POP

G4.14 Cylinder Problems: Volume, Capacity and Density

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01 — Explanation

Why does radius-vs-diameter matter so much for cylinders specifically?

The volume and curved surface area formulas, $V=\pi r^2h$ and $CSA=2\pi rh$ , both use the radius — but exam questions frequently give the diameter instead. Because $r$ is squared in the volume formula, substituting the diameter directly doesn't just double your answer, it makes it four times too big ( $r=d/2 \Rightarrow r^2=d^2/4$ ). Make it a habit: the moment you see a cylinder question, write $r=d\div2$ before touching any formula.

02 — Explanation

What's given on the formula sheet, and what do I have to supply myself?

The volume ( $\pi r^2h$ ) and curved surface area ( $2\pi rh$ ) are both provided. The total surface area is not — you must recall that it's the curved surface plus the two circular ends:

TSA = 2\pi r^2 + 2\pi rh

03 — Explanation

How do I find a missing height or radius (the inverse problem)?

Rearrange the volume formula algebraically rather than guessing:

h = \frac{V}{\pi r^2} \qquad r = \sqrt{\frac{V}{\pi h}}

Exam Tip

Writing the substituted equation explicitly — e.g. $\pi \times 8^2 \times h = 1208$ — secures the method mark even before you finish the arithmetic, and protects you if your final division or rounding slips.

04 — Explanation

How does a cylinder volume question turn into a capacity or density question?

These two conversions are the most common "second half" of a cylinder question:

Capacity: since $1\text{ L}=1000\text{ cm}^3$ , divide a volume in $\text{cm}^3$ by 1000 to get litres.
Density/mass: $M = D \times V$ . Watch the units of density carefully — if density is in $\text{g/cm}^3$ , your mass comes out in grams, and the question may then ask for kilograms, requiring a further division by 1000.

Example chain: a cylinder has volume $1208\text{ cm}^3$ and the material has density $1.25\text{ g/cm}^3$ . Mass $=1208\times1.25=1510$ g $=1.51$ kg. Missing the final g→kg conversion is a very common way to lose the last accuracy mark despite doing the harder calculation correctly.

05 — Explanation

How do I handle an open-top cylinder?

Exclude one circular base from the total surface area:

SA_{\text{open top}} = \pi r^2 + 2\pi rh

06 — Explanation

What precision is expected in the final answer?

Carry the full calculator value through every intermediate step (or keep answers "in terms of $\pi$ ," e.g. $80\pi$ ) and round only the final answer — usually to 3 significant figures unless the question specifies otherwise or gives an acceptable range (e.g. "answers which round to" a stated value).

Exam Tip

Don't confuse the volume formula ( $\pi r^2h$ , giving $\text{cm}^3$ ) with the curved surface area formula ( $2\pi rh$ , giving $\text{cm}^2$ ) — if your final units don't match what the question is asking for, you've used the wrong one.

07 — Worked example

Worked example

A cylindrical can has diameter 16 cm and height 24 cm. The metal it's made from has density $2.5\text{ g/cm}^3$ . Find the mass of the can in kilograms, to 2 d.p.

Radius: $r = 16\div2=8$ cm.
Volume: $V=\pi r^2h = \pi \times 8^2\times24 = 1536\pi \approx 4825.5\text{ cm}^3$ .
Mass in grams: $M=D\times V = 2.5\times4825.5=12\,063.7$ g.
Convert to kilograms: $12\,063.7\div1000=12.06$ kg (2 d.p.).