Math 8 · 8.G.C.9 · G.GMD.A.3

Volume of a cylinder

A can is round, so no stack of unit cubes fits inside it exactly. How do you find how much it holds?

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Volume of prisms and cylindersMath 7

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Why the volume formulas hold: dissection, slicing and CavalieriGeometry

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When you reach for it

A box is easy to measure because unit cubes fit it exactly: so many in a row, so many rows, so many layers. A can, a pipe, a water tank, a coin and a pillar are all round, and no arrangement of cubes fills a round shape without gaps. Yet each of these is a cylinder, a solid with two congruent, parallel circular ends joined by a curved side (see Cylinders and cones), and its volume comes from the same idea that measures a box.

Reach for this skill whenever the question is how much a round container holds, how much material a round solid or a tube contains, how tall or how wide a cylinder must be to hold a given amount, or how much a water level rises when something is dropped in. It rests on two things you already have: the volume of a prism, base area times height, and the area of a circle, \pi r^2.

This lesson takes two days. Part 1 builds the formula and uses it to find volumes from a radius, a diameter or a circumference, in any units, and turns cubic units into liters. Part 2 runs the formula backwards to find a height or a radius, compares cylinders of different sizes, and measures tubes and the volume of water pushed up by an object. Examples 1 to 4 and Exercises 1 to 11 belong to Part 1; Examples 5 to 9 and Exercises 12 to 25 belong to Part 2.

The procedure

Part 1: the formula and how to use it

Why base area times height. Stand a cylinder of radius r and height h on one of its ends, and slice it into layers one unit thick. Each layer is a flat round disc, a circle of radius r that is 1 unit deep. A layer with base area A and depth 1 holds A \cdot 1 cubic units, for the same reason a single layer of a box holds as many unit cubes as there are unit squares on its floor. So each layer holds \pi r^2 cubic units, and there are h of them.

Figure 1A cylinder of height h is h layers, each one unit thick. Each layer holds \pi r^2 \cdot 1 cubic units.

The stacking argument never used the shape of the floor, only its area. That is why a cylinder obeys the same rule as a prism: every slice parallel to the base is the same circle, all the way up, so the volume is that circle’s area repeated through the height. When h is not a whole number the last layer is a thinner disc, and its volume is still its base area times its thickness, so the total is still base area times height. Why the volume formulas hold gives a second argument, filling the circle with regular polygons of more and more sides so that a prism on that base closes in on the cylinder.

With base area \pi r^2 and height h,

V = \pi r^2 h

Find the radius first. The formula needs the radius, the distance from the center of an end to its edge. A diameter is twice the radius, so halve it. A circumference C satisfies C = 2\pi r (see Circumference), so \displaystyle r = \frac{C}{2\pi}. Using the diameter in place of the radius would make the base area \pi (2r)^2 = 4\pi r^2, four times too large, because the radius is squared.

Square the radius, then multiply by \pi. In \pi r^2 the exponent belongs to r alone, by the order of operations. The base area is \pi times r^2, not (\pi r)^2.

Multiply by the height. The height is the distance between the two ends, measured straight across from one to the other. For a cylinder standing up it is how tall it is; for a pipe or a log lying on its side it is the length.

Keep one unit throughout. Every length must be in the same unit before multiplying. A radius in centimeters times a height in meters gives a number that is neither cubic centimeters nor cubic meters. When the lengths are in centimeters, the volume is in cubic centimeters, \text{cm} \times \text{cm} \times \text{cm} = \text{cm}^3, and likewise for any unit.

Exact first, rounded last. The exact volume is a number times \pi, such as 72\pi\ \text{cm}^3. That is the answer to keep while working. Round only at the end, when a decimal is wanted, because a value rounded early carries its error into every later step.

From cubic units to liters. A liter is the volume of a cube 10 cm on each edge, so 1\ \text{L} = 1000\ \text{cm}^3 and 1\ \text{cm}^3 = 1\ \text{mL}. A cube 1 m on each edge holds 10 \times 10 \times 10 = 1000 of those liter cubes, so 1\ \text{m}^3 = 1000\ \text{L}. See Volume and capacity.

Part 2: running the formula backwards, and comparing

Finding the height. If V and r are known, V = \pi r^2 h is an equation in h alone. Divide both sides by the base area \pi r^2, which is not zero:

h = \frac{V}{\pi r^2}

Finding the radius. If V and h are known, divide both sides by \pi h to get \displaystyle r^2 = \frac{V}{\pi h}. Two numbers have that square, one positive and one negative, but a radius is a length, so only the positive square root answers the question:

r = \sqrt{\frac{V}{\pi h}}

Comparing sizes. The radius enters the formula squared and the height enters once. Multiply the radius by k and the height by m, and the new volume is

\pi (kr)^2 (mh) = k^2 m \cdot \pi r^2 h

so the volume is multiplied by k^2 m. Doubling the radius multiplies the volume by 4; doubling the height multiplies it by 2. This is the same reason a scaled area grows with the square of the scale factor (Scale factor, area and volume).

Tubes. A tube, such as a pipe, is a cylinder of outer radius R with a cylinder of inner radius r removed, both of length h. The volume of material is the volume of the whole minus the volume of the hole:

V = \pi R^2 h - \pi r^2 h = \pi (R^2 - r^2) h

Figure 2A tube is a cylinder of radius R with a cylinder of radius r taken out. Its cross section is the shaded ring.

Read the same way as before, \pi(R^2 - r^2) is the area of the ring at the end, and the tube is that ring stacked through the length.

Part of a cylinder. Water standing d deep in an upright cylindrical tank is itself a cylinder of radius r and height d, so it has volume \pi r^2 d. That turns a change in water level into a volume, which is how Example 9 measures a stone.

Worked examples

Example 1

A volume from the radius

A cylinder has radius 4 cm and height 9 cm. Find its volume exactly, and to one decimal place.

Solution

The base is a circle of radius 4 cm, so the base area is \pi \cdot 4^2 = 16\pi\ \text{cm}^2. Square first, then multiply by \pi. The volume is base area times height:

V = 16\pi \cdot 9 = 144\pi\ \text{cm}^3

That is the exact answer. Multiplying out with a calculator’s value of \pi,

V = 144\pi \approx 452.4\ \text{cm}^3

Check the size: \pi is a little more than 3, so the volume is a little more than 3 \cdot 16 \cdot 9 = 432, and 452.4 is.

Example 2

A volume from the diameter

A food can is 7.4 cm across and 11 cm tall. How much does it hold, in milliliters?

Solution

“Across” is the diameter, so the radius is half of it, r = 3.7 cm. Then

\begin{aligned} V &= \pi \cdot 3.7^2 \cdot 11 \\ &= \pi \cdot 13.69 \cdot 11 \\ &= 150.59\pi \approx 473.1\ \text{cm}^3 \end{aligned}

Since 1\ \text{cm}^3 = 1\ \text{mL}, the can holds about 473 mL.

Had the diameter 7.4 gone into the formula as if it were the radius, the base area would have been \pi (2 \cdot 3.7)^2 = 4 \cdot \pi \cdot 3.7^2, four times the true base area, and the volume four times too large. That is the reason for halving first.

Example 3

A volume from the circumference

A cylindrical bin is measured with a tape. It is 88 cm around and 30 cm tall. Find its volume, and say how many liters it holds.

Solution

The tape gives the circumference, not the radius. From C = 2\pi r,

r = \frac{88}{2\pi} = \frac{44}{\pi}\ \text{cm}

Keep that exact fraction rather than a decimal. The base area is

\pi r^2 = \pi \cdot \frac{44^2}{\pi^2} = \frac{1936}{\pi}\ \text{cm}^2

because one factor of \pi cancels. The volume is

V = \frac{1936}{\pi} \cdot 30 = \frac{58080}{\pi} \approx 18487\ \text{cm}^3

That is about 18.5 liters, since every 1000\ \text{cm}^3 is a liter.

The exact radius was kept on purpose. Rounding it first to r \approx 14.0 cm gives \pi \cdot 14^2 \cdot 30, which is 5880\pi \approx 18473\ \text{cm}^3, about 15\ \text{cm}^3 short, because the small rounding in r is squared and then multiplied by 30\pi.

Example 4

Mixed units and a capacity in liters

A rainwater tank is a cylinder of radius 80 cm and height 1.5 m. How many liters does it hold when full?

Solution

The radius and the height are in different units, so convert one before multiplying. In meters, r = 0.8 m and h = 1.5 m:

\begin{aligned} V &= \pi \cdot 0.8^2 \cdot 1.5 = \pi \cdot 0.64 \cdot 1.5 \\ &= 0.96\pi\ \text{m}^3 \approx 3.02\ \text{m}^3 \end{aligned}

Each cubic meter is 1000 liters, so the tank holds 960\pi \approx 3016 liters.

Check by working in centimeters instead, with h = 150 cm:

V = \pi \cdot 80^2 \cdot 150 = 960\,000\pi\ \text{cm}^3

and dividing by 1000\ \text{cm}^3 per liter gives 960\pi liters again. The two routes agree, as they must: the tank does not care which unit measures it.

Example 5

Finding the height

A drinks can must hold 750 mL and have radius 4.5 cm. How tall must it be inside, to one decimal place?

Solution

750 mL is 750\ \text{cm}^3, so the volume formula with the known values is

750 = \pi \cdot 4.5^2 \cdot h = 20.25\pi\, h

Divide both sides by 20.25\pi, the base area, to leave h alone:

h = \frac{750}{20.25\pi} = \frac{1000}{27\pi} \approx 11.8\ \text{cm}

(Multiplying the top and bottom of \displaystyle \frac{750}{20.25} by 4 gives \displaystyle \frac{3000}{81}, which cancels to \displaystyle \frac{1000}{27}.)

Check: \pi \cdot 20.25 \cdot 11.8 \approx 750.7, a little over 750 because the height was rounded up.

Example 6

Finding the radius

A water tank is to be 2 m tall and hold 5\ \text{m}^3. What radius does it need, to two decimal places, and how wide is it?

Solution

Put the known values into the formula. The unknown is now the radius:

5 = \pi r^2 \cdot 2

Divide both sides by 2\pi, then take the square root:

\begin{aligned} r^2 &= \frac{5}{2\pi} \approx 0.796 \\ r &= \sqrt{\frac{5}{2\pi}} \approx 0.89\ \text{m} \end{aligned}

The equation \displaystyle r^2 = \frac{5}{2\pi} also has the negative solution \displaystyle -\sqrt{\frac{5}{2\pi}}, but a radius is a length, so it is set aside. The tank is about 2 \times 0.892 \approx 1.78 m across.

Here the square root was the last step, and that is the reason to divide by 2\pi first: the square root undoes the square on r only once r^2 stands alone.

Example 7

Comparing two cans

(a) Can A has radius 3 cm and height 14 cm. Can B has twice the radius and half the height. Which holds more, and how many times as much?

(b) A cylinder’s radius is increased by 10\% and its height kept the same. By what percentage does its volume increase? What if instead the height is increased by 10\% and the radius kept?

Solution

(a) Can A holds \pi \cdot 3^2 \cdot 14 = 126\pi\ \text{cm}^3. Can B has radius 6 and height 7, so it holds \pi \cdot 6^2 \cdot 7 = 252\pi\ \text{cm}^3, twice as much. The general rule predicts this: the radius is multiplied by k = 2 and the height by \displaystyle m = \frac{1}{2}, so the volume is multiplied by \displaystyle k^2 m = 4 \cdot \frac{1}{2} = 2. Halving the height takes away less than doubling the radius adds, because the radius is squared.

(b) Increasing by 10\% multiplies by 1.1. With k = 1.1 and m = 1, the volume is multiplied by 1.1^2 = 1.21, an increase of 21\%. With k = 1 and m = 1.1, it is multiplied by 1.1, an increase of only 10\%. The same change does more through the radius than through the height.

Example 8

The metal in a tube

A steel tube is 1.5 m long. The hole through it is 5 cm across, and its wall is 0.5 cm thick. Find the volume of steel, to one decimal place.

Solution

The steel is the whole cylinder with the hole taken out. Work in centimeters, so the length is h = 150 cm. The inner radius is half the width of the hole, r = 2.5 cm. The wall runs from the edge of the hole out to the outside surface, so the outer radius is the inner radius plus the wall, R = 2.5 + 0.5 = 3 cm. Then

\begin{aligned} V &= \pi (R^2 - r^2) h \\ &= \pi (9 - 6.25) \cdot 150 \\ &= 412.5\pi \approx 1295.9\ \text{cm}^3 \end{aligned}

The area of the ring, \pi(R^2 - r^2) = 2.75\pi, is a difference of two circle areas. It is not the area of a circle whose radius is the wall thickness: that would be \pi \cdot 0.5^2 = 0.25\pi, far smaller, because the ring runs all the way round the tube.

Example 9

Measuring a stone with water

A cylindrical glass has inside radius 4 cm. A stone is lowered in until it rests on the bottom, completely under water, and the water level rises by 1.5 cm. Find the volume of the stone. If the stone has mass 200 g, find its density.

Figure 3The glass in Example 9. The water that the stone pushes up is itself a cylinder, 1.5 cm tall.

Solution

The stone takes up space that the water used to fill, so the water is pushed up by exactly the stone’s volume. The water above the old level is a cylinder of radius 4 cm and height 1.5 cm, whatever the stone’s shape, so

V = \pi \cdot 4^2 \cdot 1.5 = 24\pi \approx 75.4\ \text{cm}^3

That is the stone’s volume, measured without ever knowing its shape. Its density is mass divided by volume:

\frac{200}{24\pi} \approx 2.65\ \text{g/cm}^3

The argument needs the stone to be entirely under water. A stone poking out above the surface pushes up only the volume of its submerged part.

Boundaries

The formula is for a cylinder with circular ends, the same circle all the way through. A shape that narrows, such as a cone or a bucket wider at the top than the bottom, has slices of different sizes and is not base area times height. A cone holds one third of the cylinder with the same base and height (Volume of pyramids and cones).

The ends must be parallel. A cylinder sliced off on a slant has no single height, and \pi r^2 h does not apply to it as it stands.

A leaning cylinder uses the perpendicular height. For an oblique cylinder, one whose side slants, the volume is still \pi r^2 h, but h is the straight-up distance between the ends, not the length of the slanted side. That follows from Cavalieri’s principle: its slices are the same circles as those of an upright cylinder of the same height.

r is the radius. A diameter or a circumference has to be turned into a radius first.

One unit for every length. The formula multiplies three lengths, so they must share a unit, and the answer is in that unit cubed.

Decimals are approximations. Any decimal value of \pi is rounded, so a volume written as a decimal is approximate. The exact volume is a number times \pi (or, as in Example 3, a number divided by \pi).

Backwards gives one radius. Solving for r gives \displaystyle r^2 = \frac{V}{\pi h}, and only the positive square root is a length.

A real container has walls. What a container holds is measured with its inside radius and height; the material it is made from is measured as a tube, outside minus inside.

Where this goes next

The cylinder is the yardstick for the round solids that follow. A cone holds exactly one third of the cylinder with the same base and height (Volume of pyramids and cones), and a ball holds two thirds of the cylinder that just fits around it (Volume of a sphere); Why the volume formulas hold: dissection, slicing and Cavalieri shows why both fractions hold. Objects such as a grain silo, a cylinder with a half ball on top, are split into these pieces and added up (Composite solids and density), and choosing a cylinder to stand in for a log or a pillar is the subject of Modeling with solids: density and design.

Spinning a rectangle about one of its sides sweeps out a cylinder, which is the simplest of the Solids of revolution. In calculus, a curved solid is cut into thin discs, each a very short cylinder of volume \pi r^2 \cdot (\text{thickness}), and those are added up (Volumes with known cross sections); the The shell method instead cuts it into thin tubes, using the tube volume of Part 2. And when water pours into a cylindrical tank at a steady rate, the level rises at a steady rate too, since V = \pi r^2 h makes the volume proportional to the depth: the starting point of related rates.

Exercises

Part 1: finding the volume

Give each volume exactly, in terms of \pi, and to one decimal place unless asked otherwise.

A cylinder has radius 3 cm and height 8 cm.
A cylinder has radius 2.5 m and height 4 m.
A cylinder has diameter 12 inches and height 5 inches.
A cylinder has diameter 1.2 m and height 0.5 m. Give the decimal to two decimal places.
A cylinder has radius \displaystyle \frac{3}{4} inch and height 8 inches.
A cylinder has circumference 20\pi cm and height 15 cm.

Part 1: units and capacity

A coin is a cylinder 2.6 cm across and 1.8 mm thick. Find its volume in cubic centimeters, to two decimal places.
A cooking pot is a cylinder with inside radius 12 cm and inside height 15 cm. How many liters does it hold, to one decimal place?
A round paddling pool is 6 m across and filled to a depth of 1.2 m. Find the volume of water in cubic meters, to one decimal place, and roughly in liters.

Part 1: reasoning

Explain, by cutting a cylinder into layers one unit thick, why its volume is its base area times its height.
One cylinder has radius 4 cm and height 1 cm; another has radius 1 cm and height 4 cm. Which holds more, and how many times as much? Explain why they are not equal.

Part 2: working backwards

A cylinder has volume 360\pi\ \text{cm}^3 and radius 6 cm. Find its height.
A cylinder has volume 500\ \text{cm}^3 and radius 5 cm. Find its height, exactly and to two decimal places.
A jug in the shape of a cylinder is 12 cm tall inside and holds exactly 1 liter. Find its inside radius, exactly and to two decimal places.
A cylinder has volume 98\pi\ \text{m}^3 and height 2 m. Find its radius.
A cylinder’s height equals its radius, and its volume is 27\pi\ \text{cm}^3. Find its radius.

Part 2: scaling and comparing

A cylinder’s radius is tripled and its height is unchanged. What happens to its volume?
A cylinder’s radius is halved and its height is doubled. What happens to its volume?
A cylinder fits exactly inside a cube-shaped box with edges 8 cm long, touching the bottom, the top and all four sides. Find the cylinder’s volume, and what fraction of the box it fills.

Part 2: tubes, water and filling

A copper pipe is 2 m long, with outer radius 1.5 cm and inner radius 1.3 cm. Find the volume of copper, to one decimal place.
A cylindrical tank of radius 30 cm holds some water. A metal block of volume 4500\ \text{cm}^3 is lowered in until it is completely under water. By how much does the water level rise, exactly and to two decimal places?
A rain barrel is a cylinder of radius 0.3 m and height 0.9 m. A hose fills it at 12 liters per minute. How many liters does the barrel hold, and how long does it take to fill from empty, to one decimal place? (See Flow rate.)

Part 2: reasoning

Explain, from the formula, why doubling a cylinder’s radius multiplies its volume by 4, but doubling its height multiplies it only by 2.
A tube has outer radius R, inner radius r and length h. Show that its volume \pi(R^2 - r^2)h can be written as \displaystyle 2\pi \cdot \frac{R + r}{2} \cdot (R - r) \cdot h, and explain what that says about a tube slit along its length and unrolled flat.
Two cylinders hold the same volume, and one has twice the radius of the other. How do their heights compare? Explain.

Answers

72\pi \approx 226.2\ \text{cm}^3. \pi \cdot 3^2 \cdot 8 = 72\pi.
25\pi \approx 78.5\ \text{m}^3. \pi \cdot 2.5^2 \cdot 4 = \pi \cdot 6.25 \cdot 4 = 25\pi.
180\pi \approx 565.5\ \text{in}^3. The radius is 6 inches, so \pi \cdot 36 \cdot 5 = 180\pi.
0.18\pi \approx 0.57\ \text{m}^3. The radius is 0.6 m, so \pi \cdot 0.36 \cdot 0.5 = 0.18\pi.
\displaystyle \frac{9}{2}\pi \approx 14.1\ \text{in}^3. \displaystyle \pi \cdot \frac{9}{16} \cdot 8 = \frac{72}{16}\pi = \frac{9}{2}\pi.
1500\pi \approx 4712.4\ \text{cm}^3. 2\pi r = 20\pi gives r = 10, so \pi \cdot 100 \cdot 15 = 1500\pi.
0.3042\pi \approx 0.96\ \text{cm}^3. The radius is 1.3 cm and the thickness is 1.8\ \text{mm} = 0.18 cm, so \pi \cdot 1.69 \cdot 0.18 = 0.3042\pi.
About 6.8 liters. The volume is \pi \cdot 144 \cdot 15, which is 2160\pi \approx 6785.8\ \text{cm}^3, and 1000\ \text{cm}^3 is a liter.
10.8\pi \approx 33.9\ \text{m}^3, roughly 33\,900 liters. The radius is 3 m, so \pi \cdot 9 \cdot 1.2 = 10.8\pi, and each cubic meter is 1000 liters.
A layer one unit thick is a disc whose top and bottom are the base circle. It holds as many cubic units as the base has square units, since each square unit of base carries one unit of depth, so it holds \pi r^2. A height of h units is h such layers (the last one thinner if h is not whole, and still base area times its thickness), so the total is \pi r^2 h.
The first holds 4 times as much: \pi \cdot 4^2 \cdot 1 = 16\pi against \pi \cdot 1^2 \cdot 4 = 4\pi\ \text{cm}^3. Swapping the numbers is not neutral because the radius is squared and the height is not.
h = 10 cm. 360\pi = \pi \cdot 36 \cdot h, so \displaystyle h = \frac{360}{36}.
\displaystyle h = \frac{20}{\pi} \approx 6.37 cm. 500 = 25\pi h, so \displaystyle h = \frac{500}{25\pi}.
\displaystyle r = \sqrt{\frac{250}{3\pi}} \approx 5.15 cm. One liter is 1000\ \text{cm}^3, so 1000 = 12\pi r^2 and \displaystyle r^2 = \frac{1000}{12\pi} = \frac{250}{3\pi}; take the positive root.
r = 7 m. 98\pi = 2\pi r^2, so r^2 = 49, and a radius is positive.
r = 3 cm. With h = r the volume is \pi r^3, so r^3 = 27.
It is multiplied by 9. Here k = 3 and m = 1, so k^2 m = 9.
It is halved. Here \displaystyle k = \frac{1}{2} and m = 2, so \displaystyle k^2 m = \frac{1}{4} \cdot 2 = \frac{1}{2}.
128\pi \approx 402.1\ \text{cm}^3, which is \displaystyle \frac{\pi}{4} \approx 0.785 of the box. The cylinder has radius 4 cm and height 8 cm, so its volume is \pi \cdot 16 \cdot 8 = 128\pi; the box holds 8^3 = 512\ \text{cm}^3, and \displaystyle \frac{128\pi}{512} = \frac{\pi}{4}.
112\pi \approx 351.9\ \text{cm}^3. In centimeters the length is 200, so the volume is \pi(1.5^2 - 1.3^2) \cdot 200, which is \pi \cdot 0.56 \cdot 200 = 112\pi.
It rises \displaystyle \frac{5}{\pi} \approx 1.59 cm. The water pushed up is a cylinder of radius 30 cm and volume 4500\ \text{cm}^3, so 900\pi d = 4500 and \displaystyle d = \frac{5}{\pi}.
It holds 81\pi \approx 254.5 liters and takes about 21.2 minutes. Its volume is \pi \cdot 0.09 \cdot 0.9 = 0.081\pi\ \text{m}^3, which is 81\pi liters, and \displaystyle \frac{81\pi}{12} = 6.75\pi \approx 21.2.
In V = \pi r^2 h the radius appears as a factor twice, in r^2 = r \cdot r, and the height only once. Doubling r doubles each of the two factors, (2r)^2 = 4r^2, so the volume is multiplied by 4. Doubling h doubles one factor, so the volume is multiplied by 2.
Since R^2 - r^2 = (R + r)(R - r), the volume is \pi(R + r)(R - r)h, and writing \pi(R + r) as \displaystyle 2\pi \cdot \frac{R + r}{2} gives the form asked for. Here \displaystyle \frac{R + r}{2} is the radius halfway through the wall, so \displaystyle 2\pi \cdot \frac{R + r}{2} is the circumference measured halfway through the wall. The tube therefore has the volume of a flat slab that long, as thick as the wall, R - r, and as long as the tube, h: what the unrolled tube would be if the wall could be flattened with its middle keeping its length.
The cylinder with twice the radius is one quarter as tall. If the radii are r and 2r and the heights h_1 and h_2, equal volumes give \pi r^2 h_1 = \pi (2r)^2 h_2. The right side is 4\pi r^2 h_2, so dividing both sides by \pi r^2 gives h_1 = 4h_2. Doubling the radius makes the base four times as large, so the same volume needs only a quarter of the height.