Mathematics • Year 8 • Unit 3 • Lesson 11

Volume of Cylinders

Build fluency with V = πr²h. One fully worked example, one guided example with blanks, then eight independent problems ramping from clean radius/height to diameter conversions and missing dimensions.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. Each step has a short reason so you can see why, not just what.

Problem. Find the volume of a cylinder with radius r = 4 cm and height h = 10 cm. Give your answer to 1 decimal place.

Step 1 — Write the formula.

V = πr²h

Reason: a cylinder is a circular prism — base area (πr²) × height.

Step 2 — Identify r and h.

r = 4 cm, h = 10 cm

Reason: both are given directly — no diameter conversion needed.

Step 3 — Substitute and calculate r² first.

V = π × 4² × 10 = π × 16 × 10 = 160π

Reason: square the radius first, then multiply. Never multiply π × r × h.

Step 4 — Evaluate to a decimal.

V = 160π ≈ 160 × 3.14159 ≈ 502.7 cm³

Reason: leave as 160π for exact, use π button for decimal. Units are cm³ because r and h are in cm.

Answer: V ≈ 502.7 cm³

Stuck? Revisit lesson § Card 6 — the five steps: Formula → Identify → Square r → Multiply → Units.

2. We do — fill in the missing steps

Same shape as Section 1, with the working faded. Fill in each blank. 4 marks

Problem. Find the volume of a cylinder with radius r = 3 cm and height h = 7 cm. Give your answer to 1 decimal place.

Step 1 — Write the formula:

V = π______²______

Step 2 — Substitute r = 3 and h = 7:

V = π × ______² × ______

Step 3 — Calculate r² first:

V = π × ______ × 7 = ______π

Step 4 — Evaluate to 1 decimal place:

V ≈ ______ cm³

Stuck? Revisit lesson § Card 6 — square the radius first, then multiply by π and by h.

3. You do — independent practice

Show all working. Round decimals to 1 d.p. The first four are foundation (given r and h directly). The middle two are standard (given diameter — halve first!). The last two are extension (find a missing dimension).

Foundation — given r and h

3.1 r = 5 cm, h = 4 cm. Find V. 1 mark

3.2 r = 2 cm, h = 15 cm. Find V. 1 mark

3.3 r = 6 cm, h = 6 cm. Find V. 1 mark

3.4 r = 10 cm, h = 3 cm. Find V. 1 mark

Standard — given diameter (halve first!)

3.5 d = 8 cm, h = 5 cm. Find V. (Hint: r = d ÷ 2 = 4 cm.) 2 marks

3.6 d = 14 cm, h = 10 cm. Find V. 2 marks

Extension — find a missing dimension

3.7 A cylinder has V = 628.3 cm³ and r = 5 cm. Find its height to the nearest whole number. (Hint: rearrange to h = V ÷ (πr²).) 2 marks

3.8 A cylinder has V = 1130.97 cm³ and h = 10 cm. Find its radius. (Hint: r = √(V ÷ (πh)).) 2 marks

Stuck on 3.7 / 3.8? Revisit lesson § Card 8 — rearranging V = πr²h. For h, divide by πr². For r, divide by πh then take the square root.

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Section 2 — We do (r = 3, h = 7)

Step 1: V = πr²h.
Step 2: V = π × 3² × 7.
Step 3: V = π × 9 × 7 = 63π.
Step 4: V ≈ 197.9 cm³.

3.1 — r = 5, h = 4

V = π × 25 × 4 = 100π ≈ 314.2 cm³.

3.2 — r = 2, h = 15

V = π × 4 × 15 = 60π ≈ 188.5 cm³.

3.3 — r = 6, h = 6

V = π × 36 × 6 = 216π ≈ 678.6 cm³.

3.4 — r = 10, h = 3

V = π × 100 × 3 = 300π ≈ 942.5 cm³.

3.5 — d = 8, h = 5

r = 8 ÷ 2 = 4 cm. V = π × 16 × 5 = 80π ≈ 251.3 cm³.

3.6 — d = 14, h = 10

r = 14 ÷ 2 = 7 cm. V = π × 49 × 10 = 490π ≈ 1539.4 cm³.

3.7 — V = 628.3, r = 5. Find h.

h = V ÷ (πr²) = 628.3 ÷ (π × 25) = 628.3 ÷ 78.54 ≈ 8 cm.

3.8 — V = 1130.97, h = 10. Find r.

r = √(V ÷ (πh)) = √(1130.97 ÷ (π × 10)) = √(1130.97 ÷ 31.42) ≈ √36 = 6 cm.