POP

C5.4 Surface Area and Volume

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01 — Explanation

Which formulas are given to me, and which do I need to know?

Cambridge prints a List of Formulas on the exam paper covering the curved surface area of a cylinder and cone, the surface area of a sphere, and the volumes of a prism, pyramid, cylinder, cone and sphere. You don't need to memorise these — but you do need to correctly identify $r$ , $h$ and slant height $l$ from a diagram, and to find any flat-face areas the formula sheet leaves out, such as the circular ends of a cylinder or a pyramid's base.

Solid	Volume	Surface area
Cuboid	$V=lwh$	$SA = 2(lw+lh+wh)$
Prism	$V = \text{cross-section area}\times\text{length}$	Sum of all flat faces
Cylinder	$V=\pi r^2h$	$SA=2\pi r^2+2\pi rh$
Sphere	$V=\frac{4}{3}\pi r^3$	$SA=4\pi r^2$
Pyramid	$V=\frac{1}{3}\times\text{base area}\times h$	base area $+$ triangular faces
Cone	$V=\frac{1}{3}\pi r^2h$	$SA=\pi r^2+\pi rl$

"Prism" means any solid with a constant cross-section along its length, including a cylinder — this is why a cylinder's volume formula, $\pi r^2 h$ , is really just "cross-section area $\times$ length" with a circular cross-section.

02 — Explanation

How do I find the volume of a cone, pyramid or cylinder?

Worked example. A cone has radius $6\text{ cm}$ and perpendicular height $10\text{ cm}$ . Find its volume to 3 s.f.

V = \frac{1}{3}\pi r^2h = \frac{1}{3}\times\pi\times 6^2\times10 = 120\pi = 377\text{ cm}^3 \ (3\text{ s.f.})

Common mistake

using the slant height instead of the perpendicular height in a volume formula — a cone diagram usually labels the slant side $l$ more prominently. Volume always needs the perpendicular height, measured straight up from the base to the apex.

Worked example. A triangular prism has length $12\text{ cm}$ , and its triangular cross-section has base $5\text{ cm}$ and perpendicular height $4\text{ cm}$ . Find its volume.

\text{cross-section area} = \frac{1}{2}\times5\times4=10\text{ cm}^2 \qquad V = 10\times12=120\text{ cm}^3

03 — Explanation

How do I find the surface area of a sphere, cylinder or cone?

Worked example. Find the surface area of a sphere with radius $4.5\text{ cm}$ , in terms of $\pi$ .

SA = 4\pi r^2 = 4\times\pi\times4.5^2 = 81\pi\text{ cm}^2

The formula sheet's curved-surface-area formulas leave out any flat circular ends still present in the solid — add $\pi r^2$ for each one (two for a closed cylinder, one for a cone).

04 — Explanation

What do I do when the slant height of a cone isn't given directly?

A cone's slant height $l$ , its radius $r$ , and its perpendicular height $h$ always form a right-angled triangle, since the slant height is the hypotenuse running from the edge of the base up to the apex. If $l$ isn't stated, find it first with Pythagoras before using the surface area formula.

Worked example. A cone has radius $6\text{ cm}$ and perpendicular height $8\text{ cm}$ . Find its total surface area, in terms of $\pi$ .

l^2 = r^2+h^2 = 6^2+8^2 = 36+64=100 \qquad l = \sqrt{100}=10\text{ cm}

SA = \pi r^2 + \pi r l = \pi(6)^2+\pi(6)(10) = 36\pi+60\pi = 96\pi\text{ cm}^2

Exam tip

whenever a cone question gives the perpendicular height but the surface area formula needs the slant height (or vice versa), a quick Pythagoras step is almost always the missing link — check which one is actually needed before assuming a value is missing from the question.