11. Modeling in Two and Three dimensions

11.02 Prisms and cylinders

Lesson

Practice

11.03 Pyramids and cones

11.04 Spheres

11.05 Similarity in 2D and 3D figures

Book a Demo

United States of America

Grades 9-12

11.02 Prisms and cylinders

Lesson

Practice

Introduction

In 6th, 7th, and 8th grade, we saw how to solve real world problems with prisms and cylinders. This lesson will build on some of that knowledge and our work with nets in 6th grade to solve more mathematical problems and learn about a principle that we can use as another tool for solving problems.

Volume of non-right prisms and cylinders

Exploration

Explore the applet by moving the sliders and checking the boxes.

Loading interactive...

Move the slider to change the stack of slabs. Will the volume of the stack change? How do you know?
Move the slider to change the solid. What are two features of the solids that remain the same when moving the slider?
Will the volume of the solids change? How do you know?

Recall that a prism is a figure with two congruent, parallel, polygonal bases that are connected by rectangular faces. Cylinders are similar, but the bases are circles instead of polygons, and so they are joined by a curved surface instead of rectangles.

Oblique prism

A prism in which the bases are connected by rectangular faces that are not perpendicular to each base

A rectangular prism. The angles between the base and the faces are not right angles.

Oblique cylinder

A cylinder in which the bases are connected by a curved surface that is not perpendicular to each base

A cylinder with the height drawn from the center of the top circle to the center of the bottom circle. The angle between the height and the bottom circle is not a right angle.

When a prism or cylinder becomes oblique, its volume will remain the same as long as its height and cross section remain the same. This leads to Cavalieri's principle.

Cavalieri's principle

If two three-dimensional figures have the same height and the same cross-sectional area at every level, then they have the same volume

The following three figures each reach the same perpendicular height h and have the same base area B, so by Cavalieri's principle we know that they have the same volume:

A diagram showing a rectangular prism, a triangular prism, and a cylinder with their bases lying on a same plane. All figures have a perpendicular height of h. A plane parallel to the bases of the 3 figures is drawn. For each figure, the cross section created has an area of B.

Notice that it is the perpendicular height that is important, rather than the slant height, even if the solids are not right prisms/cylinders.

Density

The compactness of a substance. Usually measured in weight per unit of volume.\text{Density} = \dfrac{\text{Mass}}{\text{Volume}}

Examples

Example 1

A 3D printer is printing an object of stacked coins by stacking circular pieces with a thickness of 1 \text{ mm} and an area of 12.57 \text{ mm}^2. The volume of one of the circular pieces is 12.57 \text{ mm}^3.

Find the volume of the printed cylinder. Then find the volume of the printed cylinder if each coin was 3.2 \text{ mm} thick. Explain the process for finding the volume of the solid for a coin of any height.

Worked Solution

Create a strategy

If the volume of one of the circular pieces from the printer is 12.57 \text{ mm}^3, and the object has seven coins, we can multiply the volume of one of the coins by 7. We can apply this process to the object if each coin were a different thickness.

Apply the idea

By knowing the volume of one of the coins in the printed object, we can multiply that by the height of the object, which is 7 \text{ coins}, or 7 \text{ mm}:12.57 \text{ mm}^3 \cdot 7= 87.99 \text{ mm}^3

If each circular piece printed was instead 3.2 \text{ mm} thick, we could find the volume of the object by multiplying the area of the base by 22.4 \text{ mm}.

If each circular piece printed was instead h \text{ mm} thick, we could find the volume of the object by multiplying the area of the base by 7h \text{ mm}.

The 3D printer creates an oblique cylinder using the same dimensions as the circular pieces from the first solid. What is the volume of the new object?

Worked Solution

Apply the idea

The volume of the new object is the same as the volume of the first printed stack of coins, 87.99 \text{ mm}^3, by Cavalieri's principle.

Reflect and check

Since no settings for the dimensions of the stacks were changed and the only difference is an oblique cylindrical shape printed, the cross sections and volume of the new object should remain the same as the first stack of coins.

Find the volume of a 3D-printed cylindrical object if there was a stack of 12 coins, each 1 \text{ mm} thick with a surface area of 6\,358.5 \text{ mm}^2. Then, explain the process for finding the volume, V, of the solid for any number, n, of 1-millimeter thick, stacked coins with surface area B.

Worked Solution

Create a strategy

If we know the surface area of one of the coins, its height, and how many coins are in the stack, we can find the volume of the cylinder by multiplying the same way we approached part (a).

Apply the idea

Since 12 coins create a stack with a total height of 12 \text{ mm}, 6\,358.5 \text{ mm}^2 \cdot 12 \text{ mm} = 76\,302.0 \text{ mm}^3.

If any coin has a thickness of 1 \text{ mm}, then the height of any stack of coins will be 1n=n. We can find the volume, V, by multiplying the surface area of one coin by the height of the coins, so V=Bn.

Reflect and check

Note that the units for the volume of a three-dimensional solid must be in cubic units, so it makes sense that multiplying area by height will give us cubic millimeters as our units of measure.

Example 2

Consider a hexagonal child's ball pit. A layer of 64 balls will cover just the bottom of the ball pit.

The balls can be stacked 32 balls high and still leave extra room at the top for kids to jump in without spilling balls out. Explain how to find the number of balls needed to fill the pit.

Worked Solution

Create a strategy

Imagine the single layer of 64 balls added 32 times to fill the pit.

Apply the idea

By multiplying a layer of 64 balls 32 times, the pit can hold 64 \text{ balls} \times 32 \text{ layers} = 2\,048 \text{ balls}.

During cleaning, the ball pit gets pushed slightly and the entire prism ends up leaning to the side, but the hexagonal shape is preserved. Explain how the volume of balls that fit inside the ball pit will change.

Worked Solution

Create a strategy

Recall that Cavalieri's principle states that if two 3D solids are the same height and have the same cross section, they will have the same volume.

Apply the idea

Since the ball pit itself does not change, but instead becomes an oblique hexagonal prism, we can use Cavalieri's principle to determine that the same volume of balls from part (a) will fit in the ball pit after it is pushed during cleaning. This makes sense because the same number of balls will still be inside the pit.

Consider a hexagonally-shaped fish tank. Find the volume of the fish tank with the given dimensions.

Worked Solution

Create a strategy

Find the area of the bottom face of the tank, then think of the volume as stacking 1 \text{ in} of water 12 times to fill the tank.

Apply the idea

In order to find the area of the base, we need to cut the hexagonal base into two triangles and a rectangle, then find the sum of the areas of each part.

For the base of each triangle, we have

\displaystyle A	\displaystyle =	\displaystyle \dfrac{1}{2}bh	Formula for the area of a triangle
	\displaystyle =	\displaystyle \dfrac{1}{2} \left(6 \right) \left(2.5 \right)	Substitute b=6 and h=2.5
	\displaystyle =	\displaystyle 7.5 \text{ in}^2	Evaluate the multiplication

For the rectangle, the area is 6 \text{ in} \times 5 \text{ in}= 30 \text{ in}^2.

The sum of the shapes that make up the hexagonal base of the fish tank is 7.5 \text{ in}^2 + 7.5 \text{ in}^2 + 30 \text{ in}^2 = 45 \text{ in}^2.

If 1 \text{ in} of water were to cover the base of the tank, we could add the area of the covered base 12 times, or by multiplying the area of the base by the height of the tank, 12 \text{ in}. So 45 \text{ in}^2 \times 12 \text{ in} = 540 \text{ in}^3.

Reflect and check

We found the volume of the hexagonal prism by calculating the area of the base and multiplying it by the height of the prism.

We use this same process when calculating the volume of a cylinder, except we can relate the base of a cylinder, which is a circle, to a polygon with an infinite number of sides.

Find the density of the water in the fish tank from part (c) given that the water inside of it weighs 23.41 \text{ lbs}.

Worked Solution

Create a strategy

We calculated the volume of the tank as 540 \text{ in}^3, so we will use the formula for density given the mass of the water.

Apply the idea

\displaystyle \text{Density}	\displaystyle =	\displaystyle \dfrac{\text{Mass}}{\text{Volume}}	Density formula
	\displaystyle =	\displaystyle \dfrac{23.41}{540}	Substitute 23.41 \text{ lbs} and 540 \text{ in}^3
	\displaystyle =	\displaystyle 0.0434	Evaluate the division

Therefore, the density of the water in the tank is 0.0434 \, \text{lbs}/\text{in}^3.

Reflect and check

Rounding to a specified number of decimal places may be appropriate depending on the situation. We could round to two decimal places in this problem, but because the decimal itself is relatively small, rounding to three decimal places will give us a more precise answer.

Note that the density is calculated by dividing the mass (in pounds) by the volume (in cubic inches), and so the density has units of mass per volume - in this case, pounds per cubic inch.

Example 3

Find the density of a cube with side length 4 \text{ ft} and weight 300 \text{ lbs}.

Worked Solution

Create a strategy

We can first find the volume of the cube. Then we can calculate the density by dividing the given mass by the volume.

We can develop the formula for the volume of a cube by thinking about a cube as a square prism, where its base has side lengths s and stands s units of layers high.

We know that the base of the cube or square prism would be s \times s = s^2 square units, and we could multiply the area of the base by s layers, giving us a volume of s^2 \times s = s^3 cubic units of a cube.

Apply the idea

Finding the volume of the cube, we have:

\displaystyle V	\displaystyle =	\displaystyle s^3	Volume of a cube
	\displaystyle =	\displaystyle 4^3	Substitute side length
	\displaystyle =	\displaystyle 64	Simplify

So, the volume of the cube is 64 \text{ ft}^3.

We can now use this to calculate the density:

\displaystyle \text{Density}	\displaystyle =	\displaystyle \dfrac{\text{Mass}}{\text{Volume}}	Density formula
	\displaystyle =	\displaystyle \dfrac{300}{64}	Substitute known values
	\displaystyle =	\displaystyle 4.6875	Simplify

Therefore, the density of the cube is 4.6875 \, \text{lb}/\text{ft}^3.

Reflect and check

The solution given is the most precise measurement of the density, since it is the exact decimal value of the fraction. We might not need that level of precision, depending on why we might need to know the density of the cube, or what the practical application might be. In that case, we could round to the tenths or hundredths place. When a decimal is irrational and non-repeating, we will always need to make a rounding decision, based on contextual factors (precision of given measurements, rounding decsions made during calculations, practical ability to calibrate measurements, desired precision level, etc.)

Example 4

A company is designing a new filter for its air purifiers using a computer program. The designer maps out the 2D shape on a coordinate plane, then rotates the shape about the x-axis.

-5

-4

-3

-2

-1

\text{Feet, }x

-5

-4

-3

-2

-1

\text{Feet, }y

Sketch the filter after its revolution about the x-axis and label its dimensions.

Worked Solution

Create a strategy

The object will be the shape of a cylinder with another cylinder hollowed out from the center.

Apply the idea

What is the volume of the air filter?

Worked Solution

Create a strategy

Calculate the volume of the large cylinder as if it was solid. Then subtract the volume of the hollowed out cylinder in the center of the filter, which was created by revolving the 2D shape about the x-axis.

Apply the idea

The cylindrical air filter has a radius of 3 \text{ feet}, so the area of one of its bases is A=\pi \cdot 3^2 \approx 28.27 \text{ ft}^2. To find the volume of the solid, we can multiply the area of the base by the height of the cylinder, so 28.27 \text{ ft}^2 \cdot 3 \text{ ft} = 84.81 \text{ ft}^3.

The cylindrical center of the air filter has a radius of 1 \text{ foot}, so the area of one of its bases is A=\pi \cdot 1^2 \approx 3.14 \text{ ft}^2. To find the volume of the cylindrical center, we can multiply the area of the base by the height of the cylinder, so 3.14 \text{ ft}^2 \cdot 3 \text{ ft} = 9.42 \text{ ft}^3.

The volume of the filter is calculated by finding the difference between the volume of the large, solid cylinder and the volume of the center cylinder. The filter's volume is 84.81 \text{ ft}^3 - 9.42 \text{ ft}^3 = 75.39 \text{ ft}^3.

Reflect and check

Since \pi is an irrational number, we needed to make decisions about how to round it in our calculations. We should take the units we're working with and prractical considerations into account. Since the measurements are in feet and cubic feet, rounding to the hundredths place is reasonable, and likely within the capabilities of a computer design program and necessary level of accurate measurements for an air filter. We could also opt not to round, and instead use calculator software at each step, which uses more exact calculations of \pi, which will result in a more precise solution.

Once we make the decision to round to the hundredths place, that becomes the limit of our level of precision, and our final solution should not be given with more than two decimal placements.

Idea summary

The volume, V, of a prism or cylinder is calculated using the formula V=Bh, where B represents the area of the base and h represents the height of the prism.

We can use Cavalieri's principle to justify that if two three-dimensional figures have the same height and the same cross-sectional area at every level, then they have the same volume.

Surface area of prisms and cylinders

Exploration

Drag the sliders and check the boxes to explore the applet.

Loading interactive...

Create a triangular prism by changing N to 3 and dragging the sliders to close the net. How can you use the net to find the surface area of the prism?
What happens as you drag the N slider to its largest value?

A net is a diagram of the faces of a three-dimensional figure arranged in such a way that the diagram can be folded to form the three-dimensional figure. We can find the surface area of a net by calculating the sum of the lateral faces and the bases.

Lateral faces

The faces in a prism or pyramid that are not bases

A cube and its net

A cylinder and its net. The net is made up of a rectangle, and 2 identical circles. The circles are tangent to the lengths of the rectangle.

A right cylinder and its net

Note that even some solids with curved faces, such as cylinders, have nets consisting of flat, two-dimensional shapes. Additionally, a solid can be represented with multiple different, equivalent nets.

The surface area of a cylinder can be calculated identically; by adding the area of two circular bases to the product of the circumference and the perpendicular height between bases.

We can find the lateral area of a prism or a cylinder by ignoring the bases.

Lateral area

The sum of the areas of the lateral faces of a prism or pyramid, or the area of the lateral surface of a cylinder or cone

LA=Ph where P is the perimeter of the base and h is the height between bases

Examples

Example 5

Consider the can of tuna shown below:

Draw the net of the can of tuna and label its dimensions.

Worked Solution

Create a strategy

The net of a cylinder will have two circular bases and a rectangle for its lateral face.

Apply the idea

Find the area of each part of the net and find the total surface area of tin that a company must produce per can of tuna.

Worked Solution

Create a strategy

We have enough information from the can of tuna to find the area of each of the circular bases. The diameter of each base, is 2.75 \text{ in} so we use half the diameter for the radius, 1.375 \text{ in}. However, we will need the length of the can in order to find the area of the lateral face. We can calculate the length of the can by finding the circumference of a circular base.

Apply the idea

First, calculate the area of each base:

\displaystyle A	\displaystyle =	\displaystyle \pi r^2	Formula for area of a circle
	\displaystyle =	\displaystyle \pi \left(1.375 \right)^2	Substitute the length of the radius of the base
	\displaystyle =	\displaystyle 5.940	Evaluate the exponent and multiplication

The area of each base is 5.940 \text{ in}^2.

Then, calculate the circumference of a base to find the length of the can's lateral side:

\displaystyle C	\displaystyle =	\displaystyle 2 \pi r	Formula for circumference
	\displaystyle =	\displaystyle 2 \pi \left(1.375 \right)	Substitute the length of the radius of the base
	\displaystyle =	\displaystyle 8.639	Evaluate the multiplication

The length of the can is 8.639 \text{ in}.

Now, the area of the lateral side, which is a rectangle, is found by multiplying the length and width of the can. So, the area of the lateral side is 8.639 \text{ in} \times 1.125 \text{ in} = 9.719 \text{ in}^2.

Finally, we can find the surface area of the tuna can by adding the area of each base to its lateral side: 5.94 \text{ in}^2 + 5.94 \text{ in}^2 + 9.719 \text{ in}^2 = 21.599 \text{ in}^2.

Reflect and check

While we could have used fractions throughout the problem, converting the given dimensions to decimals may be easier to work with.

We rounded the decimals to 3 places, since the fractions we are using are exact to 3 decimal places. We may also want to keep that level of precision because the company will want to be as precise as possible when producing each can.

The formula for finding the surface area of a prism or cylinder is SA= 2B + Ph, where SA represents surface area, B represents the area of the base, P represents the perimeter of the base, and h represents the height of the prism or cylinder. Explain what each part of the formula for surface area represents, and relate it to finding the surface area for the can of tuna.

Worked Solution

Create a strategy

We explain why variables are multiplied and added a certain way, then relate it back to the can of tuna.

Apply the idea

The surface area formula is the sum of two products.

The first product is twice the area of the base. This makes sense because each prism or cylinder will have two identical bases, so their areas should be added together, or in this formula multiplied by 2. The can of tuna had circular bases, so we used the Area formula for a circle to find the Base area.

The second product is the result of multiplying the perimeter of the base and the height of the prism or cylinder. This is the area of the rectangular lateral side, since the perimeter of the base is also the width of the rectangle and the height of the prism is the height of the rectangle. For the can of tuna, the base was circular, so we used the formula for the Circumference of a circle as the perimeter.

By calculating the sum of these products, we have found the combined areas of each part of the net, which is the surface area of the entire can of tuna.

Example 6

Yuki is decorating a wedding cake with some icing. He wants to cover the outer facing surface of the cake with a layer of icing that is an eighth of an inch thick.

An image of a three-layer cake. Talk to your teacher for more information.

Determine how much icing Yuki will need to decorate the cake.

Worked Solution

Create a strategy

We need to determine whether the entire top of each layer needs to be iced, and then the next layer added, or if the layers are placed first, and then only the exposed portions are iced.

After finding the surface area, we will want to multiply that value by \frac{1}{8} \text{ inches}. Depending on the approach to the problem, we could end up with different solutions to how much icing Yuki needs for the cake.

Let's assume that the layers are placed, and then only exposed cake will be iced. That means we should calculate the lateral area of each layer, then calculate the tops of the tiers shown in the diagram using the areas of the circular tops.

Apply the idea

For the lateral area of each tier, we need the circumference of each.

Starting with the first tier we have C=2 \pi r \to C = 2 \pi \left(2 \text{ in} \right) \to C = 12.57 \text{ in}

So, the lateral area of the first tier is 12.57 \text{ in} \times 2 \text{ in}= 25.14 \text{ in}^2.

For the middle tier we have C=2 \pi r \to C = 2 \pi \left(4 \text{ in} \right) \to C = 25.13 \text{ in}

So, the lateral area of the middle tier is 25.13 \text{ in} \times 3 \text{ in}= 75.39 \text{ in}^2.

For the bottom tier we have C=2 \pi r \to C = 2 \pi \left(6 \text{ in} \right) \to C = 37.7 \text{ in}

So, the lateral area of the bottom tier is 37.7 \text{ in} \times 4 \text{ in}= 150.8 \text{ in}^2.

The top of the first tier will be iced, so we can find the area of the circular top. A=\pi r^2 \to A = \pi \left(2 \text{ in} \right)^2 = 12.57 \text{ in}^2

The exposed portion of the middle layer can be found by subtracting the area of the first tier's base from the area of the middle tier's base. The area of the top of the middle tier can be calculated as A=\pi r^2 \to A = \pi \left(4 \text{ in} \right)^2 =50.27 \text{ in}^2 Then, we can subtract the area of the first tier's base 50.27 \text{ in}^2 - 12.57 \text{ in}^2 = 37.7 \text{ in}^2

The exposed portion of the bottom layer can be found by subtracting the area of the middle tier's base from the area of the bottom tier's base. The area of the top of the bottom tier can be calculated as A=\pi r^2 \to A = \pi \left(6 \text{ in} \right)^2 =113.1 \text{ in}^2 Then, we can subtract the area of the middle tier's base 113.1 \text{ in}^2 - 50.27 \text{ in}^2 = 62.83 \text{ in}^2

Finally, we will add the total surface area of icing needed for the cake and multiply the total by \frac{1}{8} \text{ in} to find the amount of icing needed for Yuki's cake:25.14 \text{ in}^2 + 75.39 \text{ in}^2 + 150.8 \text{ in}^2 + 12.57 \text{ in}^2 + 37.7 \text{ in}^2 + 62.83 \text{ in}^2 = 364.43 \text{ in}^2 \\ 364.43 \text{ in}^2 \times \frac{1}{8} \text{ in} = 45.55 \text{ in}^3

Idea summary

We can calculate the surface area of any prism or cylinder using the formula

\displaystyle SA = 2B + Ph

\bm{B}

area of the base

\bm{P}

perimeter of the base

\bm{h}

perpendicular height between bases

Outcomes

A.SSE.A.1

Interpret expressions that represent a quantity in terms of its context.

A.SSE.A.1.B

Interpret complicated expressions by viewing one or more of their parts as a single entity.

A.CED.A.4

Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.

G.GMD.A.1

Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri's principle, and informal limit arguments.

G.GMD.A.3

Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.

11.02 Prisms and cylinders

Introduction

Ideas

Volume of non-right prisms and cylinders

Exploration

Examples

Example 1

Example 2

Example 3

Example 4

Surface area of prisms and cylinders

Exploration

Examples

Example 5

Example 6

Outcomes

A.SSE.A.1

A.SSE.A.1.B

A.CED.A.4

G.GMD.A.1

G.GMD.A.3

G.MG.A.1

G.MG.A.2

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