Surface Area and Volume

Study Sheet

Surface Area and Volume

Wrapping versus filling

Two Different Measures

Concept
The Big Idea
432
A 4 × 3 × 2 box: six faces outside, 24 unit cubes inside.

Volume fills the inside (cubic units). Surface area wraps the outside (square units) — add up the six faces, which come in matching pairs.

Surface area is the total area of all the faces — what you would paint; volume is the space inside — what you would fill. A net (the solid unfolded flat) shows every face once, so adding the areas of the net's pieces gives the surface area. Volume of a box is length ×\times width ×\times height.

Example
A 4 by 3 by 2 box
2×34×32×34×34×24×2
The net shows all six faces: two 4×3, two 4×2, two 3×2 (labels are the side lengths).

Volume =4⋅3⋅2=24= 4 \cdot 3 \cdot 2 = 24. Surface area =2(12+6+8)=52= 2(12 + 6 + 8) = 52.

Example
From a net: a 6×2×16 \times 2 \times 1 box
1×26×21×26×26×16×1
Three pairs of matching faces.

The net has two 6×26 \times 2 faces (1212 each), two 6×16 \times 1 faces (66 each) and two 2×12 \times 1 faces (22 each). Surface area: 2(12+6+2)=402(12 + 6 + 2) = 40 square units. Volume: 6×2×1=126 \times 2 \times 1 = 12 cubic units.

Side note
Where it goes wrong

Counting only the visible faces. A box has six faces, in three matching pairs; a net shows all six, which is why drawing one prevents the error.

Try it
Try it: a flat box

A box is 55 by 44 by 11. Find its volume and its surface area.

Show the answer

Volume: 5×4×1=205 \times 4 \times 1 = \mathbf{20} cubic units. Faces: 5×4=205 \times 4 = 20, 5×1=55 \times 1 = 5 and 4×1=44 \times 1 = 4, each twice: 2(20+5+4)=582(20 + 5 + 4) = \mathbf{58} square units.

In Plain Terms & More Examples

Concept
In plain terms

Volume fills the inside (cubic units); surface area covers the outside (square units) — add the areas of all six faces.

Example
A cube of side 3
3×33×33×33×33×33×3
Six identical 3 × 3 faces.

Volume =33=27= 3^3 = 27. Surface area =6×32=54= 6 \times 3^2 = 54.

Example
A 5 by 2 by 2 box

Volume =5⋅2⋅2=20= 5 \cdot 2 \cdot 2 = 20. Surface area =2(10+4+10)=48= 2(10 + 4 + 10) = 48.

Side note
Where it goes wrong

Using 6×6 \times one face for a box that is not a cube. For the 5×2×25 \times 2 \times 2 box, 6×10=606 \times 10 = 60 is too big: four faces are 5×2=105 \times 2 = 10, but the two ends are only 2×2=42 \times 2 = 4. Surface area =4×10+2×4=48= 4 \times 10 + 2 \times 4 = 48.

Try it
Try it: a cube of edge 4

A cube has edge 44. Find its volume and surface area.

Show the answer

Volume: 4×4×4=644 \times 4 \times 4 = \mathbf{64}. Each face is 4×4=164 \times 4 = 16, and there are six: 6×16=966 \times 16 = \mathbf{96}.

Triangular Prisms and Fractional Edges

Concept
Base area times height

Any prism's volume is (area of the base shape) ×\times (length). For a triangular prism the base is a triangle. Fractional edges use the same formulas.

Example
A wedge
43length 10
The triangle is the base; the prism is 10 of those triangles deep.

Triangle base 44, height 33, prism length 1010: base area 66, volume 6060.

Example
Surface area of the wedge

Take the wedge's triangle to be a right triangle with legs 44 and 33; its long side is then 55. Two triangle ends: 2×6=122 \times 6 = 12. Three rectangles, each 1010 long: 4×10=404 \times 10 = 40, 3×10=303 \times 10 = 30 and 5×10=505 \times 10 = 50. Total: 12+40+30+50=13212 + 40 + 30 + 50 = 132 square units.

Side note
Where it goes wrong

Treating the three rectangles as equal. In a triangular prism each rectangle is as wide as one side of the triangle, 44, 33 and 55 here, so their areas differ. Only the two triangle ends match each other.

Try it
Try it: fractional box

A box is 2122\tfrac12 by 44 by 1121\tfrac12 units. Find its volume and surface area.

Show the answer

Volume 52⋅4⋅32=15\tfrac52\cdot4\cdot\tfrac32 = \mathbf{15}. Faces: 2(10+3.75+6)=39.52(10 + 3.75 + 6) = \mathbf{39.5}.

Word Problems

Concept
Wrap or fill?

Wrapping, painting and covering ask for surface area (square units). Filling, holding and packing ask for volume (cubic units). Decide which the story wants before calculating.

The same box has both measures, and they answer different questions: the gift box below needs 13001300 cm2^2 of paper but holds 30003000 cm3^3. The units tell you which question an answer belongs to.

Example
Wrapping a gift
20 cm15 cm10 cm
Paper covers all six faces.

A box 20×15×1020\times15\times10 cm needs paper for all six faces: 2(300+200+150)=13002(300 + 200 + 150) = 1300 cm2^2.

Example
Filling the gift box

The same 20×15×1020 \times 15 \times 10 cm box holds 20×15×10=300020 \times 15 \times 10 = 3000 cm3^3, which is 33 litres.

Side note
Where it goes wrong

Answering a wrapping question with volume. "How much paper?" has an answer in cm2^2. The box's 30003000 cm3^3 is how much fits inside, and it says nothing about the paper.

Try it
Try it: painting a cube

A cube has edge 66 cm. What is its surface area, and how many 11-cm cubes fit inside?

Show the answer

Surface: 6×36=2166\times36 = \mathbf{216} cm2^2. Volume: 63=2166^3 = \mathbf{216} cubes — the same number, a coincidence special to edge 66.

Volume with Fractional Edges

Concept
Half-unit cubes

A cube with edges 12\tfrac12 unit has volume 12×12×12=18\tfrac12 \times \tfrac12 \times \tfrac12 = \tfrac18: eight of them fill one unit cube. Packing a box with these small cubes and counting gives the same volume as multiplying the fractional edges.

This is why V=ℓ×w×hV = \ell \times w \times h still works when the edges are fractions. Count the small cubes along each edge and multiply to count them all; each is worth 18\tfrac18. Each edge length is its count divided by 22, so the product of the edges is the total count divided by 2×2×2=82 \times 2 \times 2 = 8.

Example
A 112×1×121\tfrac12 \times 1 \times \tfrac12 box
1½1½
Each small cube is half a unit on every edge.

Half-unit cubes along the edges: 33, 22 and 11. That is 3×2×1=63 \times 2 \times 1 = 6 small cubes, each 18\tfrac18: volume 68=34\tfrac68 = \tfrac34. The formula agrees: 32×1×12=34\tfrac32 \times 1 \times \tfrac12 = \tfrac34.

Example
A 212×2×1122\tfrac12 \times 2 \times 1\tfrac12 box

52×2×32=304=712\tfrac52 \times 2 \times \tfrac32 = \tfrac{30}{4} = 7\tfrac12 cubic units. Counting half-unit cubes instead: 5×4×3=605 \times 4 \times 3 = 60, and 60÷8=71260 \div 8 = 7\tfrac12.

Side note
Where it goes wrong

Multiplying only the whole parts. For 212×2×1122\tfrac12 \times 2 \times 1\tfrac12, working out 2×2×1=42 \times 2 \times 1 = 4 ignores all the halves, and the true volume, 7127\tfrac12, is almost twice as big. Change to improper fractions first.

Try it
Try it: count the half-cubes

How many 12\tfrac12-inch cubes fill a box 22 in by 1121\tfrac12 in by 11 in? What is its volume?

Show the answer

Along the edges: 44, 33 and 22 half-inch cubes, so 4×3×2=244 \times 3 \times 2 = \mathbf{24} cubes. Each is 18\tfrac18 in3^3: 24÷8=324 \div 8 = \mathbf{3} in3^3. Check: 2×112×1=32 \times 1\tfrac12 \times 1 = 3.

Why It Works

A rectangular box ℓ×w×h\ell\times w\times h has two faces of each size: ℓw\ell w, ℓh\ell h, and whwh. So S=2ℓw+2ℓh+2wh=2(ℓw+ℓh+wh)S = 2\ell w + 2\ell h + 2wh = 2(\ell w + \ell h + wh). Its volume counts unit cubes: ℓw\ell w cubes per layer, hh layers, so V=ℓwhV = \ell wh. Surface area has square units and volume cubic units because two lengths were multiplied in one case and three in the other — the units are a built-in check.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.