

, TSA

462 cm^{2}

- Cost

=

correct area

\times

rate, with matching units

Measure a steel tumbler or a tin can at home and work out how many millilitres it should hold, then check with water.

Where do the surface area formulas for round solids come from?

How do you find the curved surface area, total surface area and volume of a cylinder?

How do you find the surface area and volume of a cone from its radius, height or slant height?

How do you find the surface area and volume of a sphere and a hemisphere?

How do you calculate the cost of painting, plastering or covering a solid?

What earns full marks on surface area and volume of solids?

Why is a sphere's surface area the same as the curved side of the cylinder around it?

How do volumes of solids lead into JEE Main calculus?

What must you be able to do from this part?

Related study notes

Ready to put this into practice?