Building the Volume Formula for a Cylinder: Extended Worked Examples

Position in the topic

This record is the worked-example companion to:

• rea.m08.geometry-measurement.volume.cylinder-derivation.formula-build for the unit overview
• rea.m08.geometry-measurement.volume.cylinder-derivation.formula-build.lesson-record for the derivation and core explanation
• rea.m08.geometry-measurement.volume.cylinder-volume-derivation-practice-set for independent graded practice
• rea.m08.geometry-measurement.volume.basic-formulas.worked-examples for broader volume-formula fluency across prisms and cylinders

This page does not re-teach the derivation in full. Instead, each example uses the idea behind the formula: a cylinder is a stack of equal circular layers, so

volume = (area of one circular base) x height = pi r^2 h

How to use these examples

For every problem, keep the same decision sequence:

1. Identify the circular base and the cylinder's height.
2. Decide whether the given measure is a radius or a diameter.
3. Compute or isolate the base area pi r^2.
4. Multiply by height, or undo the multiplication if a dimension is missing.
5. Check whether the units make sense: area units for the base, cubic units for the volume.

Worked Example 1: Direct substitution with radius given

Problem. Find the volume of a cylinder with radius 3 cm and height 8 cm.

Reasoning choice. This is the most direct case because the formula already wants radius and height.

Solution. V = pi r^2 h

Substitute r = 3 and h = 8:

V = pi(3)^2(8) V = pi(9)(8) V = 72pi

Approximate value:

V approx 72(3.14) = 226.08

Answer. 72pi cm^3, or about 226.08 cm^3.

Commentary. The key move is squaring only the radius, not the whole product. A common error is writing (3 x 8)^2 or pi x 3 x 2 x 8. The formula says base area first, then height.

Worked Example 2: Diameter is given, not radius

Problem. A cylinder has diameter 10 m and height 4 m. Find its volume.

Reasoning choice. The formula uses radius, so the first task is converting diameter to radius.

Solution. Radius is half the diameter:

r = 10 / 2 = 5 m

Now use the formula:

V = pi r^2 h V = pi(5)^2(4) V = pi(25)(4) V = 100pi

Approximate value:

V approx 100(3.14) = 314

Answer. 100pi m^3, or about 314 m^3.

Commentary. This example is routine mathematically, but it tests whether the student notices that 10 is the diameter. Using 10 as the radius would make the volume four times too large, because radius is squared.

Worked Example 3: Build the formula from base area first

Problem. The area of a cylinder's circular base is 49pi cm^2, and its height is 6 cm. Find the volume.

Reasoning choice. Since the base area is already known, there is no need to find the radius first.

Solution. Volume is base area times height:

V = Bh

Here B = 49pi cm^2, so

V = 49pi x 6 V = 294pi

Approximate value:

V approx 294(3.14) = 923.16

Answer. 294pi cm^3, or about 923.16 cm^3.

Commentary. This problem makes the derivation visible. The cylinder formula is not separate from V = Bh; it is the special case where B = pi r^2.

Worked Example 4: Same height, different radius

Problem. Cylinder A has radius 2 cm and height 9 cm. Cylinder B has radius 4 cm and the same height. Find both volumes and compare them.

Reasoning choice. This problem is about structure, not just arithmetic. Because height is unchanged, the comparison depends on the squared radius.

Solution. For Cylinder A:

V_A = pi(2)^2(9) = pi(4)(9) = 36pi cm^3

For Cylinder B:

V_B = pi(4)^2(9) = pi(16)(9) = 144pi cm^3

Compare:

144pi / 36pi = 4

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

Answer. Cylinder A: 36pi cm^3. Cylinder B: 144pi cm^3. Cylinder B is 4 times larger in volume.

Commentary. Doubling the radius does not double the volume when height stays the same. Because the radius is squared, doubling the radius multiplies the base area, and therefore the volume, by 4.

Worked Example 5: Find a missing height

Problem. A cylinder has volume 200pi cm^3 and radius 5 cm. Find its height.

Reasoning choice. When a dimension is missing, keep the formula intact and solve for the unknown instead of guessing.

Solution. Start with

V = pi r^2 h

Substitute known values:

200pi = pi(5)^2 h 200pi = 25pi h

Divide both sides by 25pi:

h = 200pi / 25pi = 8

Answer. 8 cm

Commentary. The pi cancels cleanly. This is a good reminder that exact forms like 200pi are often easier to work with than decimal approximations.

Worked Example 6: Find a missing radius

Problem. A cylinder has volume 144pi m^3 and height 9 m. Find its radius.

Reasoning choice. Radius is inside a square, so expect to solve for r^2 first and then take a square root.

Solution. Start with

V = pi r^2 h

Substitute known values:

144pi = pi r^2 (9) 144pi = 9pi r^2

Divide both sides by 9pi:

r^2 = 144pi / 9pi = 16

Take the square root:

r = 4

We use the positive root because a radius is a length.

Answer. 4 m

Commentary. Students sometimes stop at r^2 = 16. But 16 is not the radius; it is the square of the radius. The final step matters.

Worked Example 7: Mixed units must be fixed first

Problem. A can is shaped like a cylinder with radius 4 cm and height 0.25 m. Find its volume in cm^3.

Reasoning choice. Volume calculations require compatible units before substitution.

Solution. Convert height to centimetres:

0.25 m = 25 cm

Now use the formula:

V = pi r^2 h V = pi(4)^2(25) V = pi(16)(25) V = 400pi

Approximate value:

V approx 400(3.14) = 1256

Answer. 400pi cm^3, or about 1256 cm^3.

Commentary. If the height had been left as metres, the expression would mix cm^2 and m, which does not produce a clean cubic unit. Unit consistency is part of the mathematics, not an afterthought.

Worked Example 8: Reverse reasoning from a filling situation

Problem. A cylindrical water container has radius 7 cm. It currently holds 539pi cm^3 of water. Assuming the water forms a cylinder inside the container, how deep is the water?

Reasoning choice. The water itself is a cylinder, so the same formula applies. The unknown is the height of the water layer.

Solution. Use

V = pi r^2 h

Substitute known values:

539pi = pi(7)^2 h 539pi = 49pi h

Divide by 49pi:

h = 539pi / 49pi = 11

Answer. 11 cm

Commentary. This is a useful modelling idea: if the cross-section stays circular and the sides are straight, the amount filled can still be modelled by the cylinder formula.

Worked Example 9: Compare exact and approximate answers carefully

Problem. A pillar is a cylinder with radius 1.5 m and height 12 m. Find the volume exactly and approximately.

Reasoning choice. Decimal radii are common in measurement problems. The exact answer should still be written before approximating.

Solution. V = pi r^2 h V = pi(1.5)^2(12) V = pi(2.25)(12) V = 27pi

Approximate value:

V approx 27(3.14) = 84.78

Answer. 27pi m^3, or about 84.78 m^3.

Commentary. Writing 27pi first keeps the structure visible and avoids rounding too early. Early rounding can produce avoidable error in later steps.

Worked Example 10: Challenge problem with a constraint

Problem. Two cylinders have the same volume. Cylinder P has radius 3 cm and height 16 cm. Cylinder Q has radius 4 cm. Find the height of Cylinder Q.

Reasoning choice. Equal volumes allow us to set two volume expressions equal instead of finding each volume separately in decimal form.

Solution. Since the volumes are equal,

pi(3)^2(16) = pi(4)^2 h

Simplify each side:

pi(9)(16) = pi(16)h 144pi = 16pi h

Divide by 16pi:

h = 144pi / 16pi = 9

Answer. 9 cm

Commentary. This is an efficient use of structure. Because pi appears on both sides, it cancels. The problem is really comparing r^2 h values.

Common misconceptions and repairs

• Using diameter as radius. Repair: ask, "Does the formula want the distance across, or the distance from centre to edge?"
• Forgetting to square the radius. Repair: say the base is a circle, so the circle-area formula must appear inside the volume formula.
• Squaring the height too. Repair: only the base area is squared because only the circle formula contains a square.
• Mixing units. Repair: convert all lengths to one unit before substituting.
• Stopping at r^2 = .... Repair: if the question asks for radius, take the square root and keep the positive value.

Quick self-check practice

Try these after the worked examples.

1. Find the volume of a cylinder with r = 6 cm, h = 5 cm.
2. A cylinder has diameter 14 m and height 3 m. Find its volume.
3. A cylinder has volume 98pi cm^3 and radius 7 cm. Find its height.
4. A cylinder has volume 64pi cm^3 and height 4 cm. Find its radius.

Answers

1. 180pi cm^3
2. 147pi m^3
3. 2 cm
4. 4 cm

Closing note

For Grade 8 learners, the main goal is not just to get answers, but to see that every cylinder-volume problem comes back to one idea: circular base area multiplied by height. When that idea stays visible, the formula is easier to remember, easier to adapt, and harder to misuse.