Math Core

Three-dimensional problems look harder than they are. Volume is "area of the base times height" for any prism or cylinder, surface area is "add up the faces," and most tricky questions come down to picturing the solid clearly: which faces show, which cubes are hidden, and how the numbers change when the solid is scaled.

The formulas

Solid	Volume	Surface area
Rectangular box $l \times w \times h$	$lwh$	$2(lw + wh + hl)$
Cube with edge $s$	$s^3$	$6s^2$
Any prism or cylinder	$\text{base area} \times \text{height}$	two bases plus the sides
Cylinder, radius $r$ , height $h$	$\pi r^2 h$	$2\pi r^2 + 2\pi r h$
Pyramid or cone	$\dfrac{1}{3} \cdot \text{base area} \times \text{height}$	(rarely needed at this level)
Sphere, radius $r$	$\dfrac{4}{3}\pi r^3$	$4\pi r^2$

A cylinder. Unrolling the side gives a rectangle 2πr wide and h tall.

The side of a cylinder unrolls into a rectangle whose width is the circumference $2\pi r$ and whose height is $h$ . That is where $2\pi r h$ comes from. A cone holds exactly one third as much as a cylinder with the same base and height.

The space diagonal of a box (from one corner to the opposite corner, through the inside) is $\sqrt{l^2 + w^2 + h^2}$ . It comes from using the Pythagorean theorem twice: once across the bottom face, then up.

Scaling

Scale factors in 3D

If every length of a solid is multiplied by $k$ :

every length is multiplied by $k$ ,
the surface area is multiplied by $k^2$ ,
the volume is multiplied by $k^3$ .

So a cube with twice the edge length needs $4$ times as much paint and holds $8$ times as much water.

Unit cubes and painted cubes

A large cube built from $n \times n \times n$ small cubes and painted on the outside has:

$8$ corner cubes with $3$ painted faces,
$12(n - 2)$ edge cubes with $2$ painted faces,
$6(n - 2)^2$ face cubes with $1$ painted face,
$(n - 2)^3$ hidden cubes with no paint.

Don't memorize this. Picture it: the unpainted cubes form a smaller cube inside, with one layer peeled off each side.

Worked example: Volume from face areas

A rectangular box has faces with areas $12$ , $15$ and $20$ square inches. Find its volume.

Call the edges $l$ , $w$ , $h$ with $lw = 12$ , $wh = 15$ , $hl = 20$ . Multiply all three: $(lwh)^2 = 12 \cdot 15 \cdot 20 = 3600$ , so the volume is $lwh = 60$ cubic inches. (The edges are $4$ , $3$ and $5$ .)

Worked example: A painted cube

A $4 \times 4 \times 4$ cube is painted on the outside, then cut into $64$ unit cubes. How many unit cubes have exactly two painted faces?

Cubes with two painted faces lie along the edges but not at the corners. Each of the $12$ edges has $4 - 2 = 2$ such cubes, for $24$ in all.

Check the full count: $8$ corners, $24$ edge cubes, $6 \cdot 2^2 = 24$ face cubes and $2^3 = 8$ hidden cubes. $8 + 24 + 24 + 8 = 64$ .

Worked example: Water displacement

A cylinder with radius $6$ is partly filled with water. A solid ball with radius $3$ is dropped in and sinks completely under the water. How much does the water level rise?

The ball pushes up water equal to its own volume, $\dfrac{4}{3}\pi \cdot 27 = 36\pi$ . That water forms a thin cylinder with radius $6$ and height $x$ , with volume $36\pi x$ . So $36\pi x = 36\pi$ , and the water rises $x = 1$ unit.

Worked example: The space diagonal

Find the length of the space diagonal of a $6 \times 2 \times 3$ box.

A 6 by 2 by 3 box with a space diagonal.

$\sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$ .

Common mistake

When cubes are glued together, the faces that touch are hidden, and each glued joint hides two faces (one on each cube). Don't just add up the surface areas of the separate pieces.

Tip

For a solid made of stacked blocks, count surface area by views: look from the top, bottom, front, back, left and right, and add the areas you see. Each view you see from one direction appears again from the opposite direction.