Volume and Surface Area: Free Response

The floor of the crate is a rectangle $9$ cm by $7$ cm. Each cube has a $1$ cm square face resting on the floor, so one cube sits on each square centimetre of it, and a single layer holds one cube for each of those squares:

$9 \times 7 = 63 \text{ cubes}.$

Every cube is $1$ cm tall, so each layer raises the pile by $1$ cm. The crate is $4$ cm from floor to rim, so exactly $4$ layers fit, and each is an identical copy of the first:

$63 \times 4 = 252 \text{ cubes}.$

No cube belongs to two layers and none is left over, so the count is exact. The crate is filled by $252$ cubes of side $1$ cm, and that is what its volume is:

$V = 9 \times 7 \times 4 = 252 \text{ cm}^3.$

The unit is cubic centimetres because what has been counted is cubes. Adding the three measurements instead would give $20$, which is a length and measures nothing about the space inside.

Nothing about the crate has changed except which face is underneath. The face now in contact with the table is $7$ cm by $4$ cm, so one layer of cubes lying on it holds

$7 \times 4 = 28 \text{ cubes}.$

The measurement that is left, $9$ cm, is now the direction the layers stack in, so there are $9$ layers:

$28 \times 9 = 252 \text{ cubes}.$

The total is the same $252$ cubes as before, and so the volume is the same $252 \text{ cm}^3$. The two fillings disagree about what a layer is and about how many layers there are, $63$ in $4$ layers against $28$ in $9$, and agree exactly about the crate.

Look at what the two fillings actually computed. The first multiplied $9$ by $7$ to get a layer and then by $4$ to count the layers. The second multiplied $7$ by $4$ and then by $9$. The same three numbers appear in both, in a different order, and the order of the factors in a product does not change the product:

$9 \times 7 \times 4 = 7 \times 4 \times 9 = 252.$

That is the arithmetic reason. The reason underneath it is that both fillings are describing the same crate. Each cube position in the crate can be pinned down by saying which row it is in, which column, and how high up it is, and turning the crate onto another face does not change how many such positions there are; it only changes which of those three labels is read last. So the two countings are two routes through one collection, and a collection has one size.

Now the unit. The crate is $9$ cm in one direction, $7$ cm in a second, and $4$ cm in a third, and every one of those factors is a length in centimetres. Multiplying two of them measures a flat region, which is why an area is in square centimetres. Multiplying all three measures a solid region, so the answer carries three factors of centimetres and is written $\text{cm}^3$. The raised $3$ is a count of directions, not decoration: it says the measurement spans three of them.

The wall is not a shape with an area formula of its own, but it comes apart into two that are. Cut straight across it at the level of the eaves, where the upright sides stop and the roof begins.

Below the cut is a rectangle $10$ m wide and $6$ m tall:

$10 \times 6 = 60 \text{ m}^2.$

Above the cut is a triangle. Its base is the cut itself, $10$ m long, and its perpendicular height is the $3$ m rise from the eaves to the ridge, so

$\frac{1}{2} \times 10 \times 3 = 15 \text{ m}^2.$

The two pieces overlap nowhere and between them cover the whole wall, so their areas add:

$60 + 15 = 75 \text{ m}^2.$

Note that the sloping edges were never measured. The triangle's area needs its base and the perpendicular height only, and both of those were given.

A slice $1$ m thick standing on the end wall is the same idea as one layer of unit cubes standing on the floor of a box. Its face has an area of $75$ square metres, and it is $1$ m thick, so it encloses

$75 \times 1 = 75 \text{ m}^3.$

Every slice parallel to the end wall is an identical copy of that wall, so every slice encloses the same $75 \text{ m}^3$. The barn is $14$ m from one end wall to the other, so $14$ slices reach across it with nothing left over:

$75 \times 14 = 1050 \text{ m}^3.$

That is the rule that volume is base area times height, with the barn lying on its side: the base is the end wall and the height is the $14$ m run of the building. The unit is cubic metres because a square-metre area has been carried through a length in metres.

The objection is worth taking seriously, because it is true that whole cubic metres cannot be packed under a sloping roof. The answer is to cut the barn into pieces that the box rule already covers.

Cut the whole building horizontally at eaves level. The lower piece is an ordinary rectangular prism, $10$ m by $6$ m by $14$ m, and the box rule applies to it directly:

$10 \times 6 \times 14 = 840 \text{ m}^3.$

The upper piece is the roof. Cut it once more, vertically down the line of the ridge, and each half is a solid whose end is a right-angled triangle with legs $5$ m across and $3$ m up, running $14$ m along the barn. Two copies of such a solid fit together to make a box: turn one of them upside down and slide it against the other along the sloping face, and the two right-angled triangular ends make a $5$ m by $3$ m rectangle. That box is

$5 \times 3 \times 14 = 210 \text{ m}^3,$

so each half of the roof is half of that, $105 \text{ m}^3$, and the two halves together are $210 \text{ m}^3$. Nothing here needed a cube to be cut.

Adding the two pieces of the building gives

$840 + 210 = 1050 \text{ m}^3,$

which is exactly what base area times length gave, since $75 \times 14 = 1050$. The rule survives the objection.

What the argument actually leaned on is worth naming. It needed the barn to have identical cross-sections parallel to its ends all the way through, so that one slice stands for every slice, and it needed the area of that cross-section to be known. It never needed the cross-section to be a rectangle, and it never needed whole unit cubes to fit inside it. Where cubes have to be cut, the pieces along the slope match up exactly as the pieces of area do in the base, which is why the same number that measures the base area measures the volume of a slice one unit thick. That is also why the identical argument works for a cylinder, whose base is a circle that no arrangement of unit squares tiles either.

The slit runs straight from base to rim, so it is a straight edge $25$ cm long, and it becomes both the left and the right edge of the flat piece. The rim and the base circle become the other two edges, and once flattened they lie straight. So the piece is a rectangle, $25$ cm from bottom edge to top edge.

Its width is the measurement that matters here. The metal wrapped once around the rim, so laid flat it stretches exactly as far as the distance around that rim, which is the circumference:

$C = 2\pi r = 2 \times 3.14 \times 17 = 106.76 \text{ cm}.$

A rectangle's area is width times height:

$106.76 \times 25 = 2669 \text{ cm}^2.$

So the curved side alone takes $2669$ square centimetres of sheet.

The base and the lid are circles of radius $17$ cm. One of them has area

$\pi r^2 = 3.14 \times 17^2 = 3.14 \times 289 = 907.46 \text{ cm}^2,$

so the pair contribute $2 \times 907.46 = 1814.92 \text{ cm}^2$. Every piece cut from the sheet is now accounted for, so the total is the unrolled side plus the two ends:

$S = 2669 + 1814.92 = 4483.92 \text{ cm}^2.$

What the tin holds is a different question with a different answer. Filling it means stacking the base circle through the height, so its capacity is the base area times the height:

$V = 907.46 \times 25 = 22686.5 \text{ cm}^3.$

The units differ because the two quantities count different things. The sheet metal is a collection of flat pieces, and a flat piece is measured by multiplying two lengths, so it is in square centimetres. The contents are a solid region, counted in cubes of side $1$ cm, which takes three lengths, so it is in cubic centimetres. A workshop that quoted $4483.92$ for the contents, or $22686.5$ for the metal, would have confused a wrapping with a filling.

Start with what the apprentice measured. The distance from one point of the rim to the point opposite, straight through the middle, is the diameter, and here it is $34$ cm because the radius is $17$ cm. That measurement crosses the tin. The metal does not cross the tin; it travels round the outside of it.

So the width of the piece has to be the distance a point travels going once round the rim, which is the circumference. The two are not close:

$106.76 = 3.14 \times 34.$

Going round is about $3.14$ times as far as going straight across, which is exactly what $\pi$ means. The apprentice's piece would wrap less than a third of the way around before running out, leaving most of the tin open.

Measure the loss. The piece cut is

$34 \times 25 = 850 \text{ cm}^2,$

while the side needs $2669 \text{ cm}^2$, so it is short by

$2669 - 850 = 1819 \text{ cm}^2.$

The mistake is not a slip in arithmetic. It is a piece of geometry: unrolling a curved surface preserves the length of the rim, and the rim was never $34$ cm long.

The volume multiplies the three measurements together:

$V = 10 \times 6 \times 5 = 300 \text{ cm}^3.$

The surface area adds the six faces, which come in three matching pairs. One face of each kind measures

$10 \times 6 = 60, \qquad 10 \times 5 = 50, \qquad 6 \times 5 = 30,$

all in square centimetres, and each kind occurs twice:

$S = 2(60 + 50 + 30) = 280 \text{ cm}^2.$

The first result is cubic and the second is square, because one counts space and the other covers it.

Doubling every edge turns $10$, $6$ and $5$ into $20$, $12$ and $10$. Its volume is

$V = 20 \times 12 \times 10 = 2400 \text{ cm}^3,$

and its faces are $20 \times 12 = 240$, $20 \times 10 = 200$ and $12 \times 10 = 120$ square centimetres, so

$S = 2(240 + 200 + 120) = 1120 \text{ cm}^2.$

Compare each with the original by dividing:

$2400 \div 300 = 8, \qquad 1120 \div 280 = 4.$

So the same doubling multiplies one measure by $8$ and the other by $4$. Both boxes are the same shape, and one instruction was applied to both measures, yet the two multiples are not equal.

Everything follows from counting how many lengths are multiplied together in each measure.

A volume is three measurements multiplied together. Doubling every edge doubles each of the three factors, so the product is multiplied by

$2 \times 2 \times 2 = 8.$

A face is flat, so its area is only two measurements multiplied together. Doubling both of them multiplies that face by $2 \times 2 = 4$.

Check that face by face rather than trusting it once. The three kinds of face were $60$, $50$ and $30$ square centimetres, and they became $240$, $200$ and $120$:

$240 = 4 \times 60, \qquad 200 = 4 \times 50, \qquad 120 = 4 \times 30.$

Every face grew by the same factor of $4$, and the box still has six of them, so the total had no choice but to grow by $4$ as well. That is why the surface area multiple can be argued without adding anything up.

Now replace the doubling by a factor $k$. Nothing in the argument used the number $2$; it used only that every length was multiplied by the same thing. So the volume is multiplied by $k \times k \times k$ and the surface area by $k \times k$:

$V \to k^3 V, \qquad S \to k^2 S.$

The advice for the workshop is that scaling up pays for itself: the big box takes four times the sheet metal but holds eight times as much, so every square centimetre of metal is working twice as hard as it did on the small box.

Unfold the planter and see what is actually there. It has a base and four walls, and no lid, so five faces in all.

The base is the full footprint, $12$ cm by $8$ cm:

$12 \times 8 = 96 \text{ cm}^2.$

The two long walls each run the full length and rise the full depth, $12$ cm by $5$ cm, so $60 \text{ cm}^2$ each. The two end walls are $8$ cm by $5$ cm, so $40 \text{ cm}^2$ each.

Add the five faces:

$96 + 2(60) + 2(40) = 96 + 120 + 80 = 296 \text{ cm}^2.$

So the planter takes $296$ square centimetres of metal. Every piece counted was flat, which is why the answer is in square centimetres however deep the planter is.

Call the unknown depth $h$ centimetres and go round the tray face by face, exactly as with the planter.

The base is $15$ cm by $6$ cm, an area of $90 \text{ cm}^2$, and it does not involve the depth at all. Each long wall is $15$ cm by $h$ cm, so the pair contribute $30h$. Each end wall is $6$ cm by $h$ cm, so that pair contribute $12h$. Collecting the terms,

$90 + 30h + 12h = 90 + 42h.$

That expression is the metal in a tray of depth $h$. The sheet used was $258 \text{ cm}^2$, so

$90 + 42h = 258.$

Subtract the base area from both sides, leaving the walls alone: $42h = 168$. Then divide by $42$:

$h = \frac{168}{42} = 4.$

The tray is $4$ cm deep. Check it by rebuilding the total: the base is $90$, the long walls are $2 \times 60 = 120$, the ends are $2 \times 24 = 48$, and $90 + 120 + 48 = 258$ as required.

The formula is not wrong, it is answering a different question. Applied to these measurements it gives

$S = 2(96 + 60 + 40) = 392 \text{ cm}^2,$

which is the metal in a sealed box $12$ cm by $8$ cm by $5$ cm. It counts six faces because a closed box has six. The planter has five: everything except the top, which is a $12$ cm by $8$ cm rectangle of area $96 \text{ cm}^2$. Taking that one face away gives

$392 - 96 = 296 \text{ cm}^2,$

the same figure as counting the five faces directly. The lesson is that a surface area is the faces a solid actually has, and a formula is only a shortcut for one particular list of faces.

The soil is a separate matter. The planter holds

$12 \times 8 \times 5 = 480 \text{ cm}^3$

filled level with the rim, and leaving the top open does not change that by a single cubic centimetre. A lid is a piece of material laid across the opening, not a piece of the space underneath it, so removing it changes what the container is made of and not what it can contain. That is also the practical reason the two measures are quoted separately: a change to the design can move one of them and leave the other exactly where it was.