7. Modeling in Three Dimensions
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Chapter 3
7.
Modeling in Three Dimensions
This lesson covers the essentials of three-dimensional modeling, focusing on its practical applications. It teaches how to use geometric shapes like cylinders, cones, and spheres for calculating volumes and surface areas. These skills prove useful in various scenarios, such as determining the capacity of a juice glass at a fair or the water volume an aquarium can hold. The lesson even explores complex calculations like estimating the number of grains of sand a hand can hold. Overall, the aim is to equip individuals with the mathematical tools needed for making informed decisions in everyday situations.
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| | 8 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Modeling in Three Dimensions
Slide of 8
Many objects used in daily life can be modeled as three-dimensional figures. In this lesson, geometric figures will be used to describe some daily life objects.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Comparing Aquarium Volumes
Dominika wants to buy a new aquarium for her fish. She is interested in two types of aquariums that have equal linear measurements. 
Aquarium A is a right cylinder with a diameter of 10 feet and a height of 5 feet. Additionally, a right cone sits inside of it. Aquarium B is a hemisphere with a diameter of 10 feet. Help Dominika answer the following questions.
a What is the volume of Aquarium A? If necessary, round the answer to the nearest cubic foot.
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b For each aquarium, what is the area of the water’s surface when filled to a height of h feet? Write the answer in terms of h and π.
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c Use the answers found above to find the volume of the other aquarium.
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Example
Comparing Aquarium Volumes II
Recall how the formula for the volume of a sphere is proven. The same thought process used in the proof can be applied to solve the challenge.
Two types of aquariums attract the attention of Dominica: Aquarium A and Aquarium B.
a What is the volume of Aquarium A? Round the answer to the nearest cubic foot.
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b For each aquarium, what is the area of the water’s surface when filled to a height of h feet? Write the answer in terms of h and π.
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c Use the answers found above and find the volume of the other aquarium.
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Hint
a Subtract the volume of the cone from the volume of the cylinder.
b Examine how the cross-sections that are parallel to the bases change.
c What does the Cavalieri's Principle states?
Solution
a By subtracting the volume of the cone from the volume of the cylinder, the volume of the Aquarium A can be found.
The difference between V1 and V2 will give the number of cubic feet of water that Aquarium A can hold.
Aquarium A can hold about 262 cubic feet of water.
Aquarium AVA=CylinderV1−Cone∣V2
Recall the formulas for the volumes of a cylinder and a cone. | Volume of the Cylinder | Volume of the Cone | |
|---|---|---|
| Formula | V1=πr2h | V2=31πr2h |
Both the cone and the cylinder forming Aquarium A have a 10 foot diameter, and therefore both have a radius of 5 feet. With that in mind, substitute r=5 and h=5.
| Volume of the Cylinder | Volume of the Cone | |
|---|---|---|
| Formula | V1=πr2h | V2=31πr2h |
| Substitute Values | V1=π(5)25 | V2=31π(5)25 |
| Calculate | V1=125π | V2=3125π |
VA=V1−V2
SubstituteII
V1=125π, V2=3125π
VA=125π−3125π
▼
Evaluate right-hand side
NumberToFrac
a=33⋅a
VA=3375π−3125π
SubFrac
Subtract fractions
VA=3250π
UseCalc
Use a calculator
VA=261.799387…
RoundInt
Round to nearest integer
VA≈262
b Begin by examining the cross-sections of each aquarium. Then, write an expression for the area of the water’s surface at height h. 
With this in mind, consider the vertical and horizontal cross-sections of the aquarium.
As seen on the diagram, the vertical cross-sections of the water — the shaded triangles — are isosceles right triangles, and the horizontal cross-sections form two concentric circles. Using the horizontal cross-sections, the area of the water’s surface at height h will be found. To do so, subtract the area of the inner circle Ainner from the area of the outer circle Aouter.
When Aquarium A is filled to a height of h feet, the area of the water’s surface is 10πh−πh2 square feet. 
The diagram shows a right triangle ABC. Its hypotenuse and length of one of its legs is known in terms of h. Therefore, using the Pythagorean Theorem, BC can also be expressed in terms of h.
Now, considering the horizontal cross-sections of Aquarium B, BC will be the radius of the inner circle. 
In this instance, the area of the water’s surface at height h is the area of a circle with a radius of 10h–h2.
For Aquarium B, the area of the water’s surface at height h is the same as the other aquarium, 10πh−πh2 square feet.
Aquarium A
Since the cone inside the aquarium is a right cone, its vertex is directly above the center of its base. Furthermore, its height and radius have the same length. Therefore, an isosceles right triangle inside the cone can be formed, as indicated in the diagram.Aouter−Ainner
SubstituteValues
Substitute values
π(5)2−π(5−h)2
▼
Evaluate
CalcPow
Calculate power
π(25)−π(5−h)2
CommutativePropMult
Commutative Property of Multiplication
25π−π(5−h)2
ExpandNegPerfectSquare
(a−b)2=a2−2ab+b2
25π−π(25−10h+h2)
Distr
Distribute -π
25π−25π+10πh−πh2
SubTerm
Subtract term
10πh−πh2
Aquarium B
Examine the vertical cross-sections of the hemisphere.AB2+BC2=AC2
SubstituteII
AB=5−h, AC=5
(5−h)2+BC2=52
▼
Solve for BC
SubEqn
LHS−(5−h)2=RHS−(5−h)2
BC2=52−(5−h)2
CalcPow
Calculate power
BC2=25−(5−h)2
ExpandNegPerfectSquare
(a−b)2=a2−2ab+b2
BC2=25−(25−10h+h2)
Distr
Distribute -1
BC2=25−25+10h−h2
SubTerm
Subtract term
BC2=10h−h2
SqrtEqn
LHS=RHS
BC2=10h−h2
SqrtPowToNumber
a2=a
BC=10h−h2
c Recall what the Cavalieri Principle states.
According to the Cavalieri Principle, the volumes of water must be equal when both aquariums are filled. Therefore, no further calculations are necessary — Aquarium B will hold about 262 cubic feet of water.
|
Cavalieri Principle |
|
Two solids with the same height and the same cross-sectional area at every altitude have the same volume. |
Both aquariums have a height of 5 feet, and the area of the water’s surface when filled to a height of h feet is the same for each aquarium.
| Aquarium A | Aquarium B | |
|---|---|---|
| Height (ft) | 5 | |
| Cross-Sectional Area (ft2) | 10πh−πh2 | |
Example
Finding the Length of a Toilet Paper Roll
Tiffaniqua wants to calculate the length of the a toilet paper roll. Hey! It is on a great sale, Okay. She draws a diagram and denotes the thickness of the paper, the inner radius, and the outer radius by t, r, and R, respectively.
a Write an expression for the length L of the paper roll in terms of t, r, and R.
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b Find the length of the paper roll if t=0.03, r=2.5, and R=5.5, all measured in centimeters.
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Solution
a Consider the following horizontal cross-section of the toilet roll. The cross-section consists of concentric circles.
The area of the shaded region A can be calculated in two ways. It can be expressed as the area of the circle of radius R minus the area of the circle of radius r.
The area of the front face is the product of t and L. That product is also equal to A. Therefore, t⋅L can be substituted for A in the derived formula for the area of the shaded region.
Note that there might be several alternative ways to find the equation. Only one possible way was shown here.
A=πR2−πr2
Alternatively, it can be expressed as the area of the front face of the long thin rectangular prism, which is created when the paper is unrolled. b By substituting t=0.03, r=2.5, and R=5.5 into the previously derived equation, the length of the paper towel will be calculated.
The length of the paper is about 2513 centimeters.
Example
Cylindrical Soda Can
A cylindrical soda can is made of aluminum. It is 6 inches high and its bases have a radius of approximately 1.2 inches.
Give a go at answering the following set of questions. If necessary, round the answer to two decimal places.
a Find the surface area of the soda can.
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b The density of aluminum is approximately 2.7 grams per cubic centimeter. If the mass of the soda can is approximately 21 grams, how many cubic centimeters of aluminum does it contain?
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c Suppose that the thickness of the soda can is uniform throughout its body. Estimate the soda can's thickness.
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Hint
a Use the formula for the surface area of a cylinder.
b The density of a substance is equal to its mass divided by its volume.
c How might the surface area and volume of the soda can be related?
Solution
a Since the soda can is a cylinder, the formula for the surface area of a cylinder will be used.
S=2πrh+2πr2
Here, r and h are the radius and the height of the cylinder, respectively. The radius is 1.2 inches and the height is 6 inches. Substitute these values in the formula and evaluate it.
Therefore, the surface area of the soda can is about 54.29 square inches. b Recall that the density of a substance is equal to its mass divided by its volume. In other words, the volume of a substance is its mass divided by its density.
d=Vm⇔V=dm
Since the mass m of the aluminum can and the density d of aluminum are given, the volume can be calculated. Substitute m=21 and d=2.7 into the equation.
The soda can is made of about 7.78 cubic centimeters of aluminum. c To determine the thickness of the soda can, its surface area and volume are good to use.
Now, by substituting V=7.78 and S=352.885, t can be found.
This means that the thickness of the soda can is about 0.02 centimeters.
Surface Area54.29in2Volume7.78cm3
Let t denote the thickness. Then, the volume of the aluminum part of the soda can will be equal to the surface area of the soda can multiplied by its thickness. Therefore, to find t, the the aluminum part's volume should be divided by the soda can's surface area.
V=S⋅t⇔t=SV
However, since the amount of aluminum is found in cubic centimeters and the surface area of the soda can in square inches, a conversion factor must be used. Use in26.5cm2 to convert square inches to square centimeters.
54.29in2⋅in26.5cm2
▼
Simplify
MoveLeftFacToNum
a⋅cb=ca⋅b
in254.29in2⋅6.5cm2
CrossCommonFac
Cross out common factors
in254.29in2⋅6.5cm2
SimpQuot
Simplify quotient
54.29⋅6.5cm2
Multiply
Multiply
352.885cm2
Example
Choosing Type of Glass
By modeling real-life objects using geometric shapes, various characteristics of the objects can be determined. These characteristics can then be compared to make inferences which could impact real decisions to be made.
Emily is attending a fair and wants to sell 1.5 liters of homemade orange juice she is naming Oranjya Thirsty. She needs to decide the type of glass she will use to serve the juice — a cocktail glass or a Collins glass.
A cocktail glass is a type of glass that has an inverted cone bowl. The cone bowl's height is 5.8 centimeters and the radius of its base is 5.4 centimeters. A collins glass is a cylindrical glass with a height of 9 centimeters and a radius of 3.2 centimeters. Help Emily make a decision by answering the following questions.
a How many cocktail glasses of orange juice can she sell?
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b How many Collins glasses of orange juice can she sell?
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c If Emily chooses cocktail glasses, she will sell each for 6 dollars. If the Collins glasses are chosen, each will be sold for 10 dollars. Which glass type selection makes Emily more money?
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Hint
a Use the formula for the volume of a cone to determine how many liters of orange juice a cocktail glass can hold.
b Use the formula for the volume of a cylinder to determine how many liters of orange juice a Collins glass can hold.
c Use the answers found in the previous parts.
Solution
a First, the volume of the bowl of a cocktail glass will be calculated. Then, it will be used to find the number of glasses. The bowl of a cocktail glass can be modeled by a cone as shown.
The height of the cone bowl and the radius of its base are 5.8 and 5.4 centimeters, respectively. Substitute these values into the volume formula of a cone.
This means that a cocktail glass can hold approximately 177 cubic centimeters orange juice. Conversion needs to be made. The conversion factor 1000cm31L will convert cubic centimeters into liters.
V=31πr2h
SubstituteII
r=5.4, h=5.8
V=31π(5.4)25.8
▼
Evaluate right-hand side
CalcPow
Calculate power
V=31π(29.16)5.8
Multiply
Multiply
V=31169.128π
MoveRightFacToNumOne
b1⋅a=ba
V=3169.128π
UseCalc
Use a calculator
V=177.110427…
RoundInt
Round to nearest integer
V≈177
177cm3⋅1000cm31L=0.177L
Finally, the number of cocktail glasses can be calculated.
Therefore, 1.5 liters of orange juice fully fills 8 cocktail glasses. b Similarly, start by finding the volume of the cylindrical glass.
Its height is 9 centimeters and its base radius is 3.2 centimeters. Substitute these values into the formula.
This means that the volume of a collins glass is about 290 cubic centimeters. After converting its unit of measurement into liters, it is 0.29 liters. Now, the volume of a carton of orange juice can be divided by the volume of a collins glass.
0.291.5=5.172413…
Therefore, 1.5 liters of orange juice fully fills 5 collins glasses. c Emily can sell 8 cocktail glasses or 5 collins glasses. Knowing the prices, the revenue from each sale can be calculated.
As a result, Emily should choose the Collins glass, as the revenue from this selection is greater.
| Type of Glass | |
|---|---|
| Cocktail Glass | Collins Glass |
| 8⋅6 | 5⋅10 |
| $48 | $50 |
Example
Estimating How Many Grains of Sand a Hand Can Hold
With the help of geometric modeling, any number of objects can be approximated regardless of whether they are super large or tiny minuscule grains of sand.
Take, for example, Ramsha's situation. She is looking through photos from her trip to the beach to post on her social media page. A photo that shows her holding sand sparks her curiosity. She wonders how many individual grains of sand is she holding. Ramsha thinks she can model a grain of sand using a sphere. She then assumes that each grain has a diameter of 0.06 centimeters.
Ramsha figures she can hold 300 grams of sand in her hands. If the density of sand is approximately 1.52 grams per cubic centimeter, help Ramsha approximate the number of grains of sand in her hands. Write the answer in scientific notation.
Take, for example, Ramsha's situation. She is looking through photos from her trip to the beach to post on her social media page. A photo that shows her holding sand sparks her curiosity. She wonders how many individual grains of sand is she holding. Ramsha thinks she can model a grain of sand using a sphere. She then assumes that each grain has a diameter of 0.06 centimeters.
Answer
About 1.7×106
Hint
The formula for the volume of a sphere is V=34πr3, where r is the radius of the sphere.
Solution
To find the number N of grains of sand in Ramsha's hands, the mass M of the sand in her hands should be divided by the mass m of a grain.
The volume of a grain is about 1.1×10-4 cubic centimeters.
The number of grains of sand is approximately 1.7×106, or about 1.7 million.
N=mM
Recall that the density of a substance is equal to its mass divided by its volume. In other words, the mass of a substance is its density times its volume. d=Vm⇔m=d⋅V
Since the density of a grain of sand is given, the volume of a grain of sand will be calculated first. Finding the Volume of a Grain
The radius of a grain is 20.06=0.03 centimeters. Use the formula for the volume of a sphere to find the volume of a grain.V=34πr3
Substitute
r=0.03
V=34π(0.03)3
▼
Evaluate right-hand side
CalcPow
Calculate power
V=34π(0.000027)
CommutativePropMult
Commutative Property of Multiplication
V=340.000027π
MoveRightFacToNum
ca⋅b=ca⋅b
V=30.000108π
UseCalc
Use a calculator
V=0.000113…
RoundDec
Round to 5 decimal place(s)
V≈0.00011
Write in scientific notation
V≈1.1×10-4
Finding the Mass of a Grain
The density of a grain is 1.52g/cm3 and its volume is 1.1×10-4 cubic centimeters. By multiplying these values, the mass of a grain can be found.cm31.52 g⋅1.1×10-4 cm3
▼
Evaluate
MoveRightFacToNum
ca⋅b=ca⋅b
cm31.52 g⋅1.1×10-4 cm3
CancelCommonFac
Cancel out common factors
cm31.52 g⋅1.1×10-4 cm3
SimpQuot
Simplify quotient
1.52 g⋅1.1×10-4
CommutativePropMult
Commutative Property of Multiplication
1.52⋅1.1×10-4 g
Multiply
Multiply
1.672×10-4 g
RoundDec
Round to 1 decimal place(s)
1.7×10-4 g
Finding the Number of Grains
Finally, substitute the values into the formula mentioned at the beginning to calculate the number of grains of sand.N=mM
SubstituteII
M=300g, m=1.7×10-4g
N=1.7×10-4g300g
▼
Evaluate right-hand side
CrossCommonFac
Cross out common factors
N=1.7×10-4g300g
SimpQuot
Simplify quotient
N=1.7×10-4300
WriteProdFrac
Write as a product of fractions
N=1.7300(10-41)
FracToNegExponent
am1=a-m
N=1.7300(104)
CalcQuot
Calculate quotient
N=(176.470588…)(104)
Multiply
Multiply
N=1764705.882352…
RoundSigDig
Round to 2 significant digit(s)
N≈1700000
Write in scientific notation
N≈1.7×106
the total number of stars in the universe is greater than all the grains of sand on all the beaches of the planet Earth.
Closure
Modeling the Human Eye
Research projects usually require an interdisciplinary approach. That is, people from different disciplines work together to develop and test hypothesis, run experiments, and test theories.
Biologists, for example, can work with mathematicians to model a part of an organism. By doing so, researchers can predict how these parts function, grow, and change. For example, the human eye was able to be modeled as a sphere. Move the slider to rotate the eye.
Diego is making a cube out of 64 meters of copper thread which he is melting down. The thread is cylinder-shaped with a diameter of 16 millimeters. What will the cube's side s be in centimeters? Provide the amount in exact form.
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To determine the cube's side, we must first find its volume. This is the same as the volume of the copper thread. From the exercise, we know that the thread is cylinder-shaped with a diameter of 16 millimeters and a length, or height, of 64 meters.
Therefore, we must use the formula for calculating the volume of a cylinder. V=π r^2 h Notice that the radius and height have different units of measurement. Before calculating the volume of the thread, we must convert between units.
Converting Between Units
Since its easier to work with nondecimal numbers, we will convert the cylinder's height to millimeters. In one meter, we have 1000 millimeters. This gives us the following conversion factor. 1000 mm/1 m Now we can convert 64 meters to millimeters.
The height of the cylindrical thread is 64 000 millimeters.
Volume of the Thread
Since both dimensions of the thread is in the same unit, we can determine its volume. Notice that the thread's diameter is 16 millimeters, which means it has a radius of 8 millimeters.
The volume of the thread, and therefore the cube, is 4 096 000π cubic millimeters.
Side of the Cube
Recall that the volume of a cube is the cube of its side s. V=s^3 Now we can determine the side of the cube that Diego can form. To do so, we will substitute the volume of the thread into the above formula and solving for s.
The side of the cube is 160sqrt(π) millimeters, which can be converted to centimeters by multiplying the conversion factor of 1 cm10 mm. 160sqrt(π) mm * 1 cm/10 mm = 16sqrt(π) cm The cube that Diego can get at the end has sides of length 16sqrt(π) centimeters.
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Dominika is building a sauna in her backyard. The sauna's floor will be in the shape of a square and have a base area of 10 square meters. The sauna's walls will be 2 meters high.
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We know that the floor is a square with an area of 10 square meters. Therefore, we can calculate its side length by using the formula for the area of a square. A=s^2 Let's substitute the area of the floor into the equation and solve for s.
The sauna can be viewed as a composite solid, consisting of a rectangular prism and a triangular prism sitting on top of it.
From the given information, we know that the height of the sauna equals the length of sides of the sauna's base, which we calculated in the first section to be sqrt(10) meters. Since we also know the walls of the sauna are 2 meters high, we can determine the dimensions of the composite solid.
We can now find the volume of each prism, and thus the volume of the composite solid.
Volume of the Rectangular Prism
Recall that the volume of a rectangular prism is the product of its length l, width w, and height h. From the diagram, we know that l = sqrt(10), w = sqrt(10), and h= 2. Let's calculate the volume.
Volume of the Triangular Prism
The volume of the triangular prism is the product of its base area and the prism's height. V_(TP) = Bh From the diagram, we see that the triangle has a height of (sqrt(10)-2) meters and a base of sqrt(10) meters. Therefore, we can calculate the area of the base B. B = 1/2(sqrt(10))(sqrt(10)-2) We also see that the height of the prism is sqrt(10) meters. Now we can calculate the volume.
Volume of Sauna
Now we will add the volume of the the two prisms to determine the total volume of the Sauna.
The volume of the sauna will be about 26 cubic meters.
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A shop is selling cone-shaped cartons filled with cherry juice. Tiffaniqua makes a hole at the bottom of the cone and pours the juice it into a cylinder-shaped container. The cone's inner radius is 90% of the cylinder's inner radius. The cone's height is 80% of the cylinders height.
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From the exercise, we know that a shop sells cherry juice in cone-shaped cartons. Therefore, we should start by finding an expression for the volume of the cone.
Volume of the Cone
Let's recall the formula for the volume of a cone. V=1/3π r^2 h We have not been given the radius or the height of the cone. However, we do know its radius and height in relation to the cylinder. If we call the cylinder's height and radius H and R, and the cone's height and radius h and r, we can write the following equations. 90 % ofR&=r ⇔ 0.9R=r 80 % ofH&=h ⇔ 0.8H=h Now, we can write an expression for the cone's volume.
This expression shows the volume of the cone in terms of the cylinder's height H and radius R.
Volume of the Cylinder
The volume of a cylinder is calculated by using the following formula. V=π r^2 h Remember that we defined the radius and diameter of the cylinder as R and H. This means we can express the volume of the cylinder in the following way V=π R^2 H
Calculating the Percentage
We have written expressions for the volume of the cherry juice and the volume of the cylinder. We can now divide the volume of the cherry juice by the volume of the cylinder to determine what percentage of the cylinder is filled.
About 22 % of the cylinder-shaped container is filled with the cherry juice.
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Modeling in Three Dimensions
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