Geometry · G.MG.A.1 · G.MG.A.2

Modeling with solids: density and design

A log, a pipe, a roof and a can are not textbook shapes. How do you use volume and area to say how heavy, how much material, or which design is best?

You need

Composite solids and densityMath 8 Surface area of pyramids and conesGeometry

Where you are 8 in Solids

Geometry11

Solids8

▴ 3 above Cross sections of solids Nets of solids Coordinates and distance in three dimensions Surface area3

Modeling with solids: density and design

When you reach for it

A tree trunk is not a cylinder, a roof is not quite a pyramid, and a city is not a circle. But each is close enough to one that the formulas you know for cylinders, spheres, composite solids and the surface of pyramids and cones give answers that are good enough to act on. Reach for this move whenever a real object raises a question about size:

How heavy is it? That needs a volume and the density of what it is made of.
How much material wraps it? Paint, roofing, sheet metal, canvas and labels are surface area.
How crowded is it? People per square kilometer, trees per hectare: a count divided by an area.
Which shape is best? A fixed amount must be held, and one design uses less material than another.

The mathematics inside each answer is a formula you already have. What is new is the step before the formula: deciding which solid stands in for the object, which of its measures the question is really about, and what the units say.

The procedure

Choosing the solid

A model is a solid that is simple enough to have a formula and close enough to the object that the formula’s answer is useful. A log is a cylinder; a silo is a cylinder with a half sphere on top; a pipe is a cylinder with a cylinder taken out; a tent with one center pole is a pyramid. Every model throws something away: the bark’s roughness, the slight taper of the trunk, the rounded rim of a can. That is allowed exactly when what is thrown away is small next to what is kept, and it is worth saying in one line what the model ignores, because that line is what tells you later whether the answer is likely too big or too small.

Then measure what the formula needs, or turn what you can measure into it. You cannot reach the center of a tree, so you wrap a tape round it and get the circumference C. Since C = 2\pi r, the radius is

r = \frac{C}{2\pi}

which is a formula you already trust, used backwards.

Choosing the measure

The question decides whether you need volume or surface area, and mixing them up gives an answer in the wrong units.

Volume answers “how much does it hold” and “how much stuff is in it”. Mass, the amount of concrete to pour, and the water in a tank all scale with volume.
Surface area answers “how much covers it”. Paint, roofing and the metal of a thin can all scale with area. Often only some faces count: a roof has no shingles on its floor, and a tank standing on the ground is not painted underneath. Decide face by face.

Density

Density is the amount of something packed into each unit of space. For a material it is mass per unit of volume,

\rho = \frac{m}{V}

where \rho (the Greek letter rho) is the density, m the mass and V the volume. The same fraction read the other two ways gives m = \rho V and \displaystyle V = \frac{m}{\rho}. Those three lines are one fact: density is a rate, so it multiplies by volume to give mass, just as speed multiplies by time to give distance.

Some densities, at room temperature (ice just below freezing):

Material	Density in \text{g/cm}^3
water	1.00
ice	0.92
aluminum	2.70
iron	7.87
copper	8.96
lead	11.34
gold	19.3

Wood and concrete vary with the kind, so a problem about them states the density it means.

The same idea works with an area in place of a volume. Population density is people per unit of area, and a forest’s density is trees per unit of area. Whatever measures the region goes in the denominator, and the units say which: \text{people/km}^2 is per area, \text{kg/m}^3 per volume.

Units

Units travel with the numbers, and keeping them in every line is what stops a volume in \text{cm}^3 meeting a density in \text{kg/m}^3. A length conversion is used once for a length, twice for an area and three times for a volume, because a cubic meter is a cube 100 cm on each side:

\begin{aligned} 1\ \text{m}^3 &= 100 \times 100 \times 100\ \text{cm}^3 \\ &= 1{,}000{,}000\ \text{cm}^3 \end{aligned}

So 1\ \text{g/cm}^3 is 1{,}000{,}000 grams, which is 1000 kilograms, in each cubic meter:

1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3

Water is 1000\ \text{kg/m}^3, and iron is 7870\ \text{kg/m}^3. This is the same change of units for a rate as turning meters per second into kilometers per hour, with the length factor cubed.

Designing

A design question fixes one quantity and asks you to make another as small (or as large) as possible. A can must hold 500\ \text{cm}^3; the maker wants the least metal. The fixed quantity ties the dimensions together, so once the radius is chosen the height is forced. Write the quantity to be made small using one variable only, then compare candidates. A table of candidates finds the best of those you try and shows how the answer behaves; calculus later finds the exact best.

Reading the answer

The last step is to put the number back into the world. Round in the direction the question needs: paint and roofing are bought in whole tins and bundles, so round up, or the job is left unfinished; a number of objects that can be made from a bar rounds down, because a part object is no object. And compare with what you know: a log that came out at 8 tonnes or 8 grams would be a sign of a units slip.

Worked examples

Example 1

The mass of a log

A felled tree trunk is 6 m long. A tape wrapped round it reads 1.5 m. The wood has density 750\ \text{kg/m}^3. Estimate the mass of the trunk.

Figure 1The log of Example 1, modeled as a cylinder. The tape measure gives the circumference, not the radius.

Solution

Model the trunk as a cylinder, ignoring the bark’s roughness and the slight narrowing from one end to the other. The tape gives the circumference, so first find the radius:

r = \frac{C}{2\pi} = \frac{1.5}{2\pi}\ \text{m}

The volume of a cylinder is base area times length:

\begin{aligned} V = \pi r^2 h &= \pi \cdot \frac{1.5^2}{4\pi^2} \cdot 6 = \frac{13.5}{4\pi} \\ &= \frac{3.375}{\pi} \approx 1.074\ \text{m}^3 \end{aligned}

One \pi canceled, which is why it pays to keep r as a fraction rather than round it to 0.24 first. Mass is density times volume, and the units multiply as \text{kg/m}^3 \times \text{m}^3 = \text{kg}:

\begin{aligned} m = \rho V &= 750 \cdot \frac{3.375}{\pi} \\ &= \frac{2531.25}{\pi} \approx 806\ \text{kg} \end{aligned}

The trunk weighs roughly 806 kg, about four fifths of a tonne. If the tape was wrapped round the wide end of the trunk, the real trunk narrows away from it, so the model’s answer is a little too big.

Example 2

Comparing two towns

Town A sits on a roughly circular patch of land of radius 3 km and has 41{,}000 people. Town B covers a rectangle 5 km by 6 km and has 60{,}000 people. Which town is more densely populated?

Solution

Population density is people per square kilometer, so each town needs its area.

Town A: the area is \pi r^2 = \pi \cdot 3^2 = 9\pi \approx 28.27\ \text{km}^2, so its density is

\frac{41{,}000}{9\pi} \approx 1450\ \text{people/km}^2

Town B: the area is 5 \cdot 6 = 30\ \text{km}^2, so its density is

\frac{60{,}000}{30} = 2000\ \text{people/km}^2

Town B is more crowded, though it is also the bigger town. Comparing populations alone, or areas alone, would not settle the question; the ratio does. Each density is an average over the whole town and says nothing about where within it the people live.

Example 3

Which metal is it?

A solid metal ball has diameter 4 cm and mass 264 g. Use the table of densities to decide what it is most likely made of.

Solution

Model the ball as a sphere of radius 2 cm. Its volume is

V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \cdot 8 = \frac{32\pi}{3} \approx 33.51\ \text{cm}^3

Its density is mass over volume:

\begin{aligned} \rho = \frac{m}{V} &= \frac{264}{32\pi/3} = \frac{792}{32\pi} \\ &\approx 7.88\ \text{g/cm}^3 \end{aligned}

The closest value in the table is iron, 7.87\ \text{g/cm}^3; aluminum is far lighter and copper noticeably heavier. The small difference is what rounding the mass and diameter would produce. The mass alone could not have told us this, nor the size alone: a lead ball of the same size would have mass about 380 g, and a heavier iron ball would simply be bigger. Density is the property of the material, not of the particular ball.

Example 4

Roofing a pyramid roof

A garden building has a roof in the shape of a square pyramid. The base is 8 m on each side and the apex is 3 m above the center of the base. Roofing is sold in bundles that each cover 3 square meters. How many bundles are needed?

Solution

Only the four sloping faces are roofed; the square base is the ceiling of the room, not part of the roof. Each face is a triangle whose height is the slant height \ell, measured up the face, not the vertical height 3 m. The slant height is the hypotenuse of a right triangle with the vertical height and half the base as legs:

\ell = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\ \text{m}

Each triangular face has area \displaystyle \frac{1}{2} \cdot 8 \cdot 5 = 20\ \text{m}^2, so the roof has area

4 \cdot 20 = 80\ \text{m}^2

The number of bundles is \displaystyle \frac{80}{3} \approx 26.7. Twenty-six bundles cover only 78 square meters, so the roof needs 27 bundles. Here rounding to the nearest whole number and rounding up happen to agree; at 26.2 they would not, and the answer would still be 27.

Example 5

Can the crane lift the pipe?

A concrete drainage pipe is 2 m long, 1.2 m across the outside, and its wall is 0.1 m thick. Concrete of this kind has density 2400\ \text{kg/m}^3. A small crane can lift at most 1500 kg. Can it lift the pipe? If not, what is the longest pipe of this cross section it can lift?

Figure 2The pipe of Example 5. The concrete is the outer cylinder with the inner one taken away.

Solution

The pipe is a composite solid: a cylinder of outer radius 0.6 m with a cylinder of inner radius 0.6 - 0.1 = 0.5 m taken out of it, both 2 m long. Its volume is the difference:

\begin{aligned} V &= \pi(0.6)^2(2) - \pi(0.5)^2(2) \\ &= \pi(0.36 - 0.25)(2) \\ &= 0.22\pi \approx 0.691\ \text{m}^3 \end{aligned}

Its mass is

m = 2400 \cdot 0.22\pi = 528\pi \approx 1659\ \text{kg}

That is more than 1500 kg, so the crane cannot lift it.

For a pipe of length L meters, the mass is 2400 \cdot 0.11\pi \cdot L = 264\pi L kg, since the ring at the end has area 0.11\pi\ \text{m}^2. The crane can lift it when 264\pi L \le 1500, that is when

L \le \frac{1500}{264\pi} \approx 1.809\ \text{m}

Since going over the limit is what must be avoided, round down: pipe sections up to 1.80 m long can be lifted.

Example 6

Designing a can

A closed cylindrical can must hold 500\ \text{cm}^3. Which radius uses the least metal? Treat the metal as thin, so the amount of metal is the surface area.

Figure 3Three cans with the same volume. The middle one, whose height equals its diameter, uses the least metal.

Solution

The volume is fixed, and it ties the height to the radius:

\pi r^2 h = 500 \qquad\text{so}\qquad h = \frac{500}{\pi r^2}

The metal is the two circular ends plus the curved side. Substituting for h leaves one variable:

\begin{aligned} S &= 2\pi r^2 + 2\pi r h \\ &= 2\pi r^2 + 2\pi r \cdot \frac{500}{\pi r^2} \\ &= 2\pi r^2 + \frac{1000}{r} \end{aligned}

The two terms pull in opposite directions. A thin can has small ends but a tall side, so the \displaystyle \frac{1000}{r} term is large; a wide can has a short side but big ends, so 2\pi r^2 is large. Try some radii, in centimeters:

r	\displaystyle h = \frac{500}{\pi r^2}	\displaystyle S = 2\pi r^2 + \frac{1000}{r}
3	17.68	389.88
4	9.95	350.53
5	6.37	357.08
6	4.42	392.86

The least of these is at r = 4, so the best radius lies somewhere between 3 and 5. The area climbs much less from r = 4 to r = 5 than from r = 4 to r = 3, which suggests the best lies a little above 4. Narrow it down:

r	S
4.2	348.93
4.3	348.73
4.4	348.92

The best radius is about 4.3 cm, with height \displaystyle \frac{500}{\pi (4.3)^2} \approx 8.6 cm. The height is twice the radius: the can that uses least metal is exactly as tall as it is wide. Calculus confirms that this is no accident: for any fixed volume, the closed cylinder with least surface area has h = 2r.

Notice also how flat the table is near the best. Every radius from r = 4 to r = 5 uses at most about 2.4\% more metal than the best, since \displaystyle \frac{357.08}{348.73} \approx 1.024, so a maker can choose a shape that fits a hand or a shelf at very little cost in metal.

Example 7

A statue and its model

A sculptor makes a model of a statue, 30 cm tall, in solid bronze. The model has mass 2.4 kg, and gilding its surface takes 0.05 liters of gold paint. The finished statue is to be 3 m tall, the same shape, also solid bronze. Find its mass and the paint it needs.

Solution

The statue is the model scaled by a factor of \displaystyle \frac{300}{30} = 10 in every length. By scaling, every area is multiplied by 10^2 = 100 and every volume by 10^3 = 1000. Both pieces are the same bronze, so they have the same density, and mass is density times volume. The mass therefore scales like the volume:

m = 2.4 \cdot 1000 = 2400\ \text{kg}

The paint covers the surface, so it scales like the area:

0.05 \cdot 100 = 5\ \text{liters}

No formula for the statue’s shape was needed. It has no simple formula at all, which is exactly why the scaling argument is the right tool: it uses only the fact that the two shapes are similar. The jump from 2.4 kg to 2.4 tonnes, while the paint goes up only a hundredfold, is one reason a large statue may be cast hollow instead.

Boundaries

A model is only as good as what it leaves out. A cylinder for a tapering trunk, a pyramid for a roof with overhanging eaves, a circle for a ragged coastline: each answer carries the model’s error with it. State what was ignored, and expect the answer to be good to about the accuracy of the measurements, not to the last digit the calculator shows.

m = \rho V assumes one material spread evenly. A hammer with a wooden handle and a steel head needs each part’s mass worked out separately and added. A hollow object has an average density lower than its material’s, as the hollow statue does; working back from such an object’s mass and outside volume gives that average, not the density of the material it is made of.

Volume and area scale differently. Doubling every length multiplies areas by 4 and volumes by 8, so quantities that follow area (paint, heat lost through a wall, material for a thin shell) and quantities that follow volume (mass, capacity) cannot be scaled by the same factor.

Units have to agree before they multiply. A density in \text{g/cm}^3 needs a volume in \text{cm}^3. Converting a volume uses the cube of the length factor and an area its square.

Surface area is material only for thin sheets. For a thin can the metal is the area times the thickness, near enough. For a thick-walled object such as the pipe, the inside and outside surfaces are of different sizes and the material is a volume, worked out as one.

Density over an area is an average. It says how many per square kilometer over the whole region, not where they are; a town with one crowded center and empty fields has the same density as one spread evenly.

A table of candidates finds the best of those tried. It shows where the best lies and how much it matters, but the exact best needs a finer search or calculus.

Where this goes next

The pattern of the can, a fixed quantity tying two dimensions together and a second quantity to make as small as possible, is the heart of optimization in calculus, where the table is replaced by a derivative set to zero and h = 2r comes out exactly. When the dimensions of a solid change over time, for instance water draining from a cone-shaped tank, the volume formulas of this lesson become the equations of related rates. Objects whose shape has no simple formula, such as a vase or a bowl, are modeled as solids of revolution, and their volumes found by slicing them into thin pieces, as in volumes with known cross sections. And the scaling argument of the statue, why a mouse can fall from a height that would hurt an elephant, is the area and volume half of scale factor, area and volume.

Exercises

Choosing the model

Name a solid, or a combination of solids, that models each object, and one thing each model ignores. (a) a grain silo with a domed roof (b) a sharpened round pencil (c) a tent with a square floor and a single pole in the middle (d) a tennis ball
Say whether each question needs a volume, a surface area or neither, and why. (a) how much water a fish tank holds (b) how much glass the tank’s walls and base need (c) how much a solid gold bar weighs (d) how much fencing goes round a field

Mass and density

A log is a cylinder 4 m long and 0.5 m across. Its wood has density 600\ \text{kg/m}^3. Find its mass.
A gold-colored cube has edges of 2 cm and mass 154.4 g. Find its density. Could it be solid gold?
Find the mass of a rectangular block of aluminum measuring 10 cm by 5 cm by 2 cm.
Find the mass of a solid lead ball of radius 2 cm.
A concrete column is a cylinder of radius 0.25 m and height 3 m. The concrete has density 2400\ \text{kg/m}^3. Find the column’s mass.
Write the density of ice, 0.92\ \text{g/cm}^3, in \text{kg/m}^3. Then find the mass of a block of ice 0.5 m by 0.4 m by 0.3 m.

Density over an area

A county, modeled as a rectangle 40 km by 25 km, has 180{,}000 people. Find its population density.
A roughly circular island of radius 4 km has 9000 people. Find its population density, and say whether it is more or less crowded than the county in Exercise 9.

Surface area for material

A cylindrical water tank of radius 1.5 m and height 4 m stands on the ground. Its curved side and its top are to be painted. One liter of paint covers 10 square meters, and paint is sold in one-liter tins. How many tins are needed?
A tent is a square pyramid with base 3 m on each side and a center pole 2 m tall. Its four sides and its floor are made of canvas. How much canvas does it use?
A paper cup is a cone, open at the top, with radius 3 cm and height 4 cm. How much paper does it use, and how much water does it hold?

Designing under a constraint

A closed box with a square base must hold 1000\ \text{cm}^3. Find the surface area for base edges 8 cm, 10 cm and 12 cm, and say which of the three uses the least material.
A cylindrical can is to hold 1000\ \text{cm}^3, with its height equal to its diameter. Find its radius and height.
A closed storage tank must hold 8\ \text{m}^3. Compare the material needed for a cube with the material needed for a cylinder of radius 1.2 m.
A heater is rated to warm a room of up to 60\ \text{m}^3. A room is 5 m by 4.5 m with a ceiling 2.6 m high. Is the heater enough?

Scale and counting

A model of a statue, made of the same solid material as the statue, is one fifth of its height. The model needs 0.3 liters of paint and has mass 4 kg. Find the paint and the mass for the statue.
Steel balls of radius 0.5 cm are cast from a steel bar 2 cm by 2 cm by 50 cm. If no steel is wasted, how many whole balls can be made?

Reasoning

A company doubles every length of a solid plastic toy. Explain why the new toy needs 8 times the plastic but only 4 times the paint.
In Exercise 11 the answer was rounded up, and in Exercise 19 down. Explain why each direction is the right one.
A block of copper is cut in half. Explain why each half has half the mass but the same density as the whole block.
A forester models a trunk as a cylinder, using the circumference measured near the ground. Real trunks get thinner toward the top. Is the model’s mass likely too big or too small? Explain.
Show that a closed cylinder of volume V and radius r has surface area \displaystyle S = 2\pi r^2 + \frac{2V}{r}, and explain why the term \displaystyle \frac{2V}{r} makes a very thin can use a lot of metal.

Answers

(a) A cylinder with a half sphere on top; it ignores the ladder, door and the thickness of the walls. (b) A cylinder with a cone at one end; it ignores the eraser and that the sharpened wood does not end in a perfectly straight cone. (c) A square pyramid; it ignores the sag of the canvas between pole and corners. (d) A sphere; it ignores the fuzz and the seam. Other sensible models are fine if their shapes match the object.
(a) Volume: it is how much space the inside has. (b) Surface area: if the tank has an open top, that is the four walls and the base only. (c) Volume, then mass is density times volume. (d) Neither: a fence is a length, the perimeter of the field.
About 471 kg. The radius is 0.25 m, so V = \pi(0.25)^2(4) = 0.25\pi \approx 0.785\ \text{m}^3, and m = 600 \cdot 0.25\pi = 150\pi \approx 471.2 kg.
19.3\ \text{g/cm}^3. V = 2^3 = 8\ \text{cm}^3 and \displaystyle \rho = \frac{154.4}{8} = 19.3. That is the density of gold in the table, so it could be solid gold; a gold-plated cube of a lighter metal could not have this mass.
270 g. V = 10 \cdot 5 \cdot 2 = 100\ \text{cm}^3, and m = 2.70 \cdot 100 = 270.
About 380.0 g. \displaystyle V = \frac{4}{3}\pi(2)^3 = \frac{32}{3}\pi \approx 33.51\ \text{cm}^3, and \displaystyle m = 11.34 \cdot \frac{32}{3}\pi = 120.96\pi \approx 380.0 g.
About 1414 kg. V = \pi(0.25)^2(3) = 0.1875\pi\ \text{m}^3, and m = 2400 \cdot 0.1875\pi = 450\pi \approx 1413.7 kg.
920\ \text{kg/m}^3, and the block has mass 55.2 kg. Each \text{g/cm}^3 is 1000\ \text{kg/m}^3. The block is 0.5 \cdot 0.4 \cdot 0.3 = 0.06\ \text{m}^3, and 920 \cdot 0.06 = 55.2.
180\ \text{people/km}^2. The area is 40 \cdot 25 = 1000\ \text{km}^2, and \displaystyle \frac{180{,}000}{1000} = 180.
About 179\ \text{people/km}^2, very slightly less crowded than the county. The area is 16\pi \approx 50.27\ \text{km}^2, and \displaystyle \frac{9000}{16\pi} \approx 179.05.
5 tins. Side 2\pi(1.5)(4) = 12\pi, top \pi(1.5)^2 = 2.25\pi, total 14.25\pi \approx 44.77\ \text{m}^2. That needs 4.48 liters, and four tins are not enough.
24\ \text{m}^2. The slant height is \displaystyle \sqrt{2^2 + 1.5^2} = 2.5 m, the four sides are \displaystyle 4 \cdot \frac{1}{2} \cdot 3 \cdot 2.5 = 15\ \text{m}^2, and the floor is 9\ \text{m}^2.
Paper 15\pi \approx 47.1\ \text{cm}^2; it holds 12\pi \approx 37.7\ \text{cm}^3. The slant height is \displaystyle \sqrt{3^2 + 4^2} = 5, the curved surface is \pi r \ell = \pi \cdot 3 \cdot 5, and the volume is \displaystyle \frac{1}{3}\pi \cdot 3^2 \cdot 4. There is no lid, so no circle is added.
Edge 8: 628\ \text{cm}^2; edge 10: 600\ \text{cm}^2; edge 12: about 621.3\ \text{cm}^2. The least is edge 10, which is a cube. The height is \displaystyle \frac{1000}{s^2}, so \displaystyle S = 2s^2 + 4s \cdot \frac{1000}{s^2} = 2s^2 + \frac{4000}{s}.
r \approx 5.42 cm and h \approx 10.84 cm. With h = 2r, \pi r^2 (2r) = 1000, so \displaystyle r^3 = \frac{500}{\pi} and \displaystyle r = \sqrt[3]{500/\pi}.
The cube needs 24\ \text{m}^2; the cylinder about 22.38\ \text{m}^2, so the cylinder uses less. The cube has edge 2 m and six faces of 4\ \text{m}^2. The cylinder has \displaystyle h = \frac{8}{1.44\pi} \approx 1.77 m, and \displaystyle S = 2\pi(1.2)^2 + 2\pi(1.2)h = 2.88\pi + \frac{40}{3}.
Yes. The room is 5 \cdot 4.5 \cdot 2.6 = 58.5\ \text{m}^3, which is under 60\ \text{m}^3.
7.5 liters of paint and 500 kg. Every length is 5 times as long, so paint scales by 5^2 = 25 and mass by 5^3 = 125: 0.3 \cdot 25 = 7.5 and 4 \cdot 125 = 500.
381 balls. The bar is 200\ \text{cm}^3 and each ball is \displaystyle \frac{4}{3}\pi(0.5)^3 = \frac{\pi}{6} \approx 0.524\ \text{cm}^3, and \displaystyle \frac{200}{\pi/6} = \frac{1200}{\pi} \approx 381.97.
Plastic is a volume and paint covers a surface. Doubling every length multiplies each of the three dimensions of a volume by 2, giving 2^3 = 8, but each area has only two dimensions, giving 2^2 = 4.
With 4 tins, 40 square meters would be painted and almost 5 left bare, so the job needs a fifth tin even though most of it is left over. With balls, 381.97 means the steel runs out before the 382nd ball is finished, and a part of a ball is not a ball.
Cutting halves both the volume and the mass, and density is their ratio: \displaystyle \frac{m/2}{V/2} = \frac{m}{V}. Density belongs to the copper, not to how much of it there is.
Too big. The model uses the thickest part of the trunk for its whole length, so its volume, and so its mass, is more than the real trunk’s. Measuring halfway up gives a better model.
From \pi r^2 h = V, \displaystyle h = \frac{V}{\pi r^2}. The two ends have area 2\pi r^2 and the side \displaystyle 2\pi r h = 2\pi r \cdot \frac{V}{\pi r^2} = \frac{2V}{r}. As r gets small, \displaystyle \frac{2V}{r} grows without limit: a thin can must be very tall to hold V, and its side is huge.