NSWMaths Standard 2Syllabus dot point

How do you calculate the volume of prisms, cylinders, pyramids, cones, spheres and composite solids, and convert that volume into a capacity?

Calculate the volume of right prisms, cylinders, pyramids, cones, spheres and composite solids, and convert between units of volume and capacity

A focused answer to the HSC Maths Standard 2 dot point on volume and capacity. Volume of prisms and cylinders by V = Ah, the one-third for pyramids and cones, the sphere formula, composite solids by adding or subtracting parts, and the volume-to-capacity link 1 cm cubed = 1 mL, with worked Australian examples and the rounding markers expect.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-21

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What this dot point is asking

NESA wants you to find the volume of the standard solids - prisms, cylinders, pyramids, cones and spheres - and of composite solids built by joining or cutting those shapes. You then need to turn a volume into a capacity, because a volume of space ( $1$ cm $^3$ , $1$ m $^3$ ) and an amount a container holds (millilitres, litres) are two ways of measuring the same thing. The formulas are given on the HSC reference sheet, so the marks are not for memorising them. They are for choosing the right solid, substituting carefully (especially radius versus diameter), keeping $\pi$ until the last line, and converting to a capacity with the right link. Composite solids - silos, tanks, pipes, piles - are where the harder marks live, and they are exactly the real shapes the exam likes.

The answer

Every volume in this course comes from one idea: a prism (a solid with the same cross-section all the way along) has volume equal to its cross-sectional area times its height,

V = Ah.

A cylinder is just a prism with a circular cross-section, so its volume is the circle area times the height. A pyramid and a cone taper to a point, and a tapering solid holds exactly one third of the prism or cylinder on the same base. A sphere has its own formula. A composite solid is broken into these named parts, and you add the parts that are present and subtract any that are cut away. The table of formulas below is the whole toolkit.

The five volume formulas

For the standard solids (all on the HSC reference sheet):

Prism: $V = Ah$ , where $A$ is the area of the uniform cross-section and $h$ is the height (length) of the prism. A cube and a rectangular box are prisms.
Cylinder: $V = \pi r^2 h$ (a prism with a circular base of area $\pi r^2$ ).
Pyramid: $V = \tfrac{1}{3} A h$ , one third of the prism on the same base.
Cone: $V = \tfrac{1}{3}\pi r^2 h$ , one third of the cylinder on the same base.
Sphere: $V = \tfrac{4}{3}\pi r^3$ .

The " $h$ " in every formula is the perpendicular height - straight up from the base, not the slant length along a sloping face. The number $r$ is always the radius; if a question gives a diameter, halve it first.

Why pyramids and cones have the one third

A cone and the cylinder on the same circular base, with the same height, are not equal: the cylinder holds three times as much. The same is true of a pyramid and its prism. That is the reason for the $\tfrac{1}{3}$ in both formulas, and it is the single most common thing students forget. If you ever write a cone or pyramid volume that is as big as the matching cylinder or prism, you have dropped the one third.

Composite solids: add or subtract the parts

A composite solid is two or more basic solids joined together (a silo is a cylinder plus a cone; a capsule is a cylinder plus two hemispheres) or one solid with a piece removed (a pipe is a cylinder with a cylindrical hole; a drilled block is a prism minus a cylinder). The method is always the same: identify each named part, find its volume separately, then add the parts that make up the solid and subtract any hollow or cut-out part. The diagram below shows a storage shed split into its two parts.

To find the shed's volume you work out the rectangular prism ( $6 \times 4 \times 3 = 72$ m $^3$ ) and the triangular prism roof (triangle area $\tfrac{1}{2} \times 4 \times 2 = 4$ m $^2$ , times the $6$ m length, $= 24$ m $^3$ ), then add: $72 + 24 = 96$ m $^3$ . A drilled or hollow solid uses the same split but subtracts the cut-out part.

Capacity: turning a volume into litres

Capacity is the amount a container can hold. It is measured in millilitres, litres and kilolitres, and it ties to volume through one fact to commit to memory:

1 \text{ cm}^3 = 1 \text{ mL}, \qquad 1 \text{ L} = 1000 \text{ cm}^3, \qquad 1 \text{ m}^3 = 1000 \text{ L}.

That chain converts any volume into a capacity. The callout below lays it out.

So a volume of $2500$ cm $^3$ is $2500$ mL, which is $2.5$ L. A tank of $0.75$ m $^3$ holds $0.75 \times 1000 = 750$ L. Because $1$ m $^3 = 1000$ L $= 1$ kL, cubic metres and kilolitres carry the same number, which is a quick check on a large tank.

How exam questions ask about volume and capacity

The wording tells you which solid and whether a capacity is wanted:

"Find the volume of the solid" is the plain calculation: name the solid, pick its formula, substitute (halving a diameter), and keep $\pi$ until the end.
"... a cylinder / can / tank / pipe" signals $V = \pi r^2 h$ ; "... a cone / conical pile" or "... a pyramid" signals the $\tfrac{1}{3}$ .
"... how much it holds / capacity / in litres / how many litres" means convert the volume to a capacity with $1$ cm $^3 = 1$ mL or $1$ m $^3 = 1000$ L.
"... made up of / surmounted by / with a ... on top" is a composite solid to add; "... with a hole / drilled / hollow / bore" is a composite to subtract.
"How many loads / containers / how long to fill" means do the volume or capacity first, then divide by the load or rate - and usually round the count up to a whole number.
"... in terms of $\pi$ " means leave the $\pi$ in (do not press the calculator); "correct to ... decimal places" means evaluate and round at the very end.

Volumes and a capacity across four Australian contexts

Four solids: a cylinder (a rainwater tank), a cone (the $\tfrac{1}{3}$ ), a composite solid (a shed split into parts), and a volume turned into a capacity in litres. Every one follows the same plan: name the solid, write the formula, substitute (halve any diameter), keep $\pi$ to the end, then convert if a capacity is asked.

A cylinder: the volume of a rainwater tank

A cylindrical rainwater tank has a radius of $0.5$ m and a height of $1.2$ m. Find its volume, correct to two decimal places.

Name the solid and write the formula. It is a cylinder, so $V = \pi r^2 h$ .

Substitute $r = 0.5$ and $h = 1.2$ .

V = \pi \times 0.5^2 \times 1.2 = \pi \times 0.25 \times 1.2 = 0.3\pi.

Evaluate and round.

0.3\pi = 0.9424\ldots \approx 0.94 \text{ m}^3

so the tank's volume is $0.94$ m $^3$ . Carrying $0.3\pi$ rather than a rounded decimal keeps the answer exact until the final step.

A cone: remembering the one third

A cone has a radius of $6$ cm and a perpendicular height of $10$ cm. Find its volume, correct to two decimal places.

Write the cone formula. A cone is one third of the cylinder on the same base: $V = \tfrac{1}{3}\pi r^2 h$ .

Substitute $r = 6$ and $h = 10$ .

V = \tfrac{1}{3} \times \pi \times 6^2 \times 10 = \tfrac{1}{3} \times \pi \times 36 \times 10 = \tfrac{1}{3} \times 360\pi = 120\pi.

Evaluate and round.

120\pi = 376.991\ldots \approx 376.99 \text{ cm}^3

so the cone holds $376.99$ cm $^3$ . The matching cylinder would be $360\pi \approx 1130.97$ cm $^3$ , exactly three times as much, which is the sense of the $\tfrac{1}{3}$ .

A composite solid: a shed split into parts

A storage shed is a rectangular prism $4$ m wide, $3$ m high and $6$ m long, with a triangular prism roof of the same width and length rising a further $2$ m. Find its total volume.

Split into named parts. Rectangular prism (the walls) plus a triangular prism (the roof), as in the diagram above.

Rectangular prism.

V_1 = 6 \times 4 \times 3 = 72 \text{ m}^3.

Triangular prism roof. The triangle cross-section has area $\tfrac{1}{2} \times 4 \times 2 = 4$ m $^2$ , and the prism is $6$ m long:

V_2 = 4 \times 6 = 24 \text{ m}^3.

Add the parts.

V = 72 + 24 = 96 \text{ m}^3

so the shed has a volume of $96$ m $^3$ . Splitting first, computing each part, then adding is the reliable route through every composite solid.

A capacity: the shed's volume in litres, and a tank in kilolitres

Using the shed above, find its volume as a capacity in litres. Then find the capacity, in kilolitres, of a cylindrical tank of volume $4.8$ m $^3$ .

Convert the shed's volume to litres. Use $1$ m $^3 = 1000$ L:

96 \times 1000 = 96\,000 \text{ L}

so the shed encloses $96\,000$ L of space.

Convert the tank's volume to kilolitres. Since $1$ m $^3 = 1$ kL, the number is unchanged:

4.8 \text{ m}^3 = 4.8 \text{ kL}

so the tank's capacity is $4.8$ kL (which is $4800$ L).

Final answers: the rainwater tank is $0.94$ m $^3$ ; the cone is $376.99$ cm $^3$ ; the shed is $96$ m $^3$ , that is $96\,000$ L; and a $4.8$ m $^3$ tank holds $4.8$ kL.

Common traps

Dropping the one third for a pyramid or cone: A cone or pyramid is $\tfrac{1}{3}$ of the matching cylinder or prism. An answer as large as the full cylinder means you forgot the $\tfrac{1}{3}$ .
Using the diameter as the radius: The formulas use $r$ . If a tank is " $4$ m across" or has "diameter $4$ m", the radius is $2$ m, and $r^2 = 4$ , not $16$ . This is the most common lost mark on cylinder and sphere questions.
Using a slant length instead of the perpendicular height: The $h$ in every volume formula is the straight-up height, not the slope of a cone's side or a pyramid's face.
Rounding $\pi$ too early: Keep $\pi$ (or the full calculator value) until the final line; rounding partway shifts a capacity by litres, as the concrete-pipe mass showed.
Converting volume to capacity with the wrong link: It is $1$ cm $^3 = 1$ mL and $1$ m $^3 = 1000$ L. Do not square or cube these capacity links - they are already volume-to-capacity.
Adding when you should subtract (and the reverse): A hole, bore or drilled part is subtracted; a part sitting on top is added. Read whether the second solid fills space or removes it.

Exam technique

Always write the formula line before substituting; that line carries a method mark even if the arithmetic slips. Halve any diameter on sight and write " $r = \ldots$ " so the marker sees it. Keep $\pi$ symbolic through the working and round only on the final line, to the requested accuracy, with the unit attached (cm $^3$ , m $^3$ , L). For a composite solid, label each part and show " $V_1 + V_2$ " or "big $-$ hole" explicitly, because the method marks are awarded for the correct split even if a single piece is wrong. When a question asks "how many loads" or "how long to fill", do the volume or capacity first, divide by the load or rate, and round the count up to a whole number, since a part-load still needs a whole trip. A fast sanity check on size (a backyard tank is a few cubic metres, a few thousand litres) catches a diameter-for-radius slip before it costs marks.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC-style4 marksA conical pile of sand has a base radius of

3

m and a height of

2

m. (a) Find the volume of sand in the pile, correct to two decimal places. (b) A trailer carries

5

^3

of sand per load. How many full trailer loads are needed to remove all the sand?

Show worked answer →

Part (a): one mark for the correct cone formula $V = \tfrac{1}{3}\pi r^2 h$ with the right substitution $\tfrac{1}{3}\pi \times 3^2 \times 2 = 6\pi$ , and one mark for $6\pi = 18.85$ m $^3$ (2 d.p.). Part (b): one mark for dividing $18.85 \div 5 = 3.77$ , and one mark for recognising you must round UP to a whole load, giving $4$ loads. A marker docks the final mark for an answer of $3$ loads (which leaves sand behind) or for rounding $3.77$ to $4$ by ordinary rounding rather than by reasoning that $3$ loads is not enough. State the cone formula even if the arithmetic slips, since that line carries a mark.

2021 HSC-style5 marksA backyard swimming pool is a cylinder with a diameter of

4

m and a depth of

1.2

m. (a) Find the volume of the pool, correct to two decimal places. (b) Find the pool's capacity in litres when full. (c) The pool is filled to

90\%

of capacity. How many litres of water does it hold?

Show worked answer →

Part (a): one mark for halving the diameter to get $r = 2$ m (a frequent lost mark - students substitute the diameter), one mark for $V = \pi \times 2^2 \times 1.2 = 4.8\pi = 15.08$ m $^3$ . Part (b): one mark for $15.08 \times 1000 = 15\,080$ L using $1$ m $^3 = 1000$ L (full marks if they carry the unrounded $4.8\pi$ and write $15\,080$ L). Part (c): one mark for $0.90 \times 15\,080 = 13\,572$ L. Markers reward a clearly labelled radius and the unit (L) on the final line; using the diameter as the radius is the single most common error and loses the first two marks.

2023 HSC-style3 marksA rectangular metal block measures

8

cm by

5

cm by

4

cm. A cylindrical hole of radius

1.5

cm is drilled straight through the

4

cm thickness. Find the volume of metal remaining, correct to two decimal places.

Show worked answer →

One mark for the volume of the full block $8 \times 5 \times 4 = 160$ cm $^3$ ; one mark for the volume of the drilled cylinder $\pi \times 1.5^2 \times 4 = 9\pi = 28.27$ cm $^3$ (the hole's depth is the $4$ cm thickness, not a side length - a common slip); one mark for subtracting, $160 - 28.27 = 131.73$ cm $^3$ . Markers award the method marks for a correct 'block minus cylinder' set-up even if the radius or depth is misread, so the structure of the answer matters as much as the final number.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksFind the volume of a cylinder with radius

5

cm and height

10

cm, correct to two decimal places.

Show worked solution →

Use the prism rule $V = Ah$ with a circular base. A cylinder is a prism whose base is a circle of area $\pi r^2$ , so

V = \pi r^2 h.

Substitute $r = 5$ and $h = 10$ .

V = \pi \times 5^2 \times 10 = \pi \times 25 \times 10 = 250\pi

Evaluate and round.

250\pi = 785.398\ldots \approx 785.40 \text{ cm}^3

so the volume is $785.40$ cm $^3$ (to two decimal places). Keep $\pi$ until the last line so no rounding builds up.

foundation2 marksA square-based pyramid has a base

6

cm by

6

cm and a perpendicular height of

10

cm. Find its volume.

Show worked solution →

Use the pyramid rule, one third of base area times height. A pyramid is one third of the prism on the same base:

V = \tfrac{1}{3} A h.

Find the base area. The base is a square:

A = 6 \times 6 = 36 \text{ cm}^2.

Substitute and evaluate.

V = \tfrac{1}{3} \times 36 \times 10 = \tfrac{1}{3} \times 360 = 120 \text{ cm}^3

so the volume is $120$ cm $^3$ . (The answer is exact because $360$ divides by $3$ ; the one third is the whole difference between this pyramid and the box it sits inside.)

core3 marksA cone has radius

6

cm and perpendicular height

10

cm. Find its volume, correct to two decimal places.

Show worked solution →

Use the cone rule, one third of the cylinder on the same base. A cone is one third of the cylinder with the same radius and height:

V = \tfrac{1}{3} \pi r^2 h.

Substitute $r = 6$ and $h = 10$ .

V = \tfrac{1}{3} \times \pi \times 6^2 \times 10 = \tfrac{1}{3} \times \pi \times 36 \times 10 = \tfrac{1}{3} \times 360\pi = 120\pi

Evaluate and round.

120\pi = 376.991\ldots \approx 376.99 \text{ cm}^3

so the volume is $376.99$ cm $^3$ . (Compare with the foundation cylinder: same radius and height, but one third of a $360\pi$ cylinder gives $120\pi$ .)

core3 marksA storage shed has the shape of a rectangular prism with a triangular prism roof on top. The rectangular part is

4

m wide,

3

m high and

6

m long. The triangular roof has the same

4

m width and

6

m length, and rises a further

2

m to its ridge. Find the total volume of the shed.

Show worked solution →

Split the solid into named parts. The shed is a rectangular prism (the walls) plus a triangular prism (the roof). Find each volume, then add.

Volume of the rectangular prism. Use $V = Ah$ with the rectangular cross-section, or just length times width times height:

V_1 = 6 \times 4 \times 3 = 72 \text{ m}^3.

Volume of the triangular prism (the roof). Its cross-section is a triangle of base $4$ m and height $2$ m, and the prism is $6$ m long. The triangle area is

A = \tfrac{1}{2} \times 4 \times 2 = 4 \text{ m}^2,

V_2 = A \times \text{length} = 4 \times 6 = 24 \text{ m}^3.

Add the parts.

V = V_1 + V_2 = 72 + 24 = 96 \text{ m}^3

so the shed has a total volume of $96$ m $^3$ . (For any composite solid: split it, find each piece, then add the pieces that are there and subtract any that are cut out.)

core3 marksA cylindrical water tank has a radius of

0.7

m and a height of

1.5

m. Find its capacity in litres, correct to the nearest litre. (Recall

1

^3 = 1000

L.)

Show worked solution →

Find the volume in cubic metres. The tank is a cylinder, so

V = \pi r^2 h = \pi \times 0.7^2 \times 1.5 = \pi \times 0.49 \times 1.5 = 0.735\pi.

Evaluate.

0.735\pi = 2.3090\ldots \text{ m}^3

Convert volume to capacity. Using $1$ m $^3 = 1000$ L, multiply by $1000$ :

2.3090\ldots \times 1000 = 2309.07\ldots \approx 2309 \text{ L}

so the tank holds about $2309$ L. (Carry the unrounded volume into the conversion and round only at the end, so the litre figure is accurate.)

exam5 marksA grain silo is made of a cylinder of radius

2

m and height

6

m, topped by a cone of radius

2

m and height

1.5

m. (a) Find the total volume of the silo, correct to two decimal places. (b) Find the silo's capacity in kilolitres (

1

^3 = 1000

1

= 1000

L). (c) Grain is loaded at a steady

25

kilolitres per hour. How long, to the nearest minute, does it take to fill the empty silo?

Show worked solution →

Part (a) - split into a cylinder and a cone, then add.

Cylinder ( $V = \pi r^2 h$ ):

V_1 = \pi \times 2^2 \times 6 = \pi \times 4 \times 6 = 24\pi.

Cone ( $V = \tfrac{1}{3}\pi r^2 h$ ):

V_2 = \tfrac{1}{3} \times \pi \times 2^2 \times 1.5 = \tfrac{1}{3} \times \pi \times 4 \times 1.5 = 2\pi.

Total volume:

V = 24\pi + 2\pi = 26\pi = 81.681\ldots \approx 81.68 \text{ m}^3.

Part (b) - convert to a capacity. Using $1$ m $^3 = 1000$ L, the volume in litres is $26\pi \times 1000 = 81\,681\ldots$ L. Dividing by $1000$ to reach kilolitres (or using $1$ m $^3 = 1$ kL directly):

81.681\ldots \text{ kL} \approx 81.68 \text{ kL}.

Part (c) - filling time. Time is amount divided by rate. Use the unrounded capacity $26\pi$ kL:

t = \frac{26\pi}{25} = \frac{81.681\ldots}{25} = 3.267\ldots \text{ hours}.

Convert the hours to minutes by multiplying by $60$ :

3.267\ldots \times 60 = 196.0\ldots \approx 196 \text{ minutes}

so it takes about $196$ minutes, that is $3$ hours and $16$ minutes. (Working with $26\pi$ all the way avoids rounding error in the time.)

exam5 marksA length of concrete pipe is a hollow cylinder

3

m long. The outside radius is

0.5

m and the inside (the bore) has radius

0.3

m. (a) Find the volume of concrete in the pipe, correct to two decimal places. (b) Concrete has a mass of

2400

kg per cubic metre. Find the mass of the pipe, correct to the nearest kilogram.

Show worked solution →

Part (a) - subtract the hole from the solid. The concrete is the big cylinder minus the cylindrical bore. Both are $3$ m long, so

V = \pi R^2 L - \pi r^2 L = \pi L (R^2 - r^2).

Substitute $R = 0.5$ , $r = 0.3$ , $L = 3$ :

V = \pi \times 3 \times (0.5^2 - 0.3^2) = \pi \times 3 \times (0.25 - 0.09) = \pi \times 3 \times 0.16 = 0.48\pi.

Evaluate:

0.48\pi = 1.5079\ldots \approx 1.51 \text{ m}^3

so there is $1.51$ m $^3$ of concrete. (Subtracting the radii squared, $R^2 - r^2$ , is faster and more accurate than working out two volumes and subtracting.)

Part (b) - apply the density rate. Mass is volume times the mass per cubic metre. Use the unrounded volume $0.48\pi$ :

\text{mass} = 0.48\pi \times 2400 = 1.5079\ldots \times 2400 = 3619.11\ldots \approx 3619 \text{ kg}

so the pipe has a mass of about $3619$ kg. (Multiplying the rounded $1.51$ by $2400$ gives $3624$ kg, which is off by $5$ kg, so keep the exact volume until the end.)

What this dot point is asking

The answer

The five volume formulas

Why pyramids and cones have the one third

Composite solids: add or subtract the parts

Capacity: turning a volume into litres

How exam questions ask about volume and capacity

Exam-style practice questions

Practice questions

Related dot points