Composite solids and density
A grain silo is a can with a dome on top, and a pipe is a can with a hole down its middle. No single formula fits either. How do you find how much they hold, and how heavy they are?
Where you are 7 in Volume of solids
When you reach for it
Most solid objects are not one textbook shape. A grain silo is a cylinder with a dome for a roof; a candle may be a cylinder that ends in a cone; a pipe is a cylinder with a narrower cylinder missing from its middle. No single formula gives their volumes, but each is built from solids that do have formulas, and volume behaves exactly as area does in composite figures: cut a solid into pieces that do not overlap, and the volumes of the pieces add up to the volume of the whole.
Reach for it whenever a solid is “this shape on top of that shape” or “this shape with that shape taken out”. The formulas for the pieces come from earlier lessons: prisms in Volume of prisms and cylinders, cylinders in Volume of a cylinder, pyramids and cones in Volume of pyramids and cones, and balls and half balls in Volume of a sphere.
A volume says how much space an object fills. Often the real question is how heavy it is, and that needs one more fact about what it is made of: its density, the mass packed into each cubic centimeter. With a density, a volume turns into a mass, a mass turns back into a volume, and objects made of different materials can be compared.
This lesson takes two days. Part 1 splits solids into pieces and adds or subtracts their volumes. Part 2 brings in density: finding a mass from a volume, a volume from a mass, comparing objects, and weighing hollow ones. Examples 1 to 4 and Exercises 1 to 13 belong to Part 1; Examples 5 to 7 and Exercises 14 to 22 belong to Part 2.
The procedure
Part 1: splitting a solid into pieces
Why the pieces add. A volume counts how many unit cubes of space a solid fills. If the solid is cut into pieces that do not overlap and that together fill all of it, every bit of its space is counted in exactly one piece, so
This is the fact that lets the area of an L-shaped room be found as two rectangles, one dimension up. Run backwards it gives subtraction. If a solid is a larger solid with a piece removed, then the larger solid is the solid and the removed piece put back together, without overlap, so
Seeing the pieces. Look for the flat faces and circular edges where one familiar solid meets another: a dome on a round tower meets it in a circle, a roof on a shed in a rectangle. Cut along those joins in your mind. A solid with a hole in it, such as a pipe, a bowl or a block with a hole drilled through, is usually easier as subtraction: the outside shape, solid all through, minus the hole.
The formulas the pieces need. These are the volumes of the solids this lesson uses, each from its own lesson:
| Piece | Volume |
|---|---|
| prism, base area B, height h | Bh |
| cylinder, radius r, height h | \pi r^2 h |
| pyramid, base area B, height h | \displaystyle \frac{1}{3}Bh |
| cone, radius r, height h | \displaystyle \frac{1}{3}\pi r^2 h |
| ball, radius r | \displaystyle \frac{4}{3}\pi r^3 |
| hemisphere, radius r | \displaystyle \frac{2}{3}\pi r^3 |
A hemisphere is half a ball, cut by a plane through its center.
Reading each piece’s measurements from the whole. Pieces share measurements, and a shared one is often given only once. A hemisphere sitting exactly on a cylinder has the cylinder’s radius, and its height is that same radius, because it rises from its flat face to the top of the ball. A cone on a cylinder of the same width has the cylinder’s radius, and its height is measured straight up from the joining circle to its tip, not from the ground and not along its slanting side. When the total height of a solid is given, subtract the heights you know to find the one you need.
Keeping \pi until the end. Every round piece has a volume that is a number times \pi. Those add like like terms, by the distributive property: 90\pi + 18\pi = (90 + 18)\pi = 108\pi. The exact answer is 108\pi; a decimal such as 339.3 is rounded, and is written with “about” or with \approx. Round once, at the very end, since rounding each piece first can change the last digit. A piece with no \pi in it, such as a box, stays separate: 1000 - 40\pi is exact and does not combine any further.
One unit throughout. Every length must be in the same unit before anything is multiplied. The signpost of Example 3, 2.5 m tall and 6 cm across, is worked with its height as 250 cm, and the answer then comes out in \text{cm}^3.
Part 2: density
Mass per unit of volume. A block of aluminum and a block of gold of exactly the same size have very different masses. The difference lies in the material, not the size, and it is measured by density, the mass of one unit of volume:
Here \rho, the Greek letter rho, is the density, m the mass and V the volume. With mass in grams and volume in cubic centimeters, density is in grams per cubic centimeter, \text{g/cm}^3. A density of 2.7\ \text{g/cm}^3 says that every cubic centimeter of the material has mass 2.7 g.
The same fact read three ways. Multiply both sides of \displaystyle \rho = \frac{m}{V} by V, and then divide both sides by \rho, which is never zero for a real material:
The units agree with this: \text{g/cm}^3 \times \text{cm}^3 = \text{g}, since the cubic centimeters cancel, and \text{g} \div \text{g/cm}^3 = \text{cm}^3.
Some densities. Measured values depend a little on temperature and on the exact make-up of the material, so these are rounded:
| Material | Density in \text{g/cm}^3 |
|---|---|
| water | 1 |
| ice | about 0.92 |
| candle wax | about 0.9 |
| aluminum | about 2.7 |
| iron and steel | about 7.9 |
| copper | about 9.0 |
| gold | about 19.3 |
Comparing objects. For equal volumes, mass is in proportion to density, since m = \rho V with the same V: a gold block is \displaystyle \frac{19.3}{2.7} \approx 7.1 times as heavy as an aluminum block of the same size. For equal masses the denser material takes up less room, since \displaystyle V = \frac{m}{\rho}. An object put in water floats when its average density, its whole mass divided by the whole space it takes up, is less than the water’s 1\ \text{g/cm}^3, and sinks when it is more.
Hollow objects. The material of a hollow object is its shell, the outer solid minus the hollow (Part 1’s subtraction), so its mass is density times the shell volume. Air, at about 0.0012\ \text{g/cm}^3, is light enough to leave out. Mass over the outer volume is then the object’s average density, far less than its material’s.
Two materials. When the pieces of a composite solid are made of different materials, find each piece’s mass from its own density and volume, then add the masses: m = \rho_1 V_1 + \rho_2 V_2. Masses add because the pieces do not overlap; densities do not add.
Units. A kilogram is 1000 g, and 1\ \text{cm}^3 is 1 milliliter (Volume and capacity). A cubic meter is a cube 100 cm on each side, so it holds 100^3 = 1{,}000{,}000\ \text{cm}^3, and a density of 1\ \text{g/cm}^3 is one tonne in every cubic meter. Each change is a multiplication by a fraction equal to one (Conversion factors and dimensional analysis).
Worked examples
A grain silo
A grain silo is a cylinder 6 m across and 10 m tall, with a hemisphere for its roof. Find the volume of the silo, exactly and to one decimal place, and its total height.
The silo is two pieces: a cylinder below and a hemisphere above, meeting in the circle at the top of the cylinder. They do not overlap and together fill the silo, so their volumes add. The roof sits exactly on the cylinder, so both have the same radius, half of 6 m: r = 3 m.
The cylinder has height 10 m:
The hemisphere has radius 3 m:
Adding the two, 90\pi + 18\pi = 108\pi, so the silo holds
The roof rises one radius, 3 m, above the joining circle, so the silo is 10 + 3 = 13 m tall.
Check the size. A box 6 m by 6 m by 13 m would just contain the silo and holds 6 \cdot 6 \cdot 13 = 468\ \text{m}^3. The silo leaves gaps at the box’s corners and round its dome, so it must hold less, and 339.3 is less.
An ice cream cone with a scoop on top
Niamh’s ice cream cone is 12 cm tall from its tip to its rim and 5 cm across the rim. It is packed level with ice cream, and a hemisphere of ice cream as wide as the rim sits on top. How much ice cream is there?
The ice cream is a cone, point down, with a hemisphere on its rim. The rim is 5 cm across, so both pieces have radius r = 2.5 cm. The cone’s height, 12 cm, is measured straight from the tip to the center of the rim.
The cone holds
For the hemisphere, \displaystyle 2.5^3 = 15.625 = \frac{125}{8}, so it holds
To add, write 25\pi in twelfths, \displaystyle \frac{300\pi}{12}:
A hollow signpost
A road signpost is a hollow steel cylinder 2.5 m tall. Its outside diameter is 6 cm, and its wall is 0.5 cm thick. Find the volume of steel in it.
SolutionThe post is a solid cylinder with a narrower cylinder, the hollow, taken out along its whole height. So its volume is the outer cylinder minus the hollow.
First put every length in one unit. The radii are in centimeters, so the height is 2.5 \times 100 = 250 cm. The outer radius is half of 6 cm, R = 3 cm. The wall’s thickness is measured inward from the outside, so the hollow’s radius is r = 3 - 0.5 = 2.5 cm.
The steel is the difference:
which is about 2.16 liters of steel.
Working back to a missing height
Lowri pours candles in a mold shaped like a cylinder with a cone on top. Both have radius 3 cm, and the cone is 4 cm tall. She has 120\pi\ \text{cm}^3 of melted wax, about 377.0\ \text{cm}^3, and it fills the mold exactly. How tall is the cylinder part, and how tall is the whole candle?
Call the cylinder’s height h centimeters. The two pieces add to the whole candle, so the equation is “cylinder plus cone equals the wax”:
The cylinder is 9\pi h and the cone is \displaystyle \frac{1}{3}\pi \cdot 36 = 12\pi, so
Subtract 12\pi from both sides, which keeps them equal: 9\pi h = 108\pi. Divide both sides by 9\pi, which is not zero: h = 12. The factor \pi canceled, so no decimal was ever needed.
The cylinder is 12 cm tall, and with the cone on top the candle is 12 + 4 = 16 cm tall.
Check: 9\pi \cdot 12 = 108\pi, and 108\pi + 12\pi = 120\pi.
Mass from volume, and volume from mass
(a) Steel has density about 7.9\ \text{g/cm}^3. Find the mass of the signpost of Example 3.
(b) A block of ice has mass 4.6 kg. Ice has density about 0.92\ \text{g/cm}^3. Find the block’s volume. When it melts, what volume of water does it give?
Solution(a) Example 3 found the steel’s volume, 687.5\pi\ \text{cm}^3. Mass is density times volume, and grams per cubic centimeter times cubic centimeters gives grams:
That is about 17.1 kg.
(b) The density is in grams per cubic centimeter, so change the mass to grams first: 4.6 kg is 4600 g. Volume is mass divided by density:
that is 5 liters of ice. Melting changes the ice’s shape and density but not its mass, which stays 4600 g. Water has density 1\ \text{g/cm}^3, so the water takes up
400\ \text{cm}^3 less than the ice did. The same mass of a denser material takes less room.
Is the trophy solid gold?
A trophy, made up for this example: Klaus’s trophy is a solid cylinder 6 cm across and 2 cm tall, with a solid ball of the same width fixed on top. It is gold all over its surface and has mass 1450 g. Could it be solid gold, at about 19.3\ \text{g/cm}^3?
SolutionFind the volume, then the mass the trophy would have if it were gold all through, and compare.
Both pieces have radius 3 cm. The base is \pi \cdot 3^2 \cdot 2 = 18\pi, and the ball is \displaystyle \frac{4}{3}\pi \cdot 27 = 36\pi, so the trophy’s volume is
Solid gold of that volume would have mass
more than twice the trophy’s 1450 g. So the trophy is not solid gold. Its average density is
which lies between iron, about 7.9, and copper, about 9.0. It could be brass, about 8.5\ \text{g/cm}^3, under a thin layer of gold, but a density alone cannot say which metal it is.
A hollow plastic ball
A dog’s toy is a hollow plastic ball with outer radius 3 cm and a plastic wall 0.5 cm thick, with air inside. Take the plastic’s density to be 1.2\ \text{g/cm}^3, a made-up value, since plastics vary.
(a) Find the volume of plastic. (b) Find the ball’s mass. (c) Does the ball float in water?
Solution(a) The plastic is the outer ball with the inner ball of air taken out. The wall runs inward, so the inner radius is 3 - 0.5 = 2.5 cm. Then
(b) Only the plastic has mass worth counting. The air inside fills \displaystyle \frac{4}{3}\pi \cdot 15.625 \approx 65.4\ \text{cm}^3 and, at about 0.0012\ \text{g/cm}^3, has mass under a tenth of a gram. So
A solid plastic ball of the same size would have mass 1.2 \times 36\pi = 43.2\pi \approx 135.7 g, well over twice as much.
(c) In water the ball takes up the space of its whole outside, \displaystyle \frac{4}{3}\pi \cdot 27 = 36\pi\ \text{cm}^3, so its average density is
The factors of \pi cancel. That is about half the density of water, so the ball floats.
Boundaries
The pieces must not overlap, and must fill the solid. Adding works only when every bit of the solid is in exactly one piece. A ball sunk partly into a block has only part of its volume outside the block, so adding the whole ball counts the buried part twice.
A hemisphere only if the dome is half a ball. A dome that rises less than its own radius is a smaller cap of a ball, not a hemisphere, and \displaystyle \frac{2}{3}\pi r^3 does not give its volume. In the same way the cone formula needs the true height, straight up from the base to the tip.
The removed piece must lie inside. Subtracting a hole gives the material only when the hole is entirely within the larger solid. A hole that breaks out through the side removes less than its full volume.
One unit before multiplying. Lengths in meters and centimeters cannot be multiplied together; convert first. Volume conversions use the length factor cubed.
m = \rho V needs one material spread evenly. For a piece made of two materials, each part’s mass is found separately and the masses added. For a hollow object, the mass divided by the outer volume is an average density, not the density of the material.
Densities and decimals are rounded. An answer is no more precise than the measurements and densities that went into it.
Where this goes next
The same splitting finds the surface of a composite solid, except that faces where two pieces join are inside and drop out (Surface area of a sphere gives a ball’s surface). In Modeling with solids: density and design density decides what an object is made of and which design is best, in kilograms per cubic meter. Curved shapes that are not pieces of a cone, cylinder or ball, such as a vase, come from spinning a region about a line in Solids of revolution and from adding thin slices in Volumes with known cross sections.
Exercises
Part 1: adding pieces
Give each volume exactly and to one decimal place.
- A fuel tank is a cylinder of radius 2 m and length 5 m, with a hemisphere of the same radius on each end. Find its volume.
- A sharpened pencil is a cylinder of radius 0.5 cm and length 15 cm, with a cone 3 cm long at its tip. Find its volume.
- A shed is a rectangular prism 4 m long, 3 m wide and 2.5 m tall. Its roof is a triangular prism whose triangular end is 3 m wide and 1.2 m tall, running the 4 m length. Find the volume of the shed with its roof.
- A stone monument is a cube with edges 2 m long, topped by a square pyramid 1.5 m tall on the cube’s top face. Find its volume.
- A spinning top is two cones joined base to base, each of radius 3 cm. One cone is 4 cm tall and the other 5 cm. Find its volume.
- A fishing float is a hemisphere of radius 6 cm with a cone 8 cm tall standing on its flat face. Find its volume.
Part 1: taking a piece away
- A wooden napkin ring is a tube 3 cm long, with outer radius 2.5 cm and inner radius 2 cm. Find the volume of wood.
- A wooden cube with edges 10 cm long has a round hole of radius 2 cm drilled straight through it, from the middle of one face to the middle of the opposite face. Find the volume of wood left.
- A solid cylinder of radius 6 cm and height 8 cm has a cone with the same base and height hollowed out of it, the cone’s tip at the center of the top face. Find the volume that is left.
- A bowl is half of a hollow ball, with outside radius 12 cm and inside radius 11 cm. Find the volume of material in the bowl.
Part 1: working backwards
- A silo of radius 6 m is a cylinder with a hemisphere for a roof, and it holds 1008\pi\ \text{m}^3 in all. Find the height of the cylinder and the total height of the silo.
- A solid is a cylinder of radius r cm and height 2r cm with a hemisphere of radius r on top. Its volume is 72\pi\ \text{cm}^3. Find r.
Part 1: reasoning
- A cone and a hemisphere each have radius r, and the cone’s height is also r. Show that together they hold exactly as much as a cylinder of radius r and height r.
Part 2: mass, volume and density
- Find the mass of an aluminum block 5 cm by 4 cm by 3 cm.
- A gold bar has mass 1 kg. Find its volume, to one decimal place.
- A cylinder of ice has radius 10 cm and height 20 cm. Find its mass, and the volume of water, in liters, that it gives when it melts.
- Grain fills only the cylinder of the silo in Example 1; the roof is left empty. Taking wheat to be 0.75 tonnes per cubic meter, a made-up round value, find the mass of the grain, to one decimal place.
- Jacinta has 450 g of candle wax, at about 0.9\ \text{g/cm}^3. Find its volume. Is it enough for one candle like Lowri’s in Example 4, and if so how much wax is left?
Part 2: comparing and hollow objects
- Matteo has an aluminum cube with edges 3 cm long and an iron ball of radius 1.5 cm. Which is larger, and which is heavier?
- A hollow aluminum ball has outer radius 5 cm and a wall 0.5 cm thick. Find the volume of aluminum, the ball’s mass, and whether it floats in water.
Part 2: reasoning
- Explain how a hollow steel ball can float in water, although steel is nearly eight times as dense as water.
- A paperweight is a wooden cylinder with a steel cone on top. Explain why its mass is found as \rho_1 V_1 + \rho_2 V_2, and why adding the two densities and multiplying by the total volume would give a wrong answer.
Answers
- \displaystyle \frac{92\pi}{3} \approx 96.3\ \text{m}^3. The cylinder is \pi \cdot 2^2 \cdot 5 = 20\pi, and the two hemispheres make one ball, \displaystyle \frac{4}{3}\pi \cdot 2^3 = \frac{32\pi}{3}.
- 4\pi \approx 12.6\ \text{cm}^3. The cylinder is \pi \cdot 0.5^2 \cdot 15 = 3.75\pi and the cone \displaystyle \frac{1}{3}\pi \cdot 0.5^2 \cdot 3 = 0.25\pi.
- 37.2\ \text{m}^3. The box is 4 \cdot 3 \cdot 2.5 = 30; the roof’s triangular end has area \displaystyle \frac{1}{2} \cdot 3 \cdot 1.2 = 1.8, and 1.8 \cdot 4 = 7.2.
- 10\ \text{m}^3. The cube is 2^3 = 8 and the pyramid \displaystyle \frac{1}{3} \cdot 2^2 \cdot 1.5 = 2.
- 27\pi \approx 84.8\ \text{cm}^3. The cones are \displaystyle \frac{1}{3}\pi \cdot 9 \cdot 4 = 12\pi and \displaystyle \frac{1}{3}\pi \cdot 9 \cdot 5 = 15\pi.
- 240\pi \approx 754.0\ \text{cm}^3. The hemisphere is \displaystyle \frac{2}{3}\pi \cdot 216 = 144\pi and the cone \displaystyle \frac{1}{3}\pi \cdot 36 \cdot 8 = 96\pi.
- 6.75\pi \approx 21.2\ \text{cm}^3. \pi(2.5^2 - 2^2) \cdot 3 = \pi \cdot 2.25 \cdot 3.
- 1000 - 40\pi \approx 874.3\ \text{cm}^3. The hole is a cylinder 10 cm long, \pi \cdot 2^2 \cdot 10 = 40\pi.
- 192\pi \approx 603.2\ \text{cm}^3. The cylinder is \pi \cdot 36 \cdot 8 = 288\pi and the cone a third of it, 96\pi.
- \displaystyle \frac{794\pi}{3} \approx 831.5\ \text{cm}^3. \displaystyle \frac{2}{3}\pi(12^3 - 11^3) = \frac{2}{3}\pi \cdot 397.
- The cylinder is 24 m tall and the silo 30 m. The roof is \displaystyle \frac{2}{3}\pi \cdot 216 = 144\pi, so 36\pi h = 1008\pi - 144\pi = 864\pi and h = 24; the roof adds 6 m.
- r = 3. The cylinder is \pi r^2 \cdot 2r = 2\pi r^3, so the solid is \displaystyle 2\pi r^3 + \frac{2}{3}\pi r^3 = \frac{8}{3}\pi r^3 = 72\pi, giving r^3 = 27.
- The cone holds \displaystyle \frac{1}{3}\pi r^2 \cdot r = \frac{1}{3}\pi r^3 and the hemisphere \displaystyle \frac{2}{3}\pi r^3. Together they hold \displaystyle \left(\frac{1}{3} + \frac{2}{3}\right)\pi r^3 = \pi r^3, and the cylinder holds \pi r^2 \cdot r = \pi r^3, the same.
- 162 g. The volume is 5 \cdot 4 \cdot 3 = 60\ \text{cm}^3, and 2.7 \times 60 = 162.
- About 51.8\ \text{cm}^3. 1 kg is 1000 g, and \displaystyle \frac{1000}{19.3} \approx 51.81.
- 1840\pi \approx 5780.5 g, giving about 5.78 liters of water. The ice is \pi \cdot 10^2 \cdot 20 = 2000\pi\ \text{cm}^3, and 0.92 \times 2000\pi = 1840\pi; melting keeps the mass, so the water is 1840\pi\ \text{cm}^3.
- 67.5\pi \approx 212.1 tonnes. The cylinder is 90\pi\ \text{m}^3, and 0.75 \times 90\pi = 67.5\pi.
- 500\ \text{cm}^3, which is enough, with about 123.0\ \text{cm}^3 left. \displaystyle \frac{450}{0.9} = 500, and 500 - 120\pi \approx 500 - 377.0.
- The cube is larger and the ball heavier. The cube is 27\ \text{cm}^3 with mass 2.7 \times 27 = 72.9 g. The ball is \displaystyle \frac{4}{3}\pi \cdot 3.375 = 4.5\pi \approx 14.1\ \text{cm}^3, with mass 7.9 \times 4.5\pi = 35.55\pi \approx 111.7 g, about 38.8 g more.
- \displaystyle \frac{271\pi}{6} \approx 141.9\ \text{cm}^3 of aluminum, mass 121.95\pi \approx 383.1 g, and it floats. The inner radius is 4.5 cm, so the shell is \displaystyle \frac{4}{3}\pi(125 - 91.125) = \frac{4}{3}\pi \cdot 33.875. The outside is \displaystyle \frac{500\pi}{3}, so the average density is \displaystyle 121.95\pi \div \frac{500\pi}{3} \approx 0.73\ \text{g/cm}^3, less than water’s.
- Floating depends on the ball’s average density: its mass over all the space it takes up. The mass is steel’s density times the thin shell’s small volume, but the space is the whole outer ball, air included. A thin enough wall makes that average less than 1\ \text{g/cm}^3.
- The parts do not overlap, so the mass is the wood’s mass plus the steel’s, each its own density times its own volume. Adding the densities would give every cubic centimeter the mass of a wood one and a steel one together, true of none of them: (\rho_1 + \rho_2)(V_1 + V_2) = \rho_1 V_1 + \rho_2 V_2 + \rho_1 V_2 + \rho_2 V_1 has two extra terms.