Volume and surface area of 3D solids

IB Mathematics: Applications and Interpretation SL· Unit 3: Geometry and Trigonometry· 6 min read

1. Formulas for Common 3D Solids★☆☆☆☆⏱ 15 min

IB AI SL requires you to use formulas for volume and surface area of six common 3D solids. Some formulas are provided in the formula booklet, but memorizing common ones saves critical time in exams.

📘 Definition

Volume

V

The total amount of space occupied by a 3D solid, measured in cubic units ( $cm^{3}$ , $m^{3}$ etc.)

Example:

A 1 cm cube has a volume of 1 $cm^{3}$

📘 Definition

Total Surface Area

SA

The total area of all outer faces of a 3D solid, measured in square units

Example:

A 1 cm cube has a total surface area of 6 $cm^{2}$

Solid	Volume Formula	Total Surface Area Formula
Cube (side $a$ )	$a^{3}$	$6 a^{2}$
Cuboid ( $l \times w \times h$ )	$l w h$	$2 (l w + l h + w h)$
Prism ( $A$ = cross-section area, $l$ = length)	$A l$	$2 A + P l$ ( $P$ = cross-section perimeter)
Pyramid ( $A$ = base area, $h$ = height)	$\frac{1}{3} A h$	Base area + sum of lateral faces
Cone ( $r$ = radius, $s$ = slant height)	$\frac{1}{3} π r^{2} h$	$π r^{2} + π r s$
Sphere ( $r$ = radius)	$\frac{4}{3} π r^{3}$	$4 π r^{2}$

📐 Worked Example

Calculate the volume and total surface area of a right circular cone with radius 3 cm and perpendicular height 4 cm. Give your answer to 3 significant figures.

1
First calculate the slant height $s$ of the cone using Pythagoras' theorem:
$s = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \text{ cm}$
2
Calculate volume using the cone volume formula:
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 (4) = 12\pi \approx 37.7 \text{ cm}^3$
3
Calculate total surface area using the cone SA formula:
$SA = \pi r^2 + \pi r s = \pi (3)^2 + \pi (3)(5) = 24\pi \approx 75.4 \text{ cm}^2$
4
Final answer (3 s.f.): Volume = 37.7 cm³, Total Surface Area = 75.4 cm²

2. Finding Unknown Dimensions★★☆☆☆⏱ 15 min

Common exam questions give you the volume or surface area of a solid and ask you to find an unknown dimension (radius, height, side length etc.). This requires rearranging the formula to isolate the unknown variable.

📐 Worked Example

The total volume of a sphere is 100 cm³. Calculate the radius of the sphere, correct to 2 decimal places.

1
Start with the standard sphere volume formula:
$V = \frac{4}{3}\pi r^3$
2
Substitute $V = 100$ and rearrange to isolate $r^{3}$ :
$100 = \frac{4}{3}\pi r^3 \\ r^3 = \frac{100 \times 3}{4\pi} = \frac{75}{\pi} \approx 23.873$
3
Take the cube root of both sides to solve for $r$ :
$r = \sqrt[3]{23.873} \approx 2.88 \text{ cm}$

✓ Quick check

Check your understanding:

A cube has a total surface area of 150 cm². What is its side length?
- 5 cm
- 12.5 cm
- 25 cm
- 150/6 = 25 cm
Reveal answer
5 cm —
Total SA of a cube = $6 a^{2}$ , so $6 a^{2} = 150 \to a^{2} = 25 \to a = 5$ cm

3. Volume and Surface Area of Composite Solids★★★☆☆⏱ 20 min

Most extended response exam questions for this topic involve composite solids, which are made by joining two or more simpler solids. There is a key difference between calculating total volume and total surface area for these shapes.

📐 Worked Example

A solid sculpture is made by attaching a hemisphere (half a sphere) of radius 10 cm to the top of a solid cylinder of radius 10 cm and height 20 cm. Calculate the total volume of the sculpture, to 3 significant figures.

1
Add the volume of the cylinder and the volume of the hemisphere:
$V_{\text{cylinder}} = \pi r^2 h = \pi (10)^2 (20) = 2000\pi \\ V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2000}{3}\pi$
2
Calculate total volume:
$V_{\text{total}} = 2000\pi + \frac{2000}{3}\pi = \frac{8000}{3}\pi \approx 8380 \text{ cm}^3$

4. Unit Conversion for Volume and Area★★☆☆☆⏱ 10 min

Unit conversion is a frequent exam test point for this topic, because it is an area of common mistake. Remember: area is a squared unit, volume is a cubed unit, so conversion factors must match.

📐 Worked Example

Convert a volume of 2.5 m³ to cm³.

1
Start with the length conversion, then cube it for volume:
$1 \text{ m} = 100 \text{ cm} \\ 1 \text{ m}^3 = (100 \text{ cm})^3 = 1 000 000 \text{ cm}^3$
2
Multiply by 2.5 to get the final converted volume:
$2.5 \times 1 000 000 = 2 500 000 \text{ cm}^3$

5. Common Pitfalls

Wrong move:

Forgetting to subtract overlapping area when calculating composite solid surface area

Why:

Overlapping faces between joined solids are internal and not part of the outer surface

Correct move:

Subtract twice the area of any overlapping face from the total sum of individual surface areas

Wrong move:

Using the same conversion factor for area/volume as for length

Why:

Area is squared length and volume is cubed length, so conversion factors must also be squared/cubed

Correct move:

Square the length conversion factor for area, cube it for volume

Wrong move:

Forgetting the $\frac{1}{3}$ factor in pyramid and cone volume formulas

Why:

Prism and pyramid formulas are similar, but pyramids have a 1/3 factor that is often omitted

Correct move:

Memorize that all point-topped solids (pyramids, cones) have a 1/3 volume factor

Wrong move:

Using perpendicular height instead of slant height for cone surface area

Why:

The lateral surface of a cone is calculated along the slanted edge, not the perpendicular height

Correct move:

Calculate slant height with Pythagoras' theorem before substituting into the SA formula

Wrong move:

Calculating total surface area when the question asks for lateral surface area

Why:

Students often rush and miss the keyword 'lateral' in the question

Correct move:

Always read the question carefully to confirm if total or lateral surface area is required

6. Quick Reference Cheatsheet

Solid	Volume	Total Surface Area
Cube (side $a$ )	$a^{3}$	$6 a^{2}$
Cuboid ( $l, w, h$ )	$l w h$	$2 (l w + l h + w h)$
Prism ( $A, l$ )	$A l$	$2 A + P l$
Pyramid ( $A, h$ )	$\frac{1}{3} A h$	Base + lateral
Cone ( $r, s$ )	$\frac{1}{3} π r^{2} h$	$π r^{2} + π r s$
Sphere ( $r$ )	$\frac{4}{3} π r^{3}$	$4 π r^{2}$
Unit conversion	Cube length factor	Square length factor

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Volume of composite cone and cylinder
2023 · 2
Surface area of sphere given volume

What's Next

Mastering volume and surface area of 3D solids is a foundational skill for all geometry topics in IB AI SL. You will use these core skills when solving problems involving density, similar 3D solids, and 3D coordinate geometry, all of which appear regularly in both Paper 1 and Paper 2. These skills are also often combined with other topics in extended, context-based questions on Paper 2, so building accuracy here will earn you easy marks in the exam.