2. Properties of Cylinders
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Chapter 3
2.
Properties of Cylinders
This lesson focuses on the geometric aspects of cylinders, covering topics like volume, surface area, and the role of Cavalieri's Principle. Practical examples include determining the volume of a test tube for a chemistry experiment or calculating the surface area of a cylindrical object like a can. The lesson also discusses the importance of understanding the circumference of a circle when dealing with cylinders. For instance, knowing the circumference can help in calculating the surface area accurately. The Cavalieri's Principle is highlighted as a fundamental concept that applies to cylinders, stating that two solids with the same height and cross-sectional area will have the same volume. This knowledge is particularly useful in fields like engineering, architecture, and science experiments.
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| | 14 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Properties of Cylinders
Slide of 14
Challenge
Investigating the Volume of Different Cylinders
Consider the three-dimensional figure given in the diagram. By dragging the point, the figure can be skewed. However, its height and the radius of the circles that make the base and the top of the cylinder remain the same. Think about the volume — space occupied — of the solid. Does the volume change when the solid is skewed?
Discussion
Definition of a Circle
Before moving on, the definition of a circle and some of its parts will be reviewed.
A circle is the set of all the points in a plane that are equidistant from a given point. There are a few particularly notable features of a circle.
- Center - The given point from which all points of the circle are equidistant. Circles are often named by their center point.
- Radius - A segment that connects the center and any point on the circle. Its length is usually represented algebraically by r.
- Diameter - A segment whose endpoints are on the circle and that passes through the center. Its length is usually represented algebraically by d.
- Circumference - The perimeter of a circle, usually represented algebraically by C.
circle O,since it is centered at O.
Next, the formula for calculating the circumference of a circle will be discussed.
Rule
Circumference of a Circle
The circumference of a circle is calculated by multiplying its diameter by π.
C=πd
Since the diameter is twice the radius, the circumference of a circle can also be calculated by multiplying 2r by π.
C=2πr
Proof
Consider two circles and their respective diameters and circumferences.
CACB=dAdB
This proportion can be rearranged.
CACB=dAdB
▼
Rearrange equation
MultEqn
LHS⋅CA=RHS⋅CA
CB=dAdB(CA)
DivEqn
LHS/dB=RHS/dB
dBCB=dA1(CA)
MoveRightFacToNumOne
b1⋅a=ba
dBCB=dACA
dC=π⇒C=πd
Finally, the formula for calculating the area of a circle will be seen.
Rule
Area of a Circle
The area of a circle is the product of π and the square of its radius.
Proof
Informal JustificationA circle with radius r will be divided into a number of equally sized sectors. Then, the top and bottom halves of the circle will be distinguished by filling them with different colors. Because the circumference of a circle is 2πr, the arc length of each semicircle is half this value, πr.
Now, the above sectors will be unfolded. By placing the sectors of the upper hemisphere as teeth pointing downwards and the sectors of the bottom hemisphere as teeth pointing upwards, a parallelogram-like figure can be formed. As such, the area of the figure below should be the same as the circle's area.
A=πr⋅r⇔A=πr2
It has been shown that the area of a circle is the product of π and the square of its radius.
Example
Finding the Circumference and Area of a Circle
Izabella loves to use geometry in her art. Her school noticed her skills and hired her to paint the school's soccer field — for a substantial payment. The diagram shows how the field should look.
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Hint
The circumference is twice the product of π and the radius. The area is the product of π and the square of the radius.
Solution
The circumference of a circle with radius r is twice the product of π and r. Furthermore, its area is the product of π and r2. Since the radius is 6.5 meters, the circumference and area can be calculated.
| Formula | Substitute | Simplify | Approximate | |
|---|---|---|---|---|
| Circumference | C=2πr | C=2π(6.5) | C=13π | C≈40.8 |
| Area | A=πr2 | A=π(6.52) | A=42.25π | A≈132.7 |
The circumference and area of the circle located at the center of the soccer field are about 40.8 meters and 132.7 square meters, respectively.
Example
Investigating the Area of a Circle
Maya bought 20 meters of fencing with the hopes of constructing a circular dog run for her dog Opie. Because Opie is a large Saint Bernard, she wants the area of the dog run to be at least 30 square meters.
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Hint
Start by finding the radius of the circle.
Solution
Because the dog run is in the shape of a circle, its area is the product of π and the square of the radius. Therefore, to find the area of the circle, the first step is to find the radius. Since it is already known that the circumference is 20 meters, the formula for the circumference will be used to find the radius.
The radius of Opie's circular dog run is about 3.18 meters. 
With this information, the area of the dog run can be calculated.
The area of Opie's dog run is about 31.8 square meters. This is enough for him to have happy and healthy playtime!
C=2πr
Since Maya bought and used 20 meters of fencing, the circumference measures 20 meters. This value can be substituted in the above formula to find the radius r.
C=2πr
Substitute
C=20
20=2πr
▼
Solve for r
DivEqn
LHS/2=RHS/2
10=πr
DivEqn
LHS/π=RHS/π
π10=r
RearrangeEqn
Rearrange equation
r=π10
UseCalc
Use a calculator
r=3.183098…
RoundSigDig
Round to 3 significant digit(s)
r≈3.18
A=πr2
To do so, 3.18 will be substituted for r in the formula for the area of a circle.
A=πr2
Substitute
r=3.18
A=π(3.182)
▼
Evaluate right-hand side
CalcPow
Calculate power
A=π(10.1124)
Multiply
Multiply
A=31.769041
RoundSigDig
Round to 3 significant digit(s)
A≈31.8
Pop Quiz
Practice Finding the Circumference and Area of a Circle
For the following questions, approximate the answers to one decimal place. Do not include units in the answer.
Discussion
Cavalieri's Principle
Now, two-dimensional figures will be left aside to move forward into solids, which are three-dimensional shapes.
Two solids with the same height and the same cross-sectional area at every altitude have the same volume. This means that, as long as their heights are equal, skewed versions of the same solid have the same volume.
If V1 and V2 are the volumes of the above solids, then they are equal.
V1=V2
Proof
Informal JustificationThis principle will be proven by using a set of identical coins. Consider a stack in which each of these coins is placed directly on top of each other. Consider also another stack where the coins lie on top of each other, but are not aligned.
The first stack can be considered as a right cylinder. Similarly, the second stack can be considered as an oblique cylinder, which is a skewed version of the first cylinder. Because the coins are identical, the cross-sectional areas of the cylinders at the same altitude are the same.
Since the coins are identical, they have the same volume. Furthermore, since the height is the same for both stacks, they both have the same number of coins. Therefore, both stacks — cylinders — have the same volume. This reasoning is strongly based on the assumption that the face of the coins have the same area.
Discussion
Definition of a Cylinder
A cylinder is a three-dimensional figure that has two circular bases that are parallel and equal in size, connected by a curved surface.
Rule
Volume of a Cylinder
The volume of a cylinder is calculated by multiplying the area of its base by its height.
If the radius of the base of a cylinder is r, the volume can be calculated by the following formula.
V=πr2h
Proof
Informal JustificationConsider a rectangular prism and a right cylinder that have the same base area and height.
In this case, the cross-sections of the prism and the cylinder are congruent to their bases. Therefore, their cross-sectional areas at every altitude are equal.
VC=VP
Furthermore, the volume of a prism can be calculated by multiplying the area of its base by its height.
VP=Bh
By the Transitive Property of Equality, a formula for the volume of the cylinder can be written.
{VC=VPVP=Bh⇒VC=Bh
Finally, not only is B the area of the base of the prism, but also the area of the base of the cylinder. Since the base of the cylinder is a circle, its area is the product of π and the square of its radius r. Therefore, B=πr2 can be substituted in the above formula.
VC=BhsubstituteVC=πr2h
This formula applies to all cylinders because there is always a prism with the same base area and height. Also, by the Cavalieri's principle, this formula still holds true for oblique cylinders.
Example
Finding the Volume of a Cylinder
LaShay loves golfing. She is about to buy a new golf bag.
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Hint
The volume of a cylinder is the product of π, the square of its radius, and its height.
Solution
The golf bag is in the shape of a cylinder. Therefore, to calculate its volume, the formula for the volume of a cylinder can be used.
V=πr2h
SubstituteII
h=132, r=15
V=π(152)(132)
▼
Evaluate right-hand side
CalcPow
Calculate power
V=π(225)(132)
Multiply
Multiply
V=29700π
UseCalc
Use a calculator
V=93305.30181…
RoundSigDig
Round to 3 significant digit(s)
V≈93300
Example
Finding the Radius of a Cylinder
Kriz is making an experiment to complete a Chemistry project. He is using a test tube in the shape of a cylinder with a height of 75 millimeters and a volume of 20000 cube millimeters.
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Solution
The volume of a cylinder is the product of π, the square of the base's radius, and the height of the cylinder.
When solving the equation, only the principal root was considered. This is because the radius of a circle is always positive. Therefore, the radius of the cylinder's base is about 9.2 millimeters. Since the radius is greater than 9 millimeters, the tube does not suit the experiment.
V=πr2h
It is known that the height and the volume of the cylinder are 75 millimeters and 20000 cube millimeters, respectively. These two values can be substituted into the formula for volume of a cylinder, which can then be solved for r.
V=πr2h
SubstituteII
h=75, V=20000
20000=πr2(75)
▼
Solve for r
CommutativePropMult
Commutative Property of Multiplication
20000=75πr2
DivEqn
LHS/75π=RHS/75π
75π20000=r2
SqrtEqn
LHS=RHS
75π20000=r
RearrangeEqn
Rearrange equation
r=75π20000
UseCalc
Use a calculator
r=9.213177…
RoundDec
Round to 1 decimal place(s)
r≈9.2
Discussion
Surface Area of a Cylinder
Besides the radius, height, and volume, another essential characteristic of a cylinder is its surface area. The surface area of a solid is the measure of the total area that the surface of the solid occupies.
Consider a cylinder of height h and radius r.
The surface area of this cylinder is given by the following formula.
S=2πrh+2πr2
Proof
The cylinder's surface area can be seen as three separate parts that are top, bottom, and side. The area of the side has the shape of a rectangle of width h. The top and the bottom are circles of radius r.
Area ofOne CircleA=πr2Area ofTwo CirclesA=2πr2
To find the area of the rectangle, its length must be found first. The length of the rectangle that forms the lateral part of the cylinder is the circumference of the base, which is 2πr. 
Area of the RectangleA=2πrh
The surface area of the cylinder S is the sum of the areas of the rectangle and the circles. S=2πrh+2πr2
Example
Finding the Surface Area of a Cylinder
Jordan's father is giving her a new baseball bat for her birthday. To avoid damage to the bat, Papa Jordan will store the bat in a box that has the shape of a cylinder with a height of 30 inches and a radius of 1.6 inches. Since this is a present for Jordan, Papa will use wrapping paper to make it even more special.
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Hint
The area of wrapping paper Jordan's father needs is the same as the surface area of the cylinder.
Solution
To find the minimum area of paper that Jordan's father needs, the surface area S of the cylinder needs to be calculated.
S=2πrh+2πr2
Here, r and h are the radius and the height of the cylinder, respectively. The radius is 1.6 inches and the height is 30 inches. Therefore, these values can be substituted in the above formula, which can then be simplified.
Pop Quiz
Practice Finding the Surface Area and Volume of a Cylinder
Find the volume or surface area of the cylinder.
Closure
Volume of Skewed Solids
The challenge presented at the beginning of this lesson asked whether the volume of a cylinder changes if the solid is skewed.
The Cavalieri's Principle was studied in this lesson. This principle states that two solids with the same height and the same cross-sectional area at every altitude have the same volume. Therefore, if their heights are equal, skewed versions of the same solid have the same volume. This means that the volume of a cylinder does not change even if the cylinder is skewed.
Properties of Cylinders
Exercises
A can of crushed tomatoes has the shape of a cylinder with a bottom and a lid. The can has a radius of x and a height that is k times the radius.
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The surface area of a cylinder is calculated by adding the areas of the two bases, which are circles, and the lateral surface which is a rolled up rectangle whose length equals the circumference of the base.
Let's substitute the height and radius with the given dimensions.
To solve for x, we must perform inverse operations until this variable is isolated. Since x is the radius of the cylinder, it cannot be a negative. Therefore, we will disregard any negative solutions.
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A refrigerator is filled with pennies. Its dimensions are shown in the diagram. Assume that the dimensions of the refrigerator are the same as its inner measures.
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To estimate the value of the pennies that can be placed in the refrigerator, we will determine how many of them we can fit in the refrigerator. To do that we will divide the volume of the refrigerator by the volume of a penny.
Volume of the Refrigerator
The refrigerator is a rectangular prism with a square base area. To calculate its volume, we must calculate the base area and multiply by the height. We know that the base, depth, and height are b= 28, d= 28 and h= 70, respectively.
The volume of the refrigerator is 54 880 cubic inches.
Volume of a Penny
Since a penny has the shape of a cylinder, to calculate the volume of a penny we can use the formula for the volume of a cylinder. The diameter of a penny is 0.75 inches, which means that its radius is 0.75÷ 2=0.375 inches. We also know that the height of a penny is 0.05 inches.
Let's keep the volume of a penny expressed in its exact form for now.
Number of Pennies
To determine the number of pennies we can fit in the refrigerator, we divide the volume of the refrigerator by the volume of a single penny.
We can fit about 2 484 458 pennies in the refrigerator
Monetary Value
Finally, we will determine the monetary value of the pennies by multiplying the number of pennies by the value of a penny, which is $0.01. We will round the answer to the nearest thousand. 2 484 458(0.01)&= $24 844.58 & ≈ $25 000 Note that our calculation assumed that we can fill every cubic inch of the refrigerator with pennies. In practice this is not possible — since we are placing circular objects inside a square space, there will be some empty space between the pennies. Therefore, the most appropriate interval is $18 000 - $26 000.
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Tadeo bought a cylindrical paint bucket with a diameter of 12 inches and a height of 15 inches. If 231 cubic inches contain about 1 gallon of paint, how many gallons does the bucket hold? Round the answer to one decimal place.
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To determine the number of gallons the bucket holds, we must first find its volume. Since it has the form of a cylinder, we will use the formula for the volume of a cylinder. V=π r^2 h Since the diameter is 12 inches, we know that the radius is 12÷ 2= 6 inches. We also know that the height is 15 inches. Let's substitute these values in the above formula!
The volume of the paint that the bucket holds is 540π cubic inches. Since 231 cubic inches contain about 1 gallon of paint, the conversion factor that converts cubic inches to gallons is the following. 1 gallon/231 in^3 If we multiply the volume by this conversion factor, we can determine how many gallons the buckets holds.
The bucket can hold about 7.3 gallons of paint.
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In the following cylinder, the height is twice the radius.
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We know that the height h of the cylinder is twice the radius r of the base. Let's use this information to write an equation connecting h and r. h=2r Let's now recall the formula for the surface area of a cylinder, and substitute 54π for SA. SA=2 π r h+ 2 π r^2 ⇓ 54 π = 2 π r h+ 2 π r^2 By combining these equations, we obtain a system of equations. h=2r & (I) 54 π = 2 π rh + 2 π r^2 & (II) Since h is already isolated in Equation (I), it looks like the most convenient method to solve this system is the Substitution Method. Keep in mind that, since r is the radius of a cylinder, it must be positive.
The height of the cylinder is 6 feet.
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Properties of Cylinders
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