Volume and Surface Area of Prisms and Cylinders

Learn how to find the volume and surface area of a cylinder and other prisms for IGCSE Maths. This page covers the volume of a prism (cross-section area times length), the volume of a cuboid and cylinder, and how to work out surface area, including the curved surface area of a cylinder. Each method is shown with clear worked examples, with answers given as decimals and in terms of π, followed by auto-marked practice rooms so you can build confidence.

Prior Knowledge You should be able to find the area of rectangles, triangles and circles, and be comfortable substituting into a formula. If you need to revise circle area, see Circles, Arcs and Sectors.

The key idea: stack the slices

Think of a loaf of bread. Every slice is the same shape, and stacking the slices makes the whole loaf. A prism works the same way: it has the same cross-section (one slice) all along its length. So if you can find the area of one slice, the volume is just that area multiplied by how long the stack is.

The big idea \[ \text{Volume of a prism} = (\text{area of one slice}) \times (\text{length}) \]

A cuboid, a triangular prism and a cylinder are all prisms. Each one has a flat slice (its cross-section) at the end: a rectangle, a triangle, or a circle. Find the area of that slice, then multiply by the length.

Every prism is a stack of identical slicesrectangletrianglecircleVolume = area of the dotted slice × length

💡 Volume or surface area?

Volume measures the space inside, so it is in cubic units (cm³) and the formula multiplies three lengths. Surface area is the total area of the outside faces, so it is in square units (cm²) and each part multiplies two lengths. If your answer is a volume it should end in cm³; if it is a surface area it should end in cm².

Volume

For volume, the rule is the same for every prism: find the area of one slice (the cross-section), then multiply by the length.

A cuboid's slice is a rectangle, so its volume is just the three dimensions multiplied together. A cylinder's slice is a circle, so its volume is the circle area \(\pi r^2\) times the height \(h\). A triangular prism's slice is a triangle, so find \(\tfrac{1}{2}\times\text{base}\times\text{height}\) and multiply by the length.

Volume \[ \text{Any prism} = (\text{slice area}) \times \text{length} \] \[ \text{Cuboid} = w \times d \times h \qquad \text{Cylinder} = \pi r^2 h \]

Surface area

Surface area is a different skill from volume. It is the total area of all the outside faces added together. The clearest way to see every face is to open the solid out flat into its net, work out the area of each piece, then add them all up.

Cuboid: add the six faces

Start with the cuboid and its three dimensions. This one measures \(5\) by \(4\) by \(3\):

5 cm3 cm4 cm

Now open it out flat into its net: six rectangles, with opposite faces matching, so there are three pairs. The dimensions sit on the outside edges and the area of each face is written in the middle:

2015201512125 cm4 cm3 cm5 cm
Add every face
top \(= 20\)
bottom \(= 20\)
front \(= 15\)
back \(= 15\)
left \(= 12\)
right \(= 12\)
Total \(= 94 \text{ cm}^2\)

The shortcut is to add the three different faces and double, since each appears twice: \(2(20 + 15 + 12) = 2 \times 47 = 94 \text{ cm}^2\). Same answer, less writing.

Cuboid surface area \[ \text{Surface area} = 2(wd + wh + dh) \]

Triangular prism: two triangles and three rectangles

The net of a triangular prism opens into two triangles (the matching ends) and three rectangles (the sides). Each rectangle has one side equal to the length of the prism. Here is the net for a prism with a \(3, 4, 5\) right-angled triangle and length \(10\):

3×104×105×10triangletriangle345length 10

Step 1: the two triangles each have area \(\tfrac{1}{2}\times 3 \times 4 = 6\), giving \(12\) altogether.

Step 2: the three rectangles are \(3\times 10 = 30\), \(4\times 10 = 40\) and \(5\times 10 = 50\), giving \(120\).

Step 3: add everything: \(12 + 120 = 132 \text{ cm}^2\).

Cylinder: unroll the curved surface

The net of a cylinder is two circles (the ends) plus one rectangle (the curved surface). The key link is that the rectangle's width is exactly the circumference of the circle, \(2\pi r\), and its height is the height of the cylinder, \(h\). Press play (or drag the slider) to watch the net fold itself into the cylinder: the rectangle rolls into the tube and the two circles close the ends.

width = 2πr h end = πr² end = πr² the width 2πr is now the circumference: curved surface = 2πr × h press play: the rectangle rolls into the tube, the circles close the ends
Flat net

So the curved surface area is \(2\pi r \times h\), and adding the two circular ends (each \(\pi r^2\)) gives the total.

Cylinder surface area \[ \text{Curved} = 2\pi r h \qquad \text{Total} = 2\pi r h + 2\pi r^2 \]

Worked examples

💡 Example 1: cuboid volume

A cuboid measures \(8\) cm by \(4\) cm by \(5\) cm. Find its volume.

8 cm5 cm4 cm
\[ V = w \times d \times h \] \[ V = 8 \times 4 \times 5 \] \[ V = 160 \text{ cm}^3 \]
What's happening?

Multiply the three dimensions together.

Volume is in cubic units, so the answer is in cm³.

💡 Example 2: cuboid surface area

Find the total surface area of the same cuboid (\(8\) by \(4\) by \(5\) cm).

8 cm5 cm4 cm
\[ A = 2(wd + wh + dh) \] \[ = 2(8\cdot 4 + 8\cdot 5 + 4\cdot 5) \] \[ = 2(32 + 40 + 20) \] \[ = 184 \text{ cm}^2 \]
What's happening?

There are three pairs of matching faces.

Find the area of each different face, add them, then double.

💡 Example 3: cylinder volume

A cylinder has radius \(3\) cm and height \(10\) cm. Find its volume, in terms of \(\pi\) and to 3 significant figures.

3 cm10 cm
\[ V = \pi r^2 h \] \[ V = \pi \times 3^2 \times 10 \] \[ V = 90\pi \] \[ V = 283 \text{ cm}^3 \]
What's happening?

Square the radius first: \(3^2 = 9\), then multiply by the height.

Leave it as \(90\pi\) for an exact answer, or multiply out for a decimal.

💡 Example 4: cylinder surface area

A closed cylinder has radius \(5\) cm and height \(12\) cm. Find its total surface area, to 3 significant figures.

5 cm12 cm
\[ A = 2\pi r h + 2\pi r^2 \] \[ = 2\pi(5)(12) + 2\pi(5)^2 \] \[ = 120\pi + 50\pi = 170\pi \] \[ = 534 \text{ cm}^2 \]
What's happening?

The curved part is \(2\pi r h = 120\pi\).

The two circular ends add \(2\pi r^2 = 50\pi\). Add them for the total.

💡 Example 5: triangular prism

A prism has a right-angled triangular cross-section with base \(6\) cm and height \(4\) cm, and a length of \(15\) cm. Find its volume.

6 cm4 cm15 cm
\[ \text{Cross-section} = \tfrac{1}{2} \times 6 \times 4 = 12 \text{ cm}^2 \] \[ V = A \times L \] \[ V = 12 \times 15 = 180 \text{ cm}^3 \]
What's happening?

First find the area of the triangular cross-section.

Then multiply by the length of the prism. This works for any prism: find the cross-section area, then multiply by the length.

🔑 Key points

  • Volume of any prism \(= \text{cross-section area} \times \text{length}\).
  • Cuboid: \(V = w\,d\,h\); cylinder: \(V = \pi r^2 h\).
  • Curved surface area of a cylinder \(= 2\pi r h\).
  • Total surface area of a closed cylinder \(= 2\pi r h + 2\pi r^2\).
  • Volume is in cm³; surface area is in cm².
  • "In terms of \(\pi\)" means leave \(\pi\) in the answer.

⚠️ Common pitfalls

  • Using the diameter instead of the radius in \(\pi r^2 h\).
  • Forgetting the two ends when a cylinder is closed.
  • Mixing up volume (cubic units) and surface area (square units).
  • Squaring \(\pi r\) instead of just \(r\).
  • Forgetting to find the cross-section area first for a prism.
  • Rounding too early; keep full accuracy until the final line.
⇩ Jump to Practice Questions ⇩

Ready to practise? The rooms below cover cuboids, cylinder volume, cylinder surface area and prisms, with a mix of exact and decimal answers.

Next: Volume of a Frustum →

Prisms and Cylinders: Practice Room

Work from the diagram in every question, auto-marked as you type. Watch the tag: 3 s.f. wants a rounded decimal, and in terms of π wants an exact answer such as 90π. For exact answers, type pi or tap the π button next to the box; both are accepted. Questions are randomly generated, so refresh for a new set.

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