2. Properties of Cylinders
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Chapter 3
2.
Properties of Cylinders
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Why
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. 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.
Lesson Settings & Tools
| | 14 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Properties of Cylinders
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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?
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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.
The following circle can be referred to as ⊙ O, or 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=π d
This can be visualized in the following diagram.
Since the diameter is twice the radius, the circumference of a circle can also be calculated by multiplying 2r by π.
C=2π r
Proof
Consider two circles and their respective diameters and circumferences.
By the Similar Circles Theorem, all circles are similar. Therefore, their corresponding parts are proportional. C_B/C_A=d_B/d_A This proportion can be rearranged.
C_B/C_A=d_B/d_A
▼
Rearrange equation
MultEqn
LHS * C_A=RHS* C_A
C_B=d_B/d_A(C_A)
DivEqn
.LHS /d_B.=.RHS /d_B.
C_B/d_B=1/d_A(C_A)
MoveRightFacToNumOne
1/b* a = a/b
C_B/d_B=C_A/d_A
Therefore, for any two circles, the ratio of the circumference to the diameter is always the same. This means that this ratio is constant. This constant is defined as π. With this information, it can be shown that the circumference of a circle is the product between its diameter and π.
C/d=π ⇒ C=π d
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 2 π r, the arc length of each semicircle is half this value, π r.
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.
It can be noted that if the circle is divided into more and smaller sectors, then the figure will begin to look more and more like a rectangle.
Here, the shorter sides become more vertical and the longer sides become more horizontal. If the circle is divided into infinitely many sectors, the figure will become a perfect rectangle with base π r and height r. Since the area of a rectangle is the product of its height and its base, the following formula can be derived.
A = π r * r ⇔ A= π r^2
It has been shown that the area of a circle is the product of π and the square of its radius.
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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.
Izabella needs the field's measurements. She already knows the radius of the circle located at the center of the field to be 6.5 meters. For logistical reasons, she wants to find the circumference and area of this circle. Help Izabella calculate these values to one decimal place.
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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 r^2. Since the radius is 6.5 meters, the circumference and area can be calculated.
| Formula | Substitute | Simplify | Approximate | |
|---|---|---|---|---|
| Circumference | C=2π r | C=2π ( 6.5) | C=13π | C≈ 40.8 |
| Area | A=π r^2 | A=π ( 6.5^2) | 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.
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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.
With a circumference of 20 meters, is the area of the dog run 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.
C=2π r
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=2π r
Substitute
C= 20
20=2π r
▼
Solve for r
DivEqn
.LHS /2.=.RHS /2.
10=π r
DivEqn
.LHS /π.=.RHS /π.
10/π=r
RearrangeEqn
Rearrange equation
r=10/π
UseCalc
Use a calculator
r=3.183098...
RoundSigDig
Round to 3 significant digit(s)
r≈ 3.18
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. A=π r^2 To do so, 3.18 will be substituted for r in the formula for the area of a circle.
A=π r^2
Substitute
r= 3.18
A=π ( 3.18^2)
▼
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
The area of Opie's dog run is about 31.8 square meters. This is enough for him to have happy and healthy playtime!
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Pop Quiz
Practice Finding the Circumference and Area of a Circle
For the following questions, round the answers to one decimal place. Do not include units in the answer.
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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.
V_1 = V_2
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.
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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.
The axis of a cylinder is the segment that connects the center of the bases. The height of a cylinder is the perpendicular distance between the bases. The radius of the cylinder is the radius of one of the bases.
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=π r^2h
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.
By the Cavalieri's Principle, two solids with the same height and the same cross-sectional area at every altitude have the same volume. Therefore, the volume of the cylinder is the same as the volume of the prism. V_C=V_P Furthermore, the volume of a prism can be calculated by multiplying the area of its base by its height. V_P=Bh By the Transitive Property of Equality, a formula for the volume of the cylinder can be written. V_C= V_P V_P=Bh ⇒ V_C=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=π r^2 can be substituted in the above formula. V_C=Bh substitute V_C= π r^2 h 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.
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Example
Finding the Volume of a Cylinder
LaShay loves golfing. She is about to buy a new golf bag.
The the bag she is thinking about buying is a shaped like a cylinder with a height of 132 centimeters and its radius is 15 centimeters. For all of her golf clubs to fit in the bag, its volume must be at least 90 000 cubic centimeters. Therefore, she wants to calculate the volume of the bag. Help her do this, approximating the answer to three significant figures.
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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.
It is known that h= 132 and r= 15. These two values can be substituted into the formula.
V=π r^2h
SubstituteII
h= 132, r= 15
V=π ( 15^2)( 132)
▼
Evaluate right-hand side
CalcPow
Calculate power
V=π (225)(132)
Multiply
Multiply
V=29 700π
UseCalc
Use a calculator
V=93 305.30181...
RoundSigDig
Round to 3 significant digit(s)
V≈ 93 300
The volume of LaShay new golf bag is about 93 300 cubic centimeters. This is enough to keep all of her golf clubs!
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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 20 000 cube millimeters.
For this tube to suit the experiment, it radius must not be greater than 9 millimeters. Help Kriz find the radius! Approximate the answer to one decimal place.
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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.
V=π r^2 h
It is known that the height and the volume of the cylinder are 75 millimeters and 20 000 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=π r^2 h
SubstituteII
h= 75, V= 20 000
20 000=π r^2 ( 75)
▼
Solve for r
CommutativePropMult
Commutative Property of Multiplication
20 000=75π r^2
DivEqn
.LHS /75π.=.RHS /75π.
20 000/75π=r^2
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(20 000/75π)=r
RearrangeEqn
Rearrange equation
r=sqrt(20 000/75π)
UseCalc
Use a calculator
r=9.213177...
RoundDec
Round to 1 decimal place(s)
r≈ 9.2
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.
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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=2π rh+2π r^2
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.
Since the top and the bottom are congruent circles, they have the same area. cc Area of & Area of One Circle & Two Circles [0.8em] A=π r^2 & A=2π r^2 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 2π r.
Therefore, the area of the rectangle is the product of h and 2π r. Area of the Rectangle A=2π rh The surface area of the cylinder S is the sum of the areas of the rectangle and the circles.
S=2π rh+2π r^2
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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.
Ignoring any paper overlapping, what is the minimum area of paper that Jordan's father needs? Approximate the answer to one decimal place.
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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=2π rh+2π r^2
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.
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Pop Quiz
Practice Finding the Surface Area and Volume of a Cylinder
Find the volume or surface area of the cylinder.
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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.
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Properties of Cylinders
Exercise 3.1
A cylinder is inscribed in a cone. The maximum volume of the cylinder is obtained when its radius is 4 centimeters. Calculate this volume. Write the answer in exact form.
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We know that a radius of 4 maximizes the cylinder's volume. Let's substitute this value into the formula for the volume of a cylinder. V = π ( 4)^2 h ⇓ V = 16π h To calculate the maximum volume of the cylinder, we must also know its height. To do that, we can use the fact that the height is related to the radius. As the radius decreases, the height increases, and vice versa.
To understand how these variables are related, we will draw a cross-section of the cone for a radius of 4 centimeters.
Examining the diagram, we can identify two right triangles, one triangle with legs of 12 and 6 centimeters and another triangle with legs of 2 and h centimeters. Note that 12 is the height of the cone and h the height of the cylinder.
By the Angle-Angle Similarity Theorem, these two triangles are similar. Therefore, we can write a proportion. h/12=2/6 Let's solve this equation for h.
We found that the height of the cylinder h is 4 centimeters. With this information, we can finally the volume of the cylinder. V=16π h ⇒ V = 16π( 4) ⇒ V = 64π
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Solution
Dylan drills a circular hole with a radius r through the middle of the base of a cylinder with radius R. He has measured the volume of the hole to be half the volume of the original cylinder. Write r in terms of R.
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Let's start by making a sketch that describes the situation.
Let's write equations that describe the volume of the original cylinder and the volume of the hole. Cylinder:& V_C= π R^2h Hole:& V_H= π r^2h We know that the volume of the hole is half the volume of the original cylinder. In other words, if we multiply V_H by 2, then this product should equal V_C. With this information, we can obtain the following equation. V_C=2 V_H Let's substitute the expressions for the different volumes in this equation, and solve for r.
Since the radius of a cylinder is a length and lengths cannot be negative, we do not consider any negative values for r. Therefore, the radius of the hole is r= Rsqrt(2).
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Properties of Cylinders
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