Surface Area and Volume of Prisms and Cylinders
Prisms have the same cross-section all the way along their length. That single idea gives the main volume rule: volume = cross-sectional area × length. Surface area is different: it is the total area of every outside face or curved surface.
Foundation & Higher. Use perpendicular heights for volume and distinguish curved, total and exposed surface area.
Volume of a prism
Find the area of the constant cross-section first, then multiply by the perpendicular length of the prism.
V = A × l, where A is the cross-sectional area and l is the prism length.
Worked example: a triangular prism has triangular cross-section base 8 cm and perpendicular height 5 cm. Its length is 12 cm. Cross-sectional area = ½×8×5 = 20 cm², so volume = 20×12 = 240 cm³.
Surface area of a prism
Surface area means the total area of the outside. A net is often the safest way to make sure every face is counted exactly once.
For a triangular prism, add the two triangular ends and the three rectangular faces. For other prisms, the same principle applies: identify the two matching ends and every joining face.
Worked example: a cuboid measuring 8 cm by 5 cm by 3 cm has surface area 2(8×5)+2(8×3)+2(5×3) = 80+48+30 = 158 cm².
Worked example — triangular prism surface area
A right-triangular prism has a 3–4–5 cm triangle at each end and length 10 cm. Each triangular end has area ½×3×4=6 cm². Its three rectangles have areas 3×10, 4×10 and 5×10. Total surface area=2×6+30+40+50=132 cm². For a right prism this is also 2A+Pl, where P is the perimeter of the cross-section.
Cylinders
A cylinder has a constant circular cross-section. Its volume is therefore V = πr²h.
For total surface area, there are two circular ends and one curved surface. The curved surface unwraps to a rectangle whose width is the circumference 2πr and whose height is h. Therefore surface area = 2πr² + 2πrh.
Worked example: radius 4 cm and height 10 cm gives volume π×4²×10 = 160π cm³ ≈ 502.7 cm³. Total surface area = 2π×4² + 2π×4×10 = 112π cm² ≈ 351.9 cm².
Worked example — a missing cylinder radius
A cylinder has volume 180π cm³ and height 5 cm. Substitute into V=πr²h: 180π=5πr², so r²=36. Taking the positive square root gives r=6 cm. A radius cannot be negative.
Think Like a Mathematician
Do not choose a formula from the picture alone. First ask: what is the constant cross-section? For surface area, ask which surfaces are actually exposed. An open container, for example, does not include a missing lid.
Common Mistakes
- Using a sloping edge instead of the perpendicular prism length.
- Using circumference where circle area is required, or vice versa.
- Forgetting one or both circular ends when finding cylinder surface area.
- Giving area in cm³ or volume in cm².
- Rounding π too early instead of keeping the calculator value until the end.
Practice — surface area and volume
1. A prism has cross-sectional area 18 cm² and length 11 cm. Find its volume.
V = 18×11 = 198 cm³.
2. A cuboid measures 9 cm × 4 cm × 6 cm. Find its volume.
V = 9×4×6 = 216 cm³.
3. A triangular prism has cross-section base 10 cm, perpendicular height 6 cm and length 15 cm. Find its volume.
Triangle area = ½×10×6 = 30 cm². Volume = 30×15 = 450 cm³.
4. A cylinder has radius 3 cm and height 8 cm. Give its volume exactly.
V = π×3²×8 = 72π cm³.
5. The same cylinder has radius 3 cm and height 8 cm. Find its total surface area exactly.
2πr²+2πrh = 18π+48π = 66π cm².
6. A prism has volume 420 cm³ and cross-sectional area 35 cm². Find its length.
Length = 420÷35 = 12 cm.
7. A cylinder has volume 250π cm³ and height 10 cm. Find its radius.
250π = πr²×10, so r²=25 and r=5 cm.
8. An open-topped cuboid is 10 cm by 6 cm by 4 cm. Find the exposed surface area.
Base 10×6=60; two 10×4 faces=80; two 6×4 faces=48. Total = 188 cm².
Frequently Asked Questions
Why is prism volume cross-sectional area × length?
The solid can be thought of as identical cross-sections stacked through the prism length.
Why is the curved area of a cylinder 2πrh?
Unwrap the curved surface. It becomes a rectangle with width equal to the circumference 2πr and height h.
Continue Learning
Previous: Circles, Arcs and Sectors