$V_sphere = (4)/(3)π r^3, surface area = 4π r^2$

What is wrong power of r for a sphere?

Volume uses r^3 and surface area uses r^2; do not swap them.

What are mismatched units?

Volume is in cubic units and surface area in square units; convert lengths first.

Find the volume of a cube of side 5\ cm. [1 mark]

Find the volume of a cone with radius 3\ cm and height 9\ cm, taking π = 3.142. [2 marks]

A cone has radius 5\ cm and vertical height 12\ cm. Find its slant height. [2 marks]

SEAB Original-style practice: A solid cylinder has radius 4\ cm and height 10\ cm. Taking π = 3.142, find (a) its volume and (b) its total surface area, each to 1 decimal place.

(a) Volume of a cylinder = π r^2 h = 3.142 × 4^2 × 10 = 3.142 × 16 × 10 = 502.72, which is 502.7\ cm^3. (b) Total surface area = 2π r^2 + 2π r h. The two circles: 2 × 3.142 × 16 = 100.544. The curved surface: 2 × 3.142 × 4 × 10 = 251.36. Total = 351.904, which is 351.9\ cm^2. Markers reward the volume formula π r^2 h, both parts of the surface area (two ends plus the curved surface), and correct rounding.

SingaporeMathsSyllabus dot point

How do we find the volume and surface area of prisms, cylinders, cones, pyramids and spheres?

Calculate the volume and surface area of prisms, cylinders, pyramids, cones and spheres, and of composite solids

A focused answer to the O-Level E-Maths outcome on volume and surface area. The formulas for prisms, cylinders, cones, pyramids and spheres, curved and total surface area, and composite solids.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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What this dot point is asking

SEAB wants you to calculate the volume and surface area of the standard solids, prisms, cylinders, pyramids, cones and spheres, and of composite solids built from them. Knowing each formula and which dimensions it needs, including the slant height for cones, is essential.

The answer

Prisms and cylinders

A prism has a uniform cross-section, and its volume is the cross-sectional area times the length:

V_{\text{prism}} = (\text{cross-sectional area}) \times \text{length}

A cylinder is a prism with a circular cross-section, so $V = \pi r^2 h$ . Its curved surface area is $2\pi r h$ , and the total surface area adds the two circular ends: $2\pi r^2 + 2\pi r h$ .

Cones

A cone has volume one third of the cylinder with the same base and height:

V_{\text{cone}} = \frac{1}{3}\pi r^2 h

Its curved surface area is $\pi r l$ , where $l$ is the slant height, found from the radius and vertical height by Pythagoras: $l = \sqrt{r^2 + h^2}$ .

Pyramids

A pyramid's volume is one third of the base area times the vertical height:

V_{\text{pyramid}} = \frac{1}{3} \times (\text{base area}) \times h

The one-third factor it shares with the cone reflects that both taper to a point.

Spheres

A sphere of radius $r$ has:

V_{\text{sphere}} = \frac{4}{3}\pi r^3, \qquad \text{surface area} = 4\pi r^2

These two formulas are given to you in the exam, but you must know which radius power each uses.

Composite solids

Combine solids by adding volumes, or hollow one out by subtracting. For surface area, count only the faces actually on the outside, since joined faces are hidden.

Worked example

A solid is a cylinder of radius $3\ \text{cm}$ and height $7\ \text{cm}$ topped by a hemisphere of the same radius. Taking $\pi = 3.142$ , find the total volume to 1 decimal place.

Step 1: Find the cylinder's volume

V_{\text{cylinder}} = \pi r^2 h = 3.142 \times 9 \times 7 = 197.946\ \text{cm}^3

Step 2: Find the hemisphere's volume

A hemisphere is half a sphere: $\dfrac{1}{2} \times \dfrac{4}{3}\pi r^3 = \dfrac{2}{3}\pi r^3 = \dfrac{2}{3} \times 3.142 \times 27 = 56.556\ \text{cm}^3$ .

Step 3: Add the two volumes

V_{\text{total}} = 197.946 + 56.556 = 254.502\ \text{cm}^3

Step 4: Round to 1 decimal place

The total volume is $254.5\ \text{cm}^3$ .

Examples in context

Example 1. A storage tank. A fuel tank shaped as a cylinder with hemispherical ends has its capacity found by adding a cylinder volume to a full sphere (the two hemispheres). Engineers use exactly this composite calculation to specify capacity.

Example 2. An ice cream cone. A scoop modelled as a hemisphere sitting on a cone gives the total volume as a hemisphere plus a cone. The combination is a standard composite-solid exam scenario.

Try this

Q1. Find the volume of a cube of side $5\ \text{cm}$ . [1 mark]

Cue. $5^3 = 125\ \text{cm}^3$ .

Q2. Find the volume of a cone with radius $3\ \text{cm}$ and height $9\ \text{cm}$ , taking $\pi = 3.142$ . [2 marks]

Cue. $\dfrac{1}{3} \times 3.142 \times 9 \times 9 = 84.834\ \text{cm}^3$ , about $84.8\ \text{cm}^3$ .

Q3. A cone has radius $5\ \text{cm}$ and vertical height $12\ \text{cm}$ . Find its slant height. [2 marks]

Cue. $l = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\ \text{cm}$ .

Exam-style practice questions

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

Original4 marksA solid cylinder has radius

4\ \text{cm}

and height

10\ \text{cm}

. Taking

\pi = 3.142

, find (a) its volume and (b) its total surface area, each to 1 decimal place.

Show worked answer →

(a) Volume of a cylinder $= \pi r^2 h = 3.142 \times 4^2 \times 10 = 3.142 \times 16 \times 10 = 502.72$ , which is $502.7\ \text{cm}^3$ .

(b) Total surface area $= 2\pi r^2 + 2\pi r h$ . The two circles: $2 \times 3.142 \times 16 = 100.544$ . The curved surface: $2 \times 3.142 \times 4 \times 10 = 251.36$ . Total $= 351.904$ , which is $351.9\ \text{cm}^2$ .

Markers reward the volume formula $\pi r^2 h$ , both parts of the surface area (two ends plus the curved surface), and correct rounding.

Original4 marksA cone has base radius

6\ \text{cm}

and vertical height

8\ \text{cm}