UNIT 3 • CIRCLES AND VOLUME
Lesson 5: Explaining and Applying Area and Volume Formulas
Instruction
Prerequisite Skills
This lesson requires the use of the following skills:
•
calculating with fractions and decimals
•
understanding operations with exponents
•
knowing area, surface area, and some volume formulas
•
knowing what an inscribed object looks like
•
understanding radius and diameter
Introduction
There are many applications for solid geometry in the real world. In this lesson, you will apply the
formulas for the volumes of solids to solve real-world problems.
Formulas are derived in many ways, and you will learn to construct arguments for the derivations
of the volumes of solids. Specifically, you will use Cavalieri’s Principle to construct valid arguments
for the volume of a sphere as well as for other solids. You will derive as well as compare the volumes
of different solids. You will also apply the formulas for the volumes of prisms, cylinders, cones,
pyramids, and spheres to solve real-world problems.
Key Concepts
•
sphere is a three-dimensional surface that has all its points the same distance from its
A
center.
•
he volume of a sphere can be derived in several ways, including by using Cavalieri’s Principle
T
and a limit process, as shown in Example 5.
4
The formula for the volume of a sphere is V = π r 3 .
3
Many application problems can be solved using volume formulas.
•
•
•
ou can make valid comparisons of volume formulas, which can be useful in developing other
Y
volume formulas.
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CCGPS Analytic Geometry Teacher Resource
UNIT 3 • CIRCLES AND VOLUME
Lesson 5: Explaining and Applying Area and Volume Formulas
Instruction
Common Errors/Misconceptions
•
c onfusing formulas; for example, confusing volume of a sphere with surface area of
a sphere
•
performing miscalculations with the number π
•
making arithmetic errors when dealing with the formulas, especially with fractions
•
u sing incorrect notation in word problems, such as using cm2 instead of cm3 for a volume
problem
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UNIT 3 • CIRCLES AND VOLUME
Lesson 5: Explaining and Applying Area and Volume Formulas
Instruction
Guided Practice 3.5.3
Example 1
Weston has two round balloons. One balloon has a radius that is 3 times the radius of the other
balloon. How much more air will the larger balloon need than the smaller balloon?
1. Use the formula for the volume of a sphere for the smaller balloon.
4
V = πr3
3
2. F or the larger balloon, since the radius is 3 times larger, use 3r instead
of r in the volume formula.
4
V = π ( 3r )3
3
3. Simplify the formula for the volume of the larger balloon.
4
V = π ( 3r )3
3
Formula for the volume of the larger balloon
4
V = π ( 27r 3 )
3
Distribute the exponent.
V = 12π r 3
Multiply the coefficients.
4. C
ompare the coefficients of both formulas (for the smaller balloon and
for the larger balloon).
4
The coefficients are 12 and .
3
Divide to see how many times larger the volume of the larger balloon is.
 4
12 ÷   = 9
 3
The larger balloon will need 9 times as much air as the
smaller balloon.
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CCGPS Analytic Geometry Teacher Resource
UNIT 3 • CIRCLES AND VOLUME
Lesson 5: Explaining and Applying Area and Volume Formulas
Instruction
Example 2
A teenager buying some chewing gum is comparing packages of gum in order to get the most gum
possible. Each package costs the same amount. Package 1 has 20 pieces of gum shaped like spheres.
Each piece has a radius of 5 mm. Package 2 has 5 pieces of gum shaped like spheres. Each piece has a
radius of 10 mm. Which package should the teenager buy? Round to the nearest millimeter.
1. F ind the volume of a piece of gum in package 1 by using the volume
formula for a sphere.
4
V = πr3
3
Formula for volume of a sphere
4
V = π ( 5) 3
3
Substitute 5 for r.
V ≈ 524 mm3
2. M
ultiply the volume of the single piece of gum in package 1 by 20 to get
the total volume of the gum in the package.
(524)(20) = 10,480 mm3
3. T
he radius of each piece of gum in package 2 is 10 mm. Find the
volume of a piece of gum in package 2 by using the volume formula for
a sphere.
4
V = πr3
3
Formula for volume of a sphere
4
V = π (10)3
3
Substitute 10 for r.
V ≈ 4189 mm3
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UNIT 3 • CIRCLES AND VOLUME
Lesson 5: Explaining and Applying Area and Volume Formulas
Instruction
4. M
ultiply the volume of a single piece of gum in package 2 by 5 to get
the total volume of gum in the package.
(4189)(5) = 20,945 mm3
5. C
ompare the volumes of the two packages to determine the best
purchase.
10,480 < 20,945
The teenager should buy package 2. Example 3
Pictured below is a cylindrical grain silo. It can be completely filled to the top of the dome. The dome
is in the shape of a hemisphere. The height of the silo is 300 feet to the top of the dome and the radius
of the dome is 50 feet. How much grain can fit in the silo? Round to the nearest cubic foot.
50 feet
300 feet
1. F ind the height of the cylinder by subtracting the radius of the
hemisphere (which is also the same as the height of the hemisphere)
from the total height.
300 – 50 = 250 feet
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UNIT 3 • CIRCLES AND VOLUME
Lesson 5: Explaining and Applying Area and Volume Formulas
Instruction
2. C
alculate the volume of the cylinder using the formula V = π r 2 h .
Substitute 50 for r and 250 for h.
V = π(50)2 • (250) = 1,962,500 ft3
3. C
alculate the volume of the hemisphere, which is half the volume of
a sphere.
1 4

V =  πr3 
2 3

Formula for volume of a hemisphere
2
V = πr3
3
Simplify.
2
V = π ( 50)3
3
Substitute 50 for r.
V = 261,667 ft3
4. A
dd the volume of the cylinder to the volume of the hemisphere to find
the total volume of the grain silo.
1,962,500 + 261,667 = 2,224,167
The total volume of the grain silo is 2,224,167 ft3.
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UNIT 3 • CIRCLES AND VOLUME
Lesson 5: Explaining and Applying Area and Volume Formulas
Instruction
Example 4
Nisha is making two terrariums for science class. One is in the shape of a sphere and has a radius of
8 inches. The other is in the shape of a hemisphere and has a radius that is 1.5 times the radius of the
sphere, or 12 inches. Which terrarium can hold the most plant life? How can Nisha prove which one is
bigger without using the specific radius measurements? Determine a method and apply it.
1. Determine the volume of the sphere-shaped terrarium.
4
V = πr3
3
Formula for the volume of a sphere
4
V = π ( 8) 3
3
Substitute 8 for the radius.
V=
2048
3
π or ≈ 2145 in3
2. Determine the volume of the hemisphere-shaped terrarium.
A hemisphere is half a sphere, so divide the volume formula for a sphere
by 2 to determine the formula for finding the volume of a hemisphere.
4
V = πr3
3
4

V =  πr3  ÷ 2
3

2
V= π r3
3
Formula for the volume of a sphere
Divide by 2 to find the volume of the
hemisphere.
Simplify.
Use this formula to find the volume of the hemisphere.
2
V = πr3
3
Formula for the volume of a
hemisphere
2
V= π (12)3
3
Substitute 12 for the radius.
V = 1152π or ≈ 3619.11 in3
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CCGPS Analytic Geometry Teacher Resource
UNIT 3 • CIRCLES AND VOLUME
Lesson 5: Explaining and Applying Area and Volume Formulas
Instruction
3. C
ompare the volumes of the two terrariums to determine which will
hold more plant life.
2145 < 3619.11
The terrarium shaped like a hemisphere will hold more plant life.
4. D
etermine a method to prove which terrarium is bigger without using
specific radius measurements.
Nisha can prove this by using the formula for a sphere and comparing
it to the formula for a hemisphere with a radius of 1.5r. Calculate the
formula for the volume of this hemisphere.
2
V = π (1.5r )3
3
Set up the equation.
2
V = π ( 3.375r 3 )
3
Cube the term.
V=
6.75
3
πr3
Simplify.
5. Compare the two volume formulas.
4
The formula for the sphere-shaped terrarium is V = π r 3 .
3
6.75 3
πr .
The formula for the hemisphere-shaped terrarium is V =
3
Since the coefficient of the hemisphere is larger than
the coefficient of the sphere, the volume of the hemisphere
is larger. U3-254
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UNIT 3 • CIRCLES AND VOLUME
Lesson 5: Explaining and Applying Area and Volume Formulas
Instruction
Example 5
Gloria has a puzzle in the shape of a sphere similar to the diagram below. She wants to find the
volume of the spherical puzzle but does not know the formula. Each piece of the puzzle is in the shape
of a pyramid. Explain how Gloria can derive the formula for the volume of a sphere using what she
knows about this puzzle.
1
1. Recall that the formula for the volume of a pyramid is V = Bh ,
3
where B is the area of the base of the pyramid, and h is the height.
2. Add up all the volumes of the pyramids to get the volume of the sphere.
1
1
1
1
V = B1 h1 + B2 h2 + B3 h3 + ... + Bn hn , where each subscript represents
3
3
3
3
the number of the pyramid.
3. Replace the h in the formula with the radius.
Since the height extends from the outer edge of the sphere to the
center, the height is also the radius of the sphere. Substitute r for h in
the formula you determined in step 2.
1
1
1
1
V = B1 r + B2 r + B3r + ... + Bn r
3
3
3
3
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UNIT 3 • CIRCLES AND VOLUME
Lesson 5: Explaining and Applying Area and Volume Formulas
Instruction
4. Use the reverse of distribution to factor the formula in step 3.
1
Factor out the common terms, and r.
3
1
V = r ( B1 + B2 + B3 + ... + Bn )
3
1
Notice that if you distributed r over the parentheses, you would end
3
up with the formula from step 3.
5. S ince the surface area is the sum of the areas of all the bases of the
pyramids, the formula that could replace the sum in the parentheses in
step 4 is 4π r 2.
6. Replace the sum in parentheses in the formula you determined in step 5.
1
V = r ( B1 + B2 + B3 + ... + Bn ) Formula for the volume of a hemisphere
3
1
V = r ( 4π r 2 )
3
7. Simplify.
To simplify, multiply the matching variables.
4
V = πr3
3
8. Summarize your findings.
Gloria can use the volume of a pyramid and dissection
arguments to derive the volume of a sphere.
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