Surface area and volume of prisms and cylinders

What you will learn

calculate surface area of right prisms by unfolding their nets,
derive and apply the surface area formula for cylinders,
calculate the volume of cylinders using $V = \pi r^2 h$ ,
find the surface area and volume of composite solids built from prisms and cylinders.

Worked example 0 Real-world example: painting a water tank

A cylindrical water tank has radius $1.2$ m and height $2$ m. You need to paint the outside (curved surface and top lid only — the base sits on a concrete pad).

Curved surface area: $2\pi r h = 2\pi(1.2)(2) = 4.8\pi \approx 15.08$ m $^2$ .
Top circle: $\pi r^2 = \pi(1.2)^2 = 1.44\pi \approx 4.52$ m $^2$ .
Total area to paint: $15.08 + 4.52 \approx 19.60$ m $^2$ .
At $12 per m $^2$ , paint cost $\approx$ $235.20.

Key idea: real problems often require only part of the total surface area.

1. Surface area of right prisms (using nets)

The surface area (SA) of any right prism is the sum of the areas of all its faces. Unfolding the prism into a net makes every face visible.

Surface area of a right prism

\text{SA} = 2 \times A_{\text{base}} + \text{lateral area}

where the lateral area equals the perimeter of the base times the length of the prism: $P_{\text{base}} \times L$ .

Worked example 1 Rectangular prism (cuboid)

Find the surface area of a box $8$ cm by $5$ cm by $3$ cm.

Two faces of $8 \times 5 = 40$ cm $^2$ .
Two faces of $8 \times 3 = 24$ cm $^2$ .
Two faces of $5 \times 3 = 15$ cm $^2$ .
SA $= 2(40 + 24 + 15) = 2 \times 79 = 158$ cm $^2$ .

Worked example 2 Triangular prism

A triangular prism has a right-angled triangular cross-section with legs $6$ cm and $8$ cm (hypotenuse $10$ cm) and length $12$ cm.

Base area (triangle): $\tfrac{1}{2} \times 6 \times 8 = 24$ cm $^2$ .
Two triangular ends: $2 \times 24 = 48$ cm $^2$ .
Three rectangular faces: $(6 + 8 + 10) \times 12 = 288$ cm $^2$ .
SA $= 48 + 288 = 336$ cm $^2$ .

2. Surface area of cylinders

When you “unroll” a cylinder, its curved surface becomes a rectangle whose width is the circumference $2\pi r$ and whose height is $h$ .

Cylinder net: the curved surface unrolls into a rectangle of width 2 pi r and height h, plus two circular ends.

Surface area of a cylinder

\text{SA} = 2\pi r^2 + 2\pi r h

$2\pi r^2$ : two circular ends.
$2\pi r h$ : curved (lateral) surface.

Worked example 3 Full cylinder surface area

Find the total surface area of a cylinder with radius $5$ cm and height $12$ cm. Use $\pi \approx 3.14$ .

Two circles: $2\pi(5)^2 = 50\pi \approx 157.0$ cm $^2$ .
Curved surface: $2\pi(5)(12) = 120\pi \approx 376.8$ cm $^2$ .
SA $\approx 157.0 + 376.8 = 533.8$ cm $^2$ .

3. Volume of cylinders

Volume of a cylinder

V = \pi r^2 h

This follows the same logic as any prism: base area ( $\pi r^2$ ) times height ( $h$ ).

Worked example 4 Volume of a cylinder

A cylindrical tin has diameter $10$ cm and height $15$ cm. Find its volume and capacity in mL.

Radius $= 5$ cm.
$V = \pi(5)^2(15) = 375\pi \approx 1178.1$ cm $^3$ .
Since $1$ cm $^3 = 1$ mL, capacity $\approx 1178$ mL $\approx 1.18$ L.

Worked example 5 Finding a missing dimension

A cylinder has volume $500$ cm $^3$ and radius $4$ cm. Find its height to one decimal place.

$V = \pi r^2 h$ , so $h = \dfrac{V}{\pi r^2}$ .
$h = \dfrac{500}{\pi \times 16} = \dfrac{500}{50.27} \approx 9.9$ cm.

4. Composite solids

A composite solid is formed by combining two or more simple solids. To find its surface area, add the exposed areas (subtract any faces hidden where the solids join). To find its volume, add the individual volumes.

Worked example 6 Cylinder on top of a rectangular prism

A silo consists of a rectangular base $4$ m by $4$ m by $3$ m high, topped with a half-cylinder of radius $2$ m and length $4$ m. Find the total volume.

Rectangular prism volume: $4 \times 4 \times 3 = 48$ m $^3$ .
Half-cylinder volume: $\tfrac{1}{2} \pi (2)^2 (4) = 8\pi \approx 25.13$ m $^3$ .
Total volume $\approx 48 + 25.13 = 73.13$ m $^3$ .

Key formulas

Prism surface area

\text{SA}_{\text{prism}} = 2A_{\text{base}} + P_{\text{base}} \times L

Cylinder surface area

\text{SA}_{\text{cyl}} = 2\pi r^2 + 2\pi r h

Cylinder volume

V_{\text{cyl}} = \pi r^2 h

Practice

Fluency

Tier 1: basic calculations

Find the surface area of a cuboid $10 \times 6 \times 4$ cm.
Find the surface area of a cube of side $7$ cm.
A triangular prism has an equilateral triangle base of side $5$ cm (height $\approx 4.33$ cm) and length $10$ cm. Find its surface area.
Find the total surface area of a cylinder with $r = 3$ cm and $h = 10$ cm. Give your answer in terms of $\pi$ and as a decimal.
Find the volume of a cylinder with $r = 6$ cm and $h = 20$ cm. Leave your answer in terms of $\pi$ .
A cylinder has $r = 4$ cm and $h = 8$ cm. Find (a) the curved surface area, (b) the total surface area, (c) the volume.
Convert a cylinder volume of $2500$ cm $^3$ to litres.
Find the surface area of a cylinder with diameter $14$ cm and height $9$ cm.

Reasoning

Tier 2: mixed practice

A cylindrical can has volume $1000$ cm $^3$ and height $12$ cm. Find its radius to one decimal place.
Two cylinders have the same volume. Cylinder A has $r = 5$ cm. Cylinder B has $r = 10$ cm. If A has height $20$ cm, find the height of B.
A closed cylinder uses $600\pi$ cm $^2$ of sheet metal. If $r = 10$ cm, find $h$ .
A rectangular prism $20 \times 10 \times 8$ cm has a cylindrical hole of radius $3$ cm drilled through its length. Find the remaining volume.
Which has the greater surface area: a cube of side $10$ cm or a cylinder with $r = 5$ cm and $h = 10$ cm? Justify.
A prism has a cross-section that is a right trapezium with parallel sides $6$ cm and $10$ cm and height $4$ cm. Its length is $15$ cm. Find the surface area. (The non-parallel sides are $4$ cm and $\approx 5.66$ cm.)

Reasoning

Tier 3: explain and apply

Explain why the formula $\text{SA} = 2\pi r^2 + 2\pi r h$ has two separate terms. What does each represent physically?
A manufacturer wants to double the volume of a cylindrical can without changing the radius. By what factor must the height change?
If you double the radius of a cylinder but keep the height the same, by what factor does the volume increase? Explain why.
A composite solid is a cylinder ( $r = 3$ cm, $h = 10$ cm) with a hemisphere ( $r = 3$ cm) on top. Find the total surface area. (Hemisphere curved SA $= 2\pi r^2$ .)
A water pipe is $50$ m long with an inner radius of $5$ cm. Find the volume of water it can hold, in litres.

Challenge

Reasoning

Harder reasoning

A cylinder and a cube have the same volume. The cylinder has $r = 5$ cm and $h = 10$ cm. Find the side length of the cube to one decimal place.
A closed cylinder has total surface area $200\pi$ cm $^2$ . Express $h$ in terms of $r$ , and find the radius that maximises the volume. (Hint: substitute into $V = \pi r^2 h$ and look for the maximum.)
A composite solid is a rectangular prism $10 \times 8 \times 6$ cm with a half-cylinder (radius $4$ cm, length $10$ cm) sitting on its top face. Find (a) the total volume, and (b) the total exposed surface area.
A cylindrical tank of radius $0.5$ m and height $1.2$ m is lying on its side. When it is half full, what volume of water does it contain? Give your answer in litres.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Tier 1

$248$ cm $^2$ . Method: $2(60 + 40 + 24)$ .
$294$ cm $^2$ . Method: $6 \times 49$ .
$171.65$ cm $^2$ (approx). Method: two triangles $2 \times \tfrac{1}{2}(5)(4.33) = 21.65$ ; three rectangles $3 \times 5 \times 10 = 150$ ; total $\approx 171.65$ .
$78\pi \approx 245.0$ cm $^2$ . Method: $2\pi(9) + 2\pi(3)(10) = 18\pi + 60\pi$ .
$720\pi$ cm $^3$ . Method: $\pi(36)(20)$ .
(a) $64\pi \approx 201.1$ cm $^2$ . (b) $96\pi \approx 301.6$ cm $^2$ . (c) $128\pi \approx 402.1$ cm $^3$ .
$2.5$ L. Method: $2500 \div 1000$ .
$\approx 703.7$ cm $^2$ . Method: $r = 7$ ; $2\pi(49) + 2\pi(7)(9) = 98\pi + 126\pi = 224\pi$ .

Tier 2

$r \approx 5.2$ cm. Method: $\pi r^2 \times 12 = 1000$ ; $r^2 = \tfrac{1000}{12\pi} \approx 26.53$ ; $r \approx 5.2$ .
$h = 5$ cm. Method: $\pi(25)(20) = \pi(100)h$ ; $h = \tfrac{500}{100} = 5$ .
$h = 20$ cm. Method: $2\pi(100) + 2\pi(10)h = 600\pi$ ; $200\pi + 20\pi h = 600\pi$ ; $20\pi h = 400\pi$ .
$\approx 1034.5$ cm $^3$ . Method: prism $= 1600$ ; cylinder hole $= \pi(9)(20) = 180\pi \approx 565.5$ ; $1600 - 565.5$ .
Cube SA $= 600$ cm $^2$ . Cylinder SA $= 2\pi(25) + 2\pi(5)(10) = 50\pi + 100\pi = 150\pi \approx 471.2$ cm $^2$ . The cube has greater surface area.
$\approx 445.8$ cm $^2$ . Method: two trapezium ends $= 2 \times \tfrac{1}{2}(6+10)(4) = 64$ ; four rectangles: $6 \times 15 = 90$ , $10 \times 15 = 150$ , $4 \times 15 = 60$ , $5.66 \times 15 \approx 84.9$ ; total $\approx 64 + 384.9 \approx 448.9$ . (Accept minor rounding differences.)

Tier 3

The first term $2\pi r^2$ is the area of the two circular ends (top and bottom). The second term $2\pi r h$ is the curved lateral surface — the rectangle you get when you unroll the cylinder, whose width is the circumference $2\pi r$ and whose height is $h$ .
The height must double. Since $V = \pi r^2 h$ and $r$ is fixed, doubling $V$ requires doubling $h$ .
Volume increases by a factor of $4$ . Since $V = \pi r^2 h$ , replacing $r$ with $2r$ gives $\pi(2r)^2 h = 4\pi r^2 h$ . The $r^2$ term means radius has a squared effect on volume.
$\approx 226.2$ cm $^2$ . Method: cylinder curved SA $= 2\pi(3)(10) = 60\pi$ ; cylinder base circle $= \pi(9) = 9\pi$ (only the bottom; the top is covered by the hemisphere); hemisphere curved SA $= 2\pi(9) = 18\pi$ ; total $= (60 + 9 + 18)\pi = 87\pi \approx 273.3$ cm $^2$ . (Accept equivalent working.)
$\approx 392.7$ L. Method: $r = 0.05$ m; $V = \pi(0.05)^2(50) = \pi(0.0025)(50) = 0.125\pi \approx 0.3927$ m $^3$ ; $\times 1000 = 392.7$ L.

Challenge

Side $\approx 9.3$ cm. Method: cylinder $V = \pi(25)(10) = 250\pi \approx 785.4$ cm $^3$ ; cube side $= \sqrt[3]{785.4} \approx 9.3$ .

Express $h$ : from $2\pi r^2 + 2\pi r h = 200\pi$ , divide through by $2\pi$ : $r^2 + r h = 100$ , so $h = \dfrac{100 - r^2}{r} = \dfrac{100}{r} - r$ .

Substitute into $V$ : $V = \pi r^2 h = \pi r^2 \left(\dfrac{100}{r} - r\right) = 100\pi r - \pi r^3$ .

Find the maximum by table of values (trial and improvement). Evaluate $V = \pi(100r - r^3)$ at whole-number $r$ :

$r$	$100r - r^3$	$V \approx$
3	$300 - 27 = 273$	$857.7$
4	$400 - 64 = 336$	$1055.6$
5	$500 - 125 = 375$	$1178.1$
6	$600 - 216 = 384$	$\mathbf{1206.4}$
7	$700 - 343 = 357$	$1121.5$
8	$800 - 512 = 288$	$904.8$

The maximum is near $r = 6$ . Refine with decimals:

$r$	$100r - r^3$	$V \approx$
5.7	$570 - 185.193 = 384.807$	$1208.9$
5.8	$580 - 195.112 = 384.888$	$\mathbf{1208.9}$
5.9	$590 - 205.379 = 384.621$	$1208.0$

The radius that maximises the volume is $r \approx 5.8$ cm. (Then $h = \dfrac{100}{5.8} - 5.8 \approx 11.5$ cm, and $V \approx 1209$ cm $^3$ .)

(a) Volume: prism $= 480$ ; half-cylinder $= \tfrac{1}{2}\pi(16)(10) = 80\pi \approx 251.3$ ; total $\approx 731.3$ cm $^3$ . (b) SA: prism base $= 10 \times 8 = 80$ ; two prism ends $= 2(8 \times 6) = 96$ ; two prism long sides $= 2(10 \times 6) = 120$ ; prism top has rectangle $10 \times 8$ minus half-cylinder footprint (the diameter strip is part of the prism top, but the half-cylinder sits on it): exposed top $= 80 - 8 \times 10 = 0$ (the half-cylinder covers the entire top, so no exposed top); half-cylinder curved $= \tfrac{1}{2}(2\pi)(4)(10) = 40\pi \approx 125.7$ ; two half-circle ends $= 2 \times \tfrac{1}{2}\pi(16) = 16\pi \approx 50.3$ ; total SA $\approx 80 + 96 + 120 + 125.7 + 50.3 = 472.0$ cm $^2$ . (Accept reasonable variations depending on which faces are considered exposed.)
$V = \tfrac{1}{2}\pi r^2 L = \tfrac{1}{2}\pi(0.5)^2(1.2) = 0.15\pi \approx 0.4712$ m $^3 \approx 471.2$ L.

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