Math Core

Volume tells you how much a container holds. Surface area tells you how much material it takes to make or cover it: the cardboard in a box, the label on a can, the paint on a dome. It is the total area of every face or curved surface of a solid, measured in square units.

Nets

A net is a flat pattern that folds up into a solid. Unfolding a solid into a net turns a 3D question into a 2D one: the surface area is just the total area of the net.

The net of a 6 by 4 by 3 box. The four side faces unfold into one long rectangle whose length is the perimeter of the base.

Look closely at the net of the box. The four side faces unroll into a single long rectangle. Its height is the height of the box, 3, and its length runs all the way around the base: $6 + 4 + 6 + 4 = 20$ , which is the perimeter of the base. This observation works for every right prism.

The lateral area $L$ is the area of the sides only, without the bases.
The total surface area $S$ is the lateral area plus the area of both bases.

Prisms and cylinders

For a right prism with base perimeter $P$ , base area $B$ and height $h$ , the sides form a rectangle $P$ by $h$ , so

S = Ph + 2B

A cylinder works the same way. Peel off the label of a can and it flattens into a rectangle. Its height is $h$ , and its length is the distance around the circle, $2\pi r$ .

The net of a cylinder: two circles and a rectangle. The rectangle's length is the circumference of the base.

Worked example: A box and a can

a) Find the surface area of a box that is 6 in by 4 in by 3 in.

The base is 6 by 4, so $P = 20$ and $B = 24$ :

S = Ph + 2B = 20(3) + 2(24) = 60 + 48 = 108 \text{ in}^2

Check by adding the three pairs of faces: $2(6 \cdot 4) + 2(6 \cdot 3) + 2(4 \cdot 3) = 48 + 36 + 24 = 108$ . ✓

b) Find the surface area of a cylinder with radius 5 cm and height 8 cm.

S = 2\pi rh + 2\pi r^2 = 2\pi(5)(8) + 2\pi(5)^2 = 80\pi + 50\pi = 130\pi \approx 408.4 \text{ cm}^2

Pyramids and cones

A regular pyramid's lateral faces are congruent isosceles triangles. Each one has a base equal to a side $s$ of the base polygon and a height equal to the slant height $\ell$ . With $n$ faces, the lateral area is $n \cdot \tfrac{1}{2}s\ell = \tfrac{1}{2}(ns)\ell = \tfrac{1}{2}P\ell$ .

A square pyramid with base edges of 10 in. The slant height of each face is 12 in.

Worked example: A square pyramid

Find the surface area of the pyramid above.

The base is a 10 by 10 square, so $B = 100$ and $P = 40$ . The slant height is 12:

S = B + \tfrac{1}{2}P\ell = 100 + \tfrac{1}{2}(40)(12) = 100 + 240 = 340 \text{ in}^2

For a cone, cut the curved side along a slant height and flatten it. You get a sector of a circle with radius $\ell$ . Its arc wraps around the base, so the arc length is $2\pi r$ . A full circle of radius $\ell$ would have circumference $2\pi\ell$ and area $\pi\ell^2$ . The sector is the fraction $\tfrac{2\pi r}{2\pi\ell} = \tfrac{r}{\ell}$ of that circle, so its area is

\frac{r}{\ell} \cdot \pi\ell^2 = \pi r\ell

That's the lateral area of a cone. It matches the pyramid formula $\frac{1}{2}P\ell$ with the perimeter replaced by the circumference: $\frac{1}{2}(2\pi r)\ell = \pi r\ell$ .

Worked example: A cone from its height

A cone has a radius of 6 cm and a height of 8 cm. Find its total surface area.

The formula needs the slant height, not the height:

\ell = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \text{ cm}

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

Common mistake

Volume uses the height $h$ . Surface area of a pyramid or cone uses the slant height $\ell$ . Each problem tells you one of them, and you may need the Pythagorean theorem to get the other. Before you substitute, ask which length the formula wants.

Spheres

A sphere can't be unrolled into a flat net, but Archimedes found its area anyway: a sphere's surface area equals the lateral area of the cylinder that just contains it. That cylinder has radius $r$ and height $2r$ , so its lateral area is $2\pi r(2r) = 4\pi r^2$ . In other words, a sphere's surface is exactly four times the area of its great circle.

Surface area formulas

Solid	Total surface area
Right prism	$S = Ph + 2B$
Right cylinder	$S = 2\pi rh + 2\pi r^2$
Regular pyramid	$S = \frac{1}{2}P\ell + B$
Right cone	$S = \pi r\ell + \pi r^2$
Sphere	$S = 4\pi r^2$

$P$ is the perimeter of the base, $B$ is the area of the base, and $\ell$ is the slant height.

Tip

Before reaching for a formula, ask what surface the question is really about. A label covers only the lateral area of a can. A pool liner covers the bottom and sides but not the top. A solid hemisphere has a curved half-sphere plus a flat circle.