Layering a Cylinder: Worked Examples

Position in the unit

This record is the worked-example companion for the topic Layering a Cylinder in Grade 8 geometry and measurement. It assumes the learner has already met the idea that a cylinder can be viewed as identical circular layers stacked from bottom to top.

For the full conceptual development of that idea, see:

• rea.m08.geometry-measurement.volume.cylinder-derivation.layering
• rea.m08.geometry-measurement.volume.cylinder-derivation.layering-lesson

For broader cylinder-volume worked examples beyond this subtopic, see:

• rea.m08.geometry-measurement.volume.cylinder-volume-derivation-worked-examples

For independent practice after studying these examples, see:

• rea.m08.geometry-measurement.volume.cylinder-volume-derivation-practice-set

Core idea used in every example

If each layer has area A and the cylinder has height h, then:

volume = area of one layer x number/total depth of layers

For a cylinder, each layer is a circle of area pi r^2, so:

V = pi r^2 h

The point of this record is not to re-derive that formula from scratch, but to use the layering model to make each calculation sensible.

Problem-solving routine for layering questions

1. Identify the radius of the circular layer.
2. Find the area of one layer: A = pi r^2.
3. Identify the height, which tells how far the layers are stacked.
4. Multiply: V = A x h = pi r^2 h.
5. Check units: if lengths are in cm, volume is in cm^3.
6. Ask whether the answer should be exact (with pi) or approximate (decimal).

Common mistakes to watch for

• Using the diameter as the radius.
• Forgetting to square the radius.
• Multiplying circumference by height instead of base area by height.
• Writing square units (cm^2) instead of cubic units (cm^3).
• Treating a change in radius as a linear change in volume when volume depends on r^2.

---

Worked Example 1: Direct volume from radius and height

Problem. A cylinder has radius 3 cm and height 8 cm. Find its volume.

Reasoning. Since the cylinder is made of equal circular layers, first find the area of one circular layer, then multiply by the height.

Solution. Base area:

A = pi r^2 = pi(3)^2 = 9pi cm^2

Volume:

V = Ah = 9pi x 8 = 72pi cm^3

Approximate value:

72pi ≈ 226.2 cm^3

Answer. 72pi cm^3, or about 226.2 cm^3.

Commentary. The key choice was to begin with the area of one layer. That keeps the formula meaningful: each layer covers 9pi cm^2, and stacking that same layer through 8 cm gives 72pi cm^3.

---

Worked Example 2: Diameter is given instead of radius

Problem. A can is shaped like a cylinder with diameter 10 cm and height 12 cm. Find its volume.

Reasoning. The volume formula uses radius, not diameter, so convert first.

Solution. Radius:

r = 10/2 = 5 cm

Base area:

A = pi r^2 = pi(5)^2 = 25pi cm^2

Volume:

V = Ah = 25pi x 12 = 300pi cm^3

Approximate value:

300pi ≈ 942.5 cm^3

Answer. 300pi cm^3, or about 942.5 cm^3.

Commentary. The main decision here was to stop before substituting and ask, “Do I have radius or diameter?” Many errors happen because students square 10 instead of 5.

---

Worked Example 3: Decimal measurements

Problem. A cylinder has radius 2.5 m and height 4 m. Find its volume.

Reasoning. The same layering idea works with decimals. Compute the area of one circular layer carefully.

Solution. Base area:

A = pi r^2 = pi(2.5)^2 = 6.25pi m^2

Volume:

V = Ah = 6.25pi x 4 = 25pi m^3

Approximate value:

25pi ≈ 78.5 m^3

Answer. 25pi m^3, or about 78.5 m^3.

Commentary. Decimals do not change the structure of the reasoning. The only care point is squaring correctly: (2.5)^2 = 6.25, not 5.

---

Worked Example 4: Fractional radius

Problem. A cylinder has radius 1/2 ft and height 6 ft. Find its volume.

Reasoning. Fractions often look harder than they are. Keep the order: square the radius, find one-layer area, then multiply by height.

Solution. Base area:

A = pi r^2 = pi(1/2)^2 = pi/4 ft^2

Volume:

V = Ah = (pi/4) x 6 = 6pi/4 = 3pi/2 ft^3

Approximate value:

3pi/2 ≈ 4.71 ft^3

Answer. 3pi/2 ft^3, or about 4.71 ft^3.

Commentary. A good choice here is to keep the exact form until the end. That avoids rounding too early and makes the arithmetic cleaner.

---

Worked Example 5: Missing height from volume

Problem. A cylinder has volume 154pi cm^3 and radius 7 cm. Find its height.

Reasoning. If volume equals layer area times height, then height is volume divided by layer area.

Solution. Start with:

V = pi r^2 h

Substitute known values:

154pi = pi(7)^2 h

154pi = 49pi h

Divide both sides by 49pi:

h = 154pi / 49pi = 154/49 = 22/7

So:

h = 22/7 cm

Approximate value:

22/7 ≈ 3.14 cm

Answer. 22/7 cm, or about 3.14 cm.

Commentary. The useful choice was to think in reverse: if one layer has area 49pi cm^2, how tall must the stack be to make 154pi cm^3? Dividing volume by base area answers that directly.

---

Worked Example 6: Missing radius from volume and height

Problem. A cylinder has volume 200pi cm^3 and height 8 cm. Find its radius.

Reasoning. Since V = pi r^2 h, first isolate r^2, then take the square root.

Solution. Start with:

200pi = pi r^2 x 8

200pi = 8pi r^2

Divide both sides by 8pi:

r^2 = 200pi / 8pi = 25

Take the positive square root:

r = 5 cm

Answer. 5 cm.

Commentary. Radius is a length, so after finding r^2 = 25, one more step is needed. Also, in geometry we take the positive root because a radius cannot be negative.

---

Worked Example 7: Comparing two cylinders with the same height

Problem. Cylinder A has radius 2 cm and height 10 cm. Cylinder B has radius 4 cm and height 10 cm. How many times as large is the volume of Cylinder B compared with Cylinder A?

Reasoning. Because the heights are equal, the comparison depends only on the base-layer areas. Radius is squared, so doubling radius does not just double volume.

Solution. Cylinder A:

V_A = pi(2)^2(10) = 40pi cm^3

Cylinder B:

V_B = pi(4)^2(10) = 160pi cm^3

Compare:

V_B / V_A = 160pi / 40pi = 4

Answer. Cylinder B has 4 times the volume of Cylinder A.

Commentary. This example shows why the layering model matters. Each layer in Cylinder B has 4 times the area because 4^2 = 16 and 2^2 = 4. With the same stack height, the whole volume is 4 times as large.

---

Worked Example 8: Comparing the effect of changing height

Problem. A cylinder has radius 6 cm and height 5 cm. Another cylinder has the same radius but height 15 cm. Compare their volumes.

Reasoning. If the layers are the same size, tripling the number of layers triples the volume.

Solution. First cylinder:

V_1 = pi(6)^2(5) = 180pi cm^3

Second cylinder:

V_2 = pi(6)^2(15) = 540pi cm^3

Compare:

V_2 / V_1 = 540pi / 180pi = 3

Answer. The second cylinder has 3 times the volume of the first.

Commentary. Here the base layers are identical. Only the height changes, so the volume changes by the same factor as the height.

---

Worked Example 9: Real-world container question

Problem. A cylindrical water bottle has radius 3.5 cm and height 20 cm. How much water can it hold, to the nearest cubic centimetre?

Reasoning. This is still a layering problem: circular water layers fill the bottle from bottom to top.

Solution. Base area:

A = pi(3.5)^2 = 12.25pi cm^2

Volume:

V = 12.25pi x 20 = 245pi cm^3

Approximate value:

245pi ≈ 769.7 cm^3

Rounded to the nearest cubic centimetre:

770 cm^3

Answer. The bottle holds about 770 cm^3 of water.

Commentary. Since the question asks for a practical capacity, a rounded decimal answer is more useful than an exact answer with pi.

---

Worked Example 10: Challenge reverse problem with two-step reasoning

Problem. A cylinder has the same volume as a cylinder with radius 3 cm and height 16 cm. The new cylinder has height 9 cm. Find the radius of the new cylinder.

Reasoning. First find the original volume. Then use that volume with the new height to solve for the new radius.

Solution. Original cylinder volume:

V = pi(3)^2(16) = 144pi cm^3

Now let the new radius be r:

144pi = pi r^2 (9)

144pi = 9pi r^2

Divide by 9pi:

r^2 = 16

Take the positive square root:

r = 4 cm

Answer. The new cylinder has radius 4 cm.

Commentary. This problem rewards organizing the information rather than rushing. The important choice is to preserve the same volume while changing the stack shape: fewer or more layers can be balanced by changing the layer size.

---

Worked Example 11: Challenge explanation problem

Problem. Cylinder P has radius 5 cm and height h. Cylinder Q has radius 10 cm and the same height h. Explain, using the layering model, why Cylinder Q has four times the volume of Cylinder P.

Reasoning. This is an explanation question, not just a calculation question. The goal is to connect the formula to the visual meaning of layers.

Solution. Volume of Cylinder P:

V_P = pi(5)^2 h = 25pi h

Volume of Cylinder Q:

V_Q = pi(10)^2 h = 100pi h

Compare:

V_Q / V_P = 100pi h / 25pi h = 4

Layering explanation:

• Every layer in Cylinder P is a circle of area 25pi cm^2.
• Every layer in Cylinder Q is a circle of area 100pi cm^2.
• So each layer in Cylinder Q is 4 times as large.
• Since both cylinders have the same height, they have the same total stack depth.
• Therefore the whole volume is also 4 times as large.

Answer. Cylinder Q has four times the volume because its circular layers each have four times the area, while the stack height stays the same.

Commentary. This is the central power of the layering idea: it explains why the comparison works, instead of treating the formula as a rule to memorize.

---

Summary table of patterns

Situation	Useful idea
Radius and height given	Find one-layer area, then multiply by height
Diameter given	Convert to radius first
Volume and radius given	Divide by base area to find height
Volume and height given	Divide to get r^2, then square root
Same radius, different heights	Volume changes by the same factor as height
Same height, different radii	Volume changes by the square of the radius factor

Self-check questions with short answers

1. A cylinder has radius 4 cm and height 7 cm. What is its volume? Answer: pi(4)^2(7) = 112pi cm^3.

2. A cylinder has diameter 8 m and height 3 m. What is its volume? Answer: r = 4, so V = pi(4)^2(3) = 48pi m^3.

3. A cylinder has volume 81pi cm^3 and height 9 cm. What is its radius? Answer: 81pi = 9pi r^2, so r^2 = 9, hence r = 3 cm.

Misconceptions and repairs

• Misconception: “Doubling the radius doubles the volume.”
  Repair: If height stays fixed, volume depends on r^2, so doubling radius makes the volume 4 times as large.

• Misconception: “Height means slant distance.”
  Repair: For a cylinder, height is the perpendicular distance from one circular base to the other, which is exactly the stack depth of the layers.

• Misconception: “Volume formula is separate from the layering idea.”
  Repair: The formula V = pi r^2 h is the layering idea written in symbols: base area x height.

What to study next

After mastering these examples, the next step is to work independently through mixed questions and explanation tasks in rea.m08.geometry-measurement.volume.cylinder-volume-derivation-practice-set.