Math 7 · 5.MD.C.5B · 6.G.A.2

Volume of prisms and cylinders

How much does a box, a trough or a can hold, and why is it the area of the bottom times the height?

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Area of a circleMath 7 Volume with unit cubesGrade 5 Three-dimensional shapesGrades K to 6

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Cavalieri's principleGeometry Volume of a cylinderMath 8 Volume of pyramids and conesMath 8 Scale factor and perimeter, area and volumeGeometry Related ratesCalculus 1 The disk methodCalculus 1 +3 more

Where you are 7 in Volume of solids

Volume of prisms and cylinders

Cavalieri's principle Volume of a cylinder Volume of pyramids and cones ▾ Show all 7 (2 more)

When you reach for it

Sand for a sandbox, rubble in a skip, juice in a carton, batter in a cake tin: each asks how much space is inside a solid, which is how much it takes to fill it. That amount is the volume. It is measured by counting cubes. A cubic centimeter, 1\ \text{cm}^3, is the space inside a cube 1 cm along each edge, and a cubic meter, 1\ \text{m}^3, the space inside a cube 1 m along each edge.

Volume is a different question from surface area. Surface area is the wrapping, the material that covers the outside once; volume is the filling.

This lesson handles the solids in which every slice parallel to the bottom is the same shape and size as the bottom: prisms, whose base is a polygon, and cylinders, whose base is a circle. One rule works for all of them, the area of the base times the height. Part 1 builds it for prisms and Part 2 carries it to cylinders.

The procedure

Part 1: prisms

What a prism is. A prism has two faces that are the same polygon, lying in parallel planes, one directly over the other. These are its two bases, and the faces joining them are rectangles. Think of the base slid straight up through space, leaving a solid behind it. Because the base slides without turning or shrinking, every slice parallel to the base is an exact copy of the base. Three-dimensional shapes names the faces, edges and vertices of these solids.

Layers of cubes. Take a box whose base is 5 by 3 and whose height is 4, in some unit. Set unit cubes on the base: a row of 5, and 3 rows, so the bottom layer holds 5 \times 3 = 15 cubes, one on each unit square of the base. This is the count from Volume with unit cubes. Every layer above is the same, because every slice is a copy of the base, and there are 4 layers. The box holds 15 \times 4 = 60 cubes.

Figure 1The bottom layer holds 5 \times 3 = 15 unit cubes, one on each square of the base. Four such layers fill the box, 15 \times 4 = 60 cubes.

Nothing in this count needed the base to be a rectangle. If the base has area B square units, a layer 1 unit thick on top of it has volume B cubic units: each unit square of the base carries one unit cube, and each piece of a square carries the matching piece of a cube. A stack h units high is h such layers, so

V = Bh

The height need not be a whole number. A layer half a unit thick holds half as much as a full layer, so a height of 4.5 holds 4.5B, and the same goes for any other thickness.

For a box. The base is a rectangle l by w, so B = lw and V = lwh, the rule of Volume of a box. Any face of a box can serve as its base, and the product of the three edges comes out the same whichever is taken first.

Which face is the base. For any other prism, the base is the face that is copied all the way through, not whichever face happens to rest on the table. A triangular prism lying on one of its rectangles still has the two triangles as its bases, and its height is then the distance between them, measured along the table. So find the two matching parallel faces, then the area B of one of them (with Area of triangles, Area of parallelograms and trapezoids, or Area of composite figures and shaded regions for a base made of pieces), then the height h, measured at right angles to the bases.

Units. All lengths must be in the same unit before they are multiplied. The area then comes out in square units, and multiplying by one more length gives cubic units: \text{cm}^2 \times \text{cm} = \text{cm}^3.

A missing measurement. V = Bh is an equation, so when two of its three numbers are known the third follows. Dividing both sides by the same number, not zero, keeps them equal, so

h = \frac{V}{B} \qquad\qquad B = \frac{V}{h}

The height is the number of layers of area B it takes to make the volume. Once B is known, a missing side of the base comes from the base’s own area formula.

Part 2: cylinders

What a cylinder is. A cylinder is a circle slid straight up: two equal circles in parallel planes, one over the other, joined by a curved side. It is not a prism, because a circle is not a polygon, but it has the one property the layer count used. Every slice parallel to the base is the same circle.

The same rule. So the layer count goes through unchanged. A layer 1 unit thick holds as many cubic units as the base holds square units, and a circle of radius r has area \pi r^2 (see Area of a circle). With h layers,

V = Bh = \pi r^2 h

The formula uses the radius. When the diameter d is given, halve it first: \displaystyle r = \frac{d}{2}.

Exact or rounded. An answer such as 60\pi\ \text{cm}^3 is exact. A decimal answer is rounded, and it should say so and say how: 60\pi \approx 188.5\ \text{cm}^3, to one decimal place. Keep \pi as \pi through the working and round once, at the end. Rounding \pi to 3.14 early can change the digits you report, as Example 5 shows.

Capacity. How much a container holds is its capacity, often given in liters. The links are

\begin{aligned} 1\ \text{cm}^3 &= 1\ \text{mL} \\ 1000\ \text{cm}^3 &= 1\ \text{L} \\ 1\ \text{m}^3 &= 1000\ \text{L} \end{aligned}

and Converting square and cubic units shows where they come from. Capacity uses a container’s inside measurements, since its walls take up room.

Comparing containers. To say which of two containers holds more, find both volumes in the same unit and compare the numbers. A box and a can are compared by what they hold, not by how they look.

Worked examples

Examples 1 to 4 go with Part 1. Examples 5 to 7 go with Part 2.

Example 1

A cereal box

Cosimo’s cereal box is 19 cm wide, 7 cm deep and 28 cm tall. How many centimeter cubes would it take to fill it, and what is its volume?

Solution

Stand the box on its bottom, a rectangle 19 by 7. Its area is

B = 19 \times 7 = 133\ \text{cm}^2

so a layer of centimeter cubes covering the bottom holds 133 cubes, one on each square centimeter. The box is 28 cm tall, so 28 such layers fill it:

V = Bh = 133 \times 28 = 3724\ \text{cm}^3

It takes 3724 centimeter cubes, and the volume is 3724\ \text{cm}^3.

Check with a different base. Lay the box on its back, so that the front face, 19 by 28, is the base and the depth 7 is the height. Then B = 532 and V = 532 \times 7 = 3724, the same.

Example 2

A water trough with triangular ends

A water trough for animals is a triangular prism lying on its side, pointed edge down. Each end is a triangle 0.6 m across the top and 0.4 m deep, and the trough is 2.5 m long. How much water does it hold when full?

Figure 2The trough in Example 2. Its bases are the two triangular ends, so its height in V = Bh is the length 2.5 m along the ground.

Solution

The two matching parallel faces are the triangular ends, so they are the bases, even though the trough lies lengthwise. The height of the prism is the distance between them, 2.5 m, which runs along the ground.

Each end is a triangle with base 0.6 m and height 0.4 m, the depth being measured at right angles to the top edge:

B = \tfrac{1}{2} \times 0.6 \times 0.4 = 0.12\ \text{m}^2

Then

V = Bh = 0.12 \times 2.5 = 0.3\ \text{m}^3

Check. The trough fits exactly inside a box 0.6 by 0.4 by 2.5, whose volume is 0.6\ \text{m}^3. Each triangle is half of the rectangle around it, so the trough holds half the box, 0.3\ \text{m}^3.

Example 3

A skip of rubble

Aldous fills a skip level to the brim with rubble. Seen from the side, the skip is a trapezoid, 3 m along the top, 2 m along the bottom and 1.2 m deep. It is 1.6 m wide from one side to the other. How much rubble does it hold?

Figure 3The skip in Example 3. The trapezoids are the bases; the width 1.6 m between them is the height of the prism.

Solution

The ends of the skip slope, so it is not a box. But its two long sides are the same trapezoid, in parallel planes, and every slice parallel to them is that trapezoid. So the skip is a prism whose bases are the trapezoids and whose height is the width between them, 1.6 m.

A trapezoid’s area is half the sum of its parallel sides times the distance between them (see Area of parallelograms and trapezoids):

B = \tfrac{1}{2}(3 + 2) \times 1.2 = 3\ \text{m}^2

V = Bh = 3 \times 1.6 = 4.8\ \text{m}^3

The skip holds 4.8\ \text{m}^3 of rubble.

Check. The smallest box around the skip is 3 by 1.6 by 1.2, which is 5.76\ \text{m}^3. The sloping ends leave two wedges of that box empty. The top overhangs the bottom by 0.5 m at each end, so each wedge is a prism whose base is a right triangle with legs 0.5 and 1.2, of area 0.3. The two wedges together are 2 \times 0.3 \times 1.6 = 0.96\ \text{m}^3, and 5.76 - 0.96 = 4.8.

Example 4

How deep is the sand?

Delyth’s sandbox is 2.4 m long and 1.5 m wide inside. She tips in 0.72\ \text{m}^3 of sand and rakes it level. How deep is the sand? How much more sand would make it 30 cm deep?

Solution

The sand forms a box with the floor of the sandbox as its base and its depth as the unknown height h. The base is

B = 2.4 \times 1.5 = 3.6\ \text{m}^2

and V = Bh gives 0.72 = 3.6h. Dividing both sides by 3.6,

h = \frac{0.72}{3.6} = 0.2\ \text{m}

so the sand is 20 cm deep. Check: 3.6 \times 0.2 = 0.72.

The other lengths are in meters, so the 30 cm target is 0.3 m. The sand must rise another 0.1 m, which is one more layer 0.1 m thick on the same base:

3.6 \times 0.1 = 0.36\ \text{m}^3

Check another way. Sand 0.3 m deep is 3.6 \times 0.3 = 1.08\ \text{m}^3, and 1.08 - 0.72 = 0.36.

Example 5

A mug, exact and rounded

Bronwen’s mug is 7 cm across inside and 9 cm deep inside. Find its volume exactly, then to one decimal place, and say about how many milliliters it holds.

Solution

The 7 cm is the diameter, so the radius is half of it, r = 3.5 cm. The base is a circle of area \pi r^2 = 12.25\pi\ \text{cm}^2, and the mug is 9 layers of it:

V = \pi r^2 h = 12.25\pi \times 9 = 110.25\pi\ \text{cm}^3

That is the exact volume. For a decimal, multiply out with \pi as accurately as the calculator holds it, then round once:

110.25\pi = 346.36\ldots \approx 346.4\ \text{cm}^3

to one decimal place. Since 1\ \text{cm}^3 = 1\ \text{mL}, the mug holds about 346 mL, a little over a third of a liter.

Rounding \pi to 3.14 first would give 110.25 \times 3.14 = 346.185, which rounds to 346.2. The early rounding has already moved the first decimal place.

Example 6

Batter from a round tin into a square one

Fionnuala’s round cake tin is 22 cm across, and the batter in it is 5 cm deep. How much batter is there? She pours all of it into a square tin 20 cm by 20 cm instead. How deep is it there?

Solution

The batter is a cylinder of radius 11 cm and height 5 cm, its depth, not the height of the tin:

V = \pi \times 11^2 \times 5 = 605\pi \approx 1900.7\ \text{cm}^3

to one decimal place, which is about 1.90 L.

In the square tin the same batter is a box with base 20 \times 20 = 400\ \text{cm}^2 and unknown height h. Use the exact volume, so that no rounding enters yet:

h = \frac{V}{B} = \frac{605\pi}{400} = 4.751\ldots \approx 4.8\ \text{cm}

The batter is a little shallower in the square tin. That fits: the square’s area, 400, is larger than the circle’s, 121\pi \approx 380.1, so the same batter spreads thinner.

Example 7

A carton or a can

A juice carton is 7 cm by 7 cm across and 20 cm tall. A can of juice has radius 4 cm and height 19 cm. Which holds more, and by about how much? How tall would a can of radius 4 cm have to be to hold exactly as much as the carton?

Solution

Find both volumes in cubic centimeters. The carton is a box:

V_{\text{carton}} = 7 \times 7 \times 20 = 980\ \text{cm}^3

The can is a cylinder:

V_{\text{can}} = \pi \times 4^2 \times 19 = 304\pi \approx 955.0\ \text{cm}^3

The carton holds more, by 980 - 955.04\ldots \approx 25\ \text{cm}^3, about 25 mL. The can’s base is slightly larger, 16\pi \approx 50.3\ \text{cm}^2 against 49\ \text{cm}^2, but the can is a whole centimeter shorter, and the missing layer outweighs the gain.

For the matching can the radius is fixed, so B = 16\pi, and the volume must be 980:

h = \frac{980}{16\pi} = 19.49\ldots \approx 19.5\ \text{cm}

Check: 16\pi \times 19.5 \approx 980.2, just over 980 because 19.5 was rounded up.

Boundaries

Every slice must match the base. V = Bh counts identical layers, so it holds only when each slice parallel to the base is a copy of it. A pyramid or a cone narrows toward its tip, its layers shrink, and it holds less than Bh: exactly a third of it, as Volume of pyramids and cones shows. A sphere has its own formula, in Volume of a sphere.
The height is measured at right angles to the bases. For an upright solid that is the straight-up height; for one lying on its side it runs along the ground. A prism or cylinder that leans still has V = Bh with this perpendicular height, not the length of its slanting side; Cavalieri's principle gives the reason.
One unit throughout. Lengths in different units are converted before they are multiplied, and the answer is in cubic units of that one unit.
Capacity uses the inside. Outside measurements of a container with thick walls overstate what it holds. A container that is only partly filled holds a solid whose height is the depth of the filling, not the height of the container.
A rounded answer is not exact. Once \pi is replaced by a decimal the result is an approximation, written with \approx and with its rounding stated.

Where this goes next

Volume of a cylinder takes the cylinder further: finding a radius from a volume, tubes with a hole down the middle, and how far water rises when something is dropped into it. Cavalieri's principle shows why a leaning stack holds as much as an upright one, and Why the volume formulas hold: dissection, slicing and Cavalieri uses the same slicing to reach the other volume formulas. A pyramid or cone holds a third of the matching prism or cylinder (Volume of pyramids and cones), and a sphere two thirds of the cylinder that just contains it (Volume of a sphere). Solids built from several pieces, like a grain silo, are found piece by piece in Composite solids and density.

Volume also grows faster than length or area. Multiply every length of a solid by k and its base area is multiplied by k^2 and its height by k, so its volume is multiplied by k^3; Scale factor, area and volume and Scale factor and perimeter, area and volume follow this through. Later still, the The disk method slices a solid into very thin cylinders of changing radius and adds up their volumes, each one \pi r^2 times a small thickness.

Exercises

Exercises 1 to 11 go with Part 1, 12 to 20 with Part 2, and 21 to 24 draw on both.

Part 1: boxes and triangular prisms

A box is 8 cm long, 5 cm wide and 6 cm tall. How many centimeter cubes fit in its bottom layer, how many layers fill it, and what is its volume?
A triangular prism has bases that are triangles with base 9 cm and height 4 cm. The two triangles are 15 cm apart. Find the volume of the prism.
A prism’s bases are right triangles with legs 5 m and 12 m and longest side 13 m. The prism is 3 m long. Find its volume.
A prism has a base of area 27.5\ \text{cm}^2 and a height of 8 cm. Find its volume.
Find the volume of a cube whose edges are 1.5 m long.

Part 1: bases made of other shapes

A toy wooden block is shaped like a house. Each end is a rectangle 4 cm wide and 2.5 cm tall with a triangle on top, 4 cm along its bottom and 1.5 cm tall. The block is 6 cm long. Find its volume.
A prism has trapezoid bases with parallel sides 3 cm and 5 cm, which are 2 cm apart. The prism is 9 cm long. Find its volume.

Part 1: a missing measurement

A box has volume 360\ \text{cm}^3 and a base 9 cm by 8 cm. How tall is it?
A raised garden bed is 2 m long and 0.8 m wide inside. It takes 0.64\ \text{m}^3 of soil to fill it level. How deep is it?
A triangular prism has volume 84\ \text{cm}^3 and is 7 cm long. One side of each triangular base is 6 cm. What is the height of the triangle, measured to that side?
A cereal box holds 3150\ \text{cm}^3. It is 25 cm tall and 6 cm deep. How wide is it?

Part 2: cylinders

Find the volume of a cylinder with radius 5 cm and height 10 cm, exactly and to one decimal place.
A round hay bale is a cylinder lying on its side, 1.5 m across and 1.2 m long. Find its volume exactly and to one decimal place.
A thermos flask has inside radius 3.2 cm and inside height 13 cm. Find its volume to one decimal place, and its capacity in liters to two decimal places.
A round cake tin is 24 cm across, and batter fills it to a depth of 3 cm. How much batter is there, in cubic centimeters to one decimal place and in liters to two decimal places?

Part 2: capacity, missing measurements and comparing

A cylindrical sports bottle is to have inside radius 3 cm and hold 330 mL. How tall must it be inside, to one decimal place?
A hot-water tank is a cylinder with inside radius 24 cm and inside height 120 cm. How many liters does it hold, to one decimal place?
An oil drum is a cylinder with inside radius 28 cm, and it holds 200 L when full. How tall is it inside, to the nearest centimeter?
A square laundry basket is 40 cm by 40 cm across and 30 cm deep inside. A round one is 46 cm across and 30 cm deep inside. Which holds more, and by about how many liters?
Cylinder A has radius 5 cm and height 9 cm. Cylinder B has radius 4 cm and height 14 cm. Which holds more, and by how much?

Explaining

Eberhard finds the volume of a triangular prism whose bases are right triangles with legs a and b, and whose length is h, by multiplying a \times b \times h. Explain why his answer is twice the true volume, and give the true volume.
A box is 2 m long, 1 m wide and 50 cm tall. Explain why 2 \times 1 \times 50 = 100 is not its volume in cubic meters, and find its volume in cubic meters and in liters.
Explain why V = Bh still gives the right volume when the height is 2.5 units, though 2.5 layers cannot be stacked from whole unit cubes.
A pyramid and a prism have the same base and the same height. By thinking about slices parallel to the base, explain why the pyramid must hold less than Bh.

Answers

40 cubes, 6 layers, 240\ \text{cm}^3. The bottom is 8 \times 5 = 40 square centimeters, one cube on each; the box is 6 cm tall, so 6 layers, and 40 \times 6 = 240.
270\ \text{cm}^3. \displaystyle B = \frac{1}{2} \times 9 \times 4 = 18, and 18 \times 15 = 270.
90\ \text{m}^3. The legs meet at a right angle, so one is the base of the triangle and the other its height: \displaystyle B = \frac{1}{2} \times 5 \times 12 = 30, and 30 \times 3 = 90. The longest side, 13, is not needed.
220\ \text{cm}^3. V = Bh = 27.5 \times 8 = 220.
3.375\ \text{m}^3. B = 1.5^2 = 2.25, and 2.25 \times 1.5 = 3.375.
78\ \text{cm}^3. The rectangle is 4 \times 2.5 = 10 and the triangle \displaystyle \frac{1}{2} \times 4 \times 1.5 = 3, so B = 13, and 13 \times 6 = 78.
72\ \text{cm}^3. \displaystyle B = \frac{1}{2}(3 + 5) \times 2 = 8, and 8 \times 9 = 72.
5 cm. B = 9 \times 8 = 72, so \displaystyle h = \frac{360}{72} = 5.
0.4 m, which is 40 cm. B = 2 \times 0.8 = 1.6, so \displaystyle h = \frac{0.64}{1.6} = 0.4.
4 cm. \displaystyle B = \frac{84}{7} = 12, and \displaystyle \frac{1}{2} \times 6 \times t = 12 gives 3t = 12, so t = 4.
21 cm. With width w, V = w \times 6 \times 25 = 150w, so \displaystyle w = \frac{3150}{150} = 21.
250\pi \approx 785.4\ \text{cm}^3. \pi \times 5^2 \times 10 = 250\pi, and 250\pi = 785.39\ldots
0.675\pi \approx 2.1\ \text{m}^3. The radius is 0.75 m and the height of this cylinder is its length, 1.2 m, so \pi \times 0.75^2 \times 1.2 = 0.5625\pi \times 1.2 = 0.675\pi, which is 2.120\ldots
About 418.2\ \text{cm}^3, about 0.42 L. \pi \times 3.2^2 \times 13 = 133.12\pi, and 133.12\pi = 418.20\ldots, which is 418.2 mL, or 0.4182 L.
432\pi \approx 1357.2\ \text{cm}^3, about 1.36 L. The radius is 12, and \pi \times 12^2 \times 3 = 432\pi, which is 1357.16\ldots; dividing by 1000 gives liters.
About 11.7 cm. 330 mL is 330\ \text{cm}^3 and B = 9\pi, so \displaystyle h = \frac{330}{9\pi} = 11.67\ldots
About 217.1 L. V = \pi \times 24^2 \times 120 = 69\,120\pi, which is 217\,146.8\ldots\ \text{cm}^3, and dividing by 1000 gives 217.146\ldots liters.
About 81 cm. 200 L is 200\,000\ \text{cm}^3 and B = \pi \times 28^2 = 784\pi, so \displaystyle h = \frac{200\,000}{784\pi} = 81.20\ldots
The round one, by about 1.9 L. The square basket holds 40 \times 40 \times 30 = 48\,000; the round one, of radius 23, holds \pi \times 23^2 \times 30 = 15\,870\pi, which is 49\,857.0\ldots\ \text{cm}^3. The difference is 1857.0\ldots\ \text{cm}^3. Both are 30 cm deep, so only the bases matter: 529\pi \approx 1661.9 against 1600.
A, by \pi \approx 3.1\ \text{cm}^3. A holds \pi \times 5^2 \times 9 = 225\pi and B holds \pi \times 4^2 \times 14 = 224\pi. B is much taller, but A’s base, 25\pi, is larger than B’s, 16\pi, by more than enough to make up for it.
The true volume is \displaystyle \frac{1}{2}abh. The product ab is the area of the rectangle with sides a and b, and a diagonal cuts that rectangle into two copies of the right triangle, so the triangle’s area is \displaystyle \frac{1}{2}ab. Eberhard has found the volume of the box built on the rectangle, which holds two copies of the prism.
1\ \text{m}^3, which is 1000 L. The product 2 \times 1 \times 50 mixes meters with centimeters, so its answer is in neither cubic meters nor cubic centimeters. In meters the height is 0.5, and 2 \times 1 \times 0.5 = 1. Each cubic meter is 1000 liters.
Two whole layers hold 2B. For the last half layer, cut every unit cube of a layer in half across its height: each half cube has volume \displaystyle \frac{1}{2}, so a layer half a unit thick holds \displaystyle \frac{1}{2}B. Together \displaystyle 2B + \frac{1}{2}B = 2.5B, which is Bh with h = 2.5. The same cutting into thinner pieces works for any height.
Every slice of the prism parallel to the base is a copy of the base, of area B. A slice of the pyramid is a smaller copy of the base, shrinking to a point at the tip, so above the bottom every slice has area less than B. Each thin layer of the pyramid therefore holds less than the matching layer of the prism, and adding the layers, the pyramid holds less than Bh. It holds exactly \displaystyle \frac{1}{3}Bh, which Volume of pyramids and cones shows.