Volume and Surface Area

Intuition

The last two lessons stayed flat. Perimeter measured the distance around a shape and area measured the space inside it, both living on a single sheet of paper. Real objects are not flat: a cereal box, a soup can, and a brick all take up space in three directions. Two new questions come up the moment a shape has thickness. The first is “how much space does it fill?”, which is its volume. Volume is the right measure for how much cereal a box holds or how much water a tank takes. The second is “how much material covers the outside?”, which is its surface area. Surface area is the right measure for how much cardboard the box needs or how much paint coats the can. Volume counts unit cubes; surface area adds up the areas of the faces. Both follow from ideas you already have, stacking and unfolding, so you can rebuild every formula here yourself.

Learning goals

Count unit cubes for volume, in cubic units
Stack layers to get $V = lwh$ for a rectangular prism
Multiply base area by height for any prism or cylinder
Unfold a net and add the face areas for surface area
Read a cylinder's side as a rectangle whose width is the circumference
Keep cubic units for volume and square units for surface area

Volume: counting unit cubes

To measure a flat region we tiled it with unit squares. To measure a solid we do the same thing one dimension up: we fill it with unit cubes. A unit cube is a cube one unit long on every edge. A cube one centimetre on each edge has a volume of one cubic centimetre, written $1 \text{ cm}^3$ . The volume of any solid is simply the number of unit cubes it takes to fill the solid with no gaps and no overlaps.

This is why volume is always measured in cubic units. A length needs one direction and uses plain units like cm; an area needs two directions and uses square units like $\text{cm}^2$ . A volume needs three directions, and the little raised $3$ in $\text{cm}^3$ records exactly that. The exponent is not decoration. It counts how many directions the unit spans.

Volume of a rectangular prism

A rectangular prism has a length $l$ , a width $w$ , and a height $h$ , all meeting at right angles. A box and a brick are both rectangular prisms. Filling the prism with unit cubes turns into a multiplication the same way tiling a rectangle did. The one difference is that now there is an extra factor for the stacked layers.

A box 4 units long, 3 units wide, and 2 units high, filled with unit cubes. One bottom layer holds 4 times 3 equals 12 cubes; stacking 2 such layers gives 24 cubes in all, so the volume is 24 cubic units.

The figure shows a box $4$ long, $3$ wide, and $2$ high. Look at just the bottom layer of cubes. It is one cube thick and tiles the floor of the box, which is a $4$ by $3$ rectangle, so that single layer holds

4 \times 3 = 12 \text{ cubes}.

That count is exactly the area of the base. Now stack layers to fill the height. Each layer is a copy of the bottom one with the same $12$ cubes, and the box is $2$ layers tall, so the total is

12 \times 2 = 24 \text{ cubes}.

There is the whole idea: fill one layer (that is length times width), then stack height-many copies of it (that is the times height). Multiplying all three counts every cube exactly once.

Why the volume of a rectangular prism is length times width times height#

Take a box whose length is $l$ units, width is $w$ units, and height is $h$ units, with $l$ , $w$ , and $h$ whole numbers. Fill the bottom of the box with a single layer of unit cubes. The floor is a rectangle $l$ long and $w$ wide, and one cube covers one unit square of it. Therefore this bottom layer contains exactly $l \times w$ cubes, one cube for each unit square of the floor.

Now stack identical layers upward until the box is full. Each layer is one unit tall and is a copy of the bottom layer, so each holds the same $l \times w$ cubes. The box is $h$ units tall, so it takes exactly $h$ layers to reach the top, with no gaps and no overlaps.

The total number of cubes is the cubes in one layer added up once per layer:

V = \underbrace{(l \times w) + (l \times w) + \cdots + (l \times w)}_{h \text{ layers}} = (l \times w) \times h.

Every unit cube sits in exactly one layer and one spot within that layer, so the product counts each cube once. That is the volume.

So for a rectangular prism,

V = l \times w \times h.

A cube is the special box whose length, width, and height are all the same value $s$ , so its volume is

V = s \times s \times s = s^3.

This is exactly why $s^3$ is read ” $s$ cubed”. Raising a length to the third power is the volume of the cube built on that length. That is just what $s^2$ did for the area of a square. The exponent notation from chapter 6 and the geometry of a cube are the same idea seen from two sides.

Check your understanding

A box is $5$ cm long, $3$ cm wide, and $4$ cm high. What is its volume?

60 \text{ cm}^3

12 \text{ cm}^3

60 \text{ cm}^2

47 \text{ cm}^3

Volume of any prism: base area times height

A prism is a solid with two identical flat ends (the bases) joined by straight sides. Every cross-section parallel to the bases is the same shape and size all the way through, which is what makes it a prism. A box is a prism whose base is a rectangle. A triangular prism, the shape of a tent or a wedge, is a prism whose base is a triangle. The layer-and-stack argument never used the fact that the base was a rectangle, so it carries over word for word to any prism.

Slice the prism into thin layers parallel to its base. Each layer is a flat copy of the base, and stacking enough of them to reach the height $h$ rebuilds the solid. The number of unit cubes in one unit-thick layer is just the area of the base, $B$ , and there are $h$ layers, so

V = B \times h.

For a box the base is a rectangle, $B = l \times w$ , and this collapses back to $l \times w \times h$ . For a triangular prism the base is a triangle, so $B = \tfrac{1}{2} \, b \, h_\triangle$ using the triangle area from the previous lesson. The volume of that prism is then that base area times the length of the prism. The single rule volume equals base area times height covers every prism at once.

Worked example 1 Volume of a triangular prism

A tent is a triangular prism. Its triangular end has a base of $6$ m and a perpendicular height of $4$ m, and the tent runs $10$ m long. Find its volume.

First find the area of the triangular base, $B$ , using the triangle area formula:

B = \frac{1}{2} \times b \times h_\triangle = \frac{1}{2} \times 6 \times 4 = 12 \text{ m}^2.

The prism is $10$ m long, so that base area is stacked through a height (here a length) of $10$ m:

V = B \times h = 12 \times 10 = 120 \text{ m}^3.

So the tent encloses $120$ cubic metres. The base area carries the units of square metres, and the extra factor of metres from the length pushes the result up to cubic metres.

Volume of a cylinder

A cylinder, the shape of a can or a pipe, is like a prism but with a circle for its base instead of a polygon. The reasoning does not change at all. Stack thin circular disks, each a copy of the base circle, until they reach the height $h$ . Each unit-thick disk holds a number of cubes equal to the area of the circle, and there are $h$ of them. So once again the cylinder’s volume is base area times height:

V = (\text{base area}) \times h.

The base is a circle of radius $r$ , and you found its area in the last lesson: $A = \pi r^2$ . Substituting that base area gives the cylinder volume:

V = \pi r^2 h.

Notice this is the very same “base area times height” rule, with the circle’s $\pi r^2$ playing the part the rectangle’s $l \times w$ played for a box. The only new ingredient is the circle area, which you already have. As always with $\pi$ , you must use the radius: if a problem gives the diameter, halve it first.

Worked example 2 Volume of a cylinder

A can of soup is a cylinder with radius $3$ cm and height $10$ cm. Find its volume. Use $\pi \approx 3.14$ .

The base is a circle of radius $3$ cm, so find its area first:

A = \pi r^2 = 3.14 \times 3^2 = 3.14 \times 9 = 28.26 \text{ cm}^2.

Now stack that base through the height of $10$ cm:

V = A \times h = 28.26 \times 10 = 282.6 \text{ cm}^3.

So the can holds about $283$ cubic centimetres. The answer is approximate because $\pi$ was rounded to $3.14$ . The unit is cubic centimetres because squaring the radius gave square centimetres and the height added one more factor of centimetres.

Worked example 3 A cylinder given the diameter (halve first)

A water pipe is a cylinder with a diameter of $8$ cm and a length of $20$ cm. Find its volume. Use $\pi \approx 3.14$ .

The formula needs the radius, but the diameter was given, so halve it first:

r = \frac{d}{2} = \frac{8}{2} = 4 \text{ cm}.

Now find the base circle area and multiply by the length:

A = \pi r^2 = 3.14 \times 4^2 = 3.14 \times 16 = 50.24 \text{ cm}^2,

V = A \times h = 50.24 \times 20 = 1004.8 \text{ cm}^3.

So the pipe holds about $1005$ cubic centimetres. Using the diameter $8$ in place of the radius would square to $64$ instead of $16$ , making the answer four times too big. That is the same halve-first trap from the circles lesson.

Check your understanding

A cylinder has radius $5$ m and height $2$ m. What is its volume? Use $\pi \approx 3.14$ .

157 \text{ m}^3

31.4 \text{ m}^3

62.8 \text{ m}^3

157 \text{ m}^2

Surface area: unfolding the net

Volume measured the inside. Surface area measures the outside: it is the total area of all the faces of a solid. That total is the amount of material it would take to wrap or paint the solid. Because it is a sum of flat areas, surface area is measured in square units, just like ordinary area, not in cubic units. Keeping that straight is the single biggest trap of this lesson, so it is worth saying once more: volume is cubic, surface area is square.

The clean way to find surface area is to unfold the solid flat. Cut along enough edges and lay every face out on a table, and you get a flat picture called a net. The net has exactly the same faces as the solid, just spread out where you can see them. So the solid’s surface area is the plain area of that net: the sum of the face areas you already know how to compute.

Unfolding a box into its net. The six faces fold flat into a cross: top and bottom, front and back, and the two ends. Opposite faces are equal, so the six areas come in three matching pairs.

For a box the net is six rectangles that come in three matching pairs, because opposite faces of a box are identical. With length $l$ , width $w$ , and height $h$ :

The top and bottom are each $l$ by $w$ , area $l \times w$ .
The front and back are each $l$ by $h$ , area $l \times h$ .
The two ends are each $w$ by $h$ , area $w \times h$ .

Add all six. Each kind appears twice, so

S = 2(l \times w) + 2(l \times h) + 2(w \times h) = 2(lw + lh + wh).

Why the surface area of a box is $2(lw + lh + wh)$ #

A rectangular box has exactly six flat faces, and the surface area is the sum of their six areas. The faces come in three pairs of opposites. Opposite faces of a box are congruent, meaning identical in size and shape. That congruence holds because the box maintains the same cross-section all the way across in each direction.

The top and the bottom are both rectangles of size $l$ by $w$ , so together they contribute $2(l \times w)$ . The front and the back are both $l$ by $h$ , contributing $2(l \times h)$ . The left and right ends are both $w$ by $h$ , contributing $2(w \times h)$ . Every face has now been counted exactly once.

Adding the three pairs gives the total surface area:

S = 2(l \times w) + 2(l \times h) + 2(w \times h).

Factoring the common $2$ out front writes it more compactly as $S = 2(lw + lh + wh)$ . The same logic, unfold and add the face areas, works for any solid; only the shapes of the faces change.

A cube is the box where $l = w = h = s$ , so all six faces are identical squares of area $s^2$ , and

S = 6 s^2.

Worked example 4 Surface area of a box

A box is $5$ cm long, $4$ cm wide, and $3$ cm high. Find its surface area.

Find the area of one face of each kind, then double each because opposite faces match:

lw = 5 \times 4 = 20, \qquad lh = 5 \times 3 = 15, \qquad wh = 4 \times 3 = 12.

Add the three and double the total:

S = 2(20 + 15 + 12) = 2 \times 47 = 94 \text{ cm}^2.

So it takes $94$ square centimetres of material to cover the box. The unit is square centimetres because surface area adds up flat face areas, even though the box itself encloses a volume measured in cubic centimetres.

Check your understanding

A cube has an edge of $4$ cm. What is its surface area?

96 \text{ cm}^2

64 \text{ cm}^3

16 \text{ cm}^2

24 \text{ cm}^2

Surface area of a cylinder

A cylinder unfolds too, and its net reveals the formula. Cut off the two circular ends and slit the curved side from top to bottom. The two ends are circles. The curved side, once unrolled, flattens into a rectangle, and that is the only piece that needs a second look.

Unrolling a cylinder. The two ends are circles of radius r. The curved side unrolls into a rectangle whose height is the cylinder's height h and whose width is the circumference of the circle, 2 pi r, because that is how far it is around the rim.

Here is the key step. The curved side wraps around the circular rim. So when you unroll the side, its width is exactly the distance around that rim, which is the circumference $C = 2\pi r$ . Its height is just the height of the cylinder, $h$ . So the unrolled side is a rectangle of width $2\pi r$ and height $h$ , with area

(2\pi r) \times h = 2\pi r h.

The two circular ends each have area $\pi r^2$ , contributing $2 \pi r^2$ together. Adding the two ends and the rolled-out side gives the full surface area:

S = 2\pi r^2 + 2\pi r h.

Why the surface area of a cylinder is $2\pi r^2 + 2\pi r h$ #

A cylinder has three pieces of surface: a top, a bottom, and the curved side that joins them. The surface area is the sum of all three.

The top and the bottom are congruent circles of radius $r$ . The area of a circle is $\pi r^2$ , so the two ends together contribute $2\pi r^2$ .

Now picture peeling the curved side off and unrolling it onto a table. It opens out into a flat rectangle, because the cut runs straight up the side. That vertical cut edge is a straight seam of length $h$ , which is the height of the cylinder. The seam becomes the left and right edges of the rectangle, so the rectangle’s height is $h$ . The top rim and the bottom rim, which were circles, straighten out into the top and bottom edges of the rectangle. Each rim had length equal to the circumference of the circular end, $C = 2\pi r$ , so the rectangle’s width is $2\pi r$ . A rectangle’s area is width times height, so the curved side has area

(2\pi r)(h) = 2\pi r h.

Adding the two ends and the side gives the total surface area:

S = 2\pi r^2 + 2\pi r h.

The first term accounts for the two circular caps, and the second accounts for the wrapped-around side. This split explains why one piece carries $r^2$ and the other carries $r h$ .

Worked example 5 Surface area of a cylinder

A cylinder has radius $2$ cm and height $5$ cm. Find its surface area. Use $\pi \approx 3.14$ .

Find the two parts separately. The two circular ends together:

2\pi r^2 = 2 \times 3.14 \times 2^2 = 2 \times 3.14 \times 4 = 25.12 \text{ cm}^2.

The rolled-out side, a rectangle of width $2\pi r$ and height $h$ :

2\pi r h = 2 \times 3.14 \times 2 \times 5 = 62.8 \text{ cm}^2.

Add the ends and the side:

S = 25.12 + 62.8 = 87.92 \text{ cm}^2.

So it takes about $88$ square centimetres to cover the cylinder. The result is in square centimetres because every piece, the circles and the unrolled rectangle, is a flat area.

Common mistakes

Mixing up volume and surface area units. Volume counts unit cubes, so it is cubic ( $\text{cm}^3$ ); surface area sums flat faces, so it is square ( $\text{cm}^2$ ). If your volume answer ends in $\text{cm}^2$ , or your surface-area answer ends in $\text{cm}^3$ , something is wrong.
Adding the dimensions instead of multiplying. Volume of a box is $l \times w \times h$ , not $l + w + h$ . A $5$ by $3$ by $4$ box has volume $60$ , not $12$ .
Using the diameter for a cylinder. Both $V = \pi r^2 h$ and the surface area use the radius. Given the diameter, halve it first. Forgetting makes the volume four times too big, because the radius is squared.
Counting a face once instead of twice. A box has three pairs of opposite faces. Surface area is $2(lw + lh + wh)$ ; leaving off the factor of $2$ counts only half the box.
Forgetting the ends of a cylinder. The full surface area is the curved side plus both circular caps: $2\pi r h + 2\pi r^2$ . The side alone ( $2\pi r h$ ) covers a tube open at both ends, not a closed can.

Key takeaways

Volume is the number of unit cubes that fill a solid, measured in cubic units ( $\text{cm}^3$ ). Surface area is the total area of all the faces, measured in square units ( $\text{cm}^2$ ).
Rectangular prism: $V = l \times w \times h$ , by filling one $l \times w$ layer and stacking $h$ of them. A cube has $V = s^3$ .
Any prism (and a cylinder): $V = (\text{base area}) \times h$ . For a triangular prism the base is $\tfrac{1}{2} b\, h_\triangle$ ; for a cylinder the base is $\pi r^2$ , so $V = \pi r^2 h$ (use the radius).
Surface area from a net: unfold the solid flat and add the face areas. A box gives $S = 2(lw + lh + wh)$ ; a cube gives $S = 6 s^2$ .
Cylinder surface area: two circular ends plus the unrolled side, whose width is the circumference, so $S = 2\pi r^2 + 2\pi r h$ .
Cones, pyramids, and spheres also enclose volume and have their own formulas. Each of those solids needs a tool beyond this lesson, so they are left for later.

Volume and Surface Area

Learning goals

Volume: counting unit cubes

Volume of a rectangular prism

Volume of any prism: base area times height

Volume of a cylinder

Surface area: unfolding the net

Surface area of a cylinder

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)