Unlock Real-World Applications of Three-Dimensional Modeling, Geometric Shapes, and Volume Calculations

Calculate power

L = \frac{π ( 3 0 . 2 5 ) - π ( 6 . 2 5 )}{0 . 0 3}

L = \frac{3 0 . 2 5 π - 6 . 2 5 π}{0 . 0 3}

SubTerm

Subtract term

L = \frac{2 4 π}{0 . 0 3}

Use a calculator

L = 2513.274122 \dots

RoundInt

Round to nearest integer

L \approx 2513

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.

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?

c Suppose that the thickness of the soda can is uniform throughout its body. Estimate the soda can's thickness.

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 = 2 π r h + 2 π r^{2}

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.

S = 2 π r h + 2 π r^{2}

$r = 1.2$ , $h = 6$

S = 2 π (1.2) (6) + 2 π (1.2)^{2}

Evaluate right-hand side

Calculate power

S = 2 π (1.2) (6) + 2 π (1.44)

S = 14.4 π + 2.88 π

AddTerms

Add terms

S = 17.28 π

Use a calculator

S = 54.286721 \dots

Round to $2$ decimal place(s)

S \approx 54.29

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 = \frac{m}{V} \Leftrightarrow V = \frac{m}{d}

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.

V = \frac{m}{d}

$m = 21$ , $d = 2.7$

V = \frac{2 1}{2 . 7}

Evaluate right-hand side

Calculate quotient

V = 7.777777 \dots

Round to $2$ decimal place(s)

V \approx 7.78

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.

\underline{Surface Area} 54.29 in^{2} \underline{Volume} 7.78 cm^{3}

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 \cdot t \Leftrightarrow t = \frac{V}{S}

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

\frac{6 . 5 cm ^{2}}{in ^{2}}

to convert square inches to square centimeters.

54.29 in^{2} \cdot \frac{6 . 5 cm ^{2}}{in ^{2}}

Simplify

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

\frac{5 4 . 2 9 in ^{2} \cdot 6 . 5 cm ^{2}}{in ^{2}}

CrossCommonFac

Cross out common factors

\frac{5 4 . 2 9 in ^{2} \cdot 6 . 5 cm ^{2}}{in ^{2}}

SimpQuot

Simplify quotient

54.29 \cdot 6.5 cm^{2}

352.885 cm^{2}

Now, by substituting

V = 7.78

and

S = 352.885,

t

can be found.

t = \frac{V}{S}

$V = 7.78$ , $S = 352.885$

t = \frac{7 . 7 8}{3 5 2 . 8 8 5}

Evaluate right-hand side

Calculate quotient

t = 0.022046 \dots

Round to $2$ decimal place(s)

t \approx 0.02

This means that the thickness of the soda can is about

0.02

centimeters.

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 carton of orange juice and two glasses

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?

b How many Collins glasses of orange juice can she sell?

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?

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.

V = \frac{1}{3} π r^{2} h

$r = 5.4$ , $h = 5.8$

V = \frac{1}{3} π (5.4)^{2} 5.8

Evaluate right-hand side

Calculate power

V = \frac{1}{3} π (29.16) 5.8

V = \frac{1}{3} 169.128 π

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

V = \frac{1 6 9 . 1 2 8 π}{3}

Use a calculator

V = 177.110427 \dots

RoundInt

Round to nearest integer

V \approx 177

This means that a cocktail glass can hold approximately

177

cubic centimeters orange juice. Conversion needs to be made. The conversion factor

\frac{1 L}{1 0 0 0 cm ^{3}}

will convert cubic centimeters into liters.

177 cm^{3} \cdot \frac{1 L}{1 0 0 0 cm ^{3}} = 0.177 L

Finally, the number of cocktail glasses can be calculated.

\frac{1 . 5}{0 . 1 7 7}

Calculate quotient

8.474576 \dots

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.

V_{2} = π r^{2} h

$r = 3.2$ , $h = 9$

V_{2} = π (3.2)^{2} 9

Evaluate right-hand side

Calculate power

V_{2} = π (10.24) 9

V_{2} = 92.16 π

Use a calculator

V_{2} = 289.529178 \dots

RoundInt

Round to nearest integer

V_{2} \approx 290

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.

\frac{1 . 5}{0 . 2 9} = 5.172413 \dots

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.

Type of Glass
Cocktail Glass	Collins Glass
$8 \cdot 6$	$5 \cdot 10$
$$ 48$	$$ 50$

As a result, Emily should choose the Collins glass, as the revenue from this selection is greater.

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.

Answer

About $1.7 \times 1 0^{6}$

Hint

The formula for the volume of a sphere is $V = \frac{4}{3} π r^{3},$ 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.

N = \frac{M}{m}

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 = \frac{m}{V} \Leftrightarrow m = d \cdot 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

\frac{0 . 0 6}{2} = 0.03

centimeters. Use the formula for the volume of a sphere to find the volume of a grain.

V = \frac{4}{3} π r^{3}

Substitute

$r = 0.03$

V = \frac{4}{3} π (0.03)^{3}

Evaluate right-hand side

Calculate power

V = \frac{4}{3} π (0.000027)

CommutativePropMult

\CommutativePropMult

V = \frac{4}{3} 0.000027 π

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

V = \frac{0 . 0 0 0 1 0 8 π}{3}

Use a calculator

V = 0.000113 \dots

Round to $5$ decimal place(s)

V \approx 0.00011

Write in scientific notation

V \approx 1.1 \times 1 0^{- 4}

The volume of a grain is about

1.1 \times 1 0^{- 4}

cubic centimeters.

Finding the Mass of a Grain

The density of a grain is

1.52 g / cm^{3}

and its volume is

1.1 \times 1 0^{- 4}

cubic centimeters. By multiplying these values, the mass of a grain can be found.

\frac{1 . 5 2 g}{cm ^{3}} \cdot 1.1 \times 1 0^{- 4} cm^{3}

Evaluate

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

\frac{1 . 5 2 g \cdot 1 . 1 \times 1 0 ^{- 4} cm ^{3}}{cm ^{3}}

CancelCommonFac

Cancel out common factors

\frac{1 . 5 2 g \cdot 1 . 1 \times 1 0 ^{- 4} cm ^{3}}{cm ^{3}}

SimpQuot

Simplify quotient

1.52 g \cdot 1.1 \times 1 0^{- 4}

CommutativePropMult

\CommutativePropMult

1.52 \cdot 1.1 \times 1 0^{- 4} g

1.672 \times 1 0^{- 4} g

Round to $1$ decimal place(s)

1.7 \times 1 0^{- 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 = \frac{M}{m}

$M = 300 g$ , $m = 1.7 \times 1 0^{- 4} g$

N = \frac{3 0 0 g}{1 . 7 \times 1 0 ^{- 4} g}

Evaluate right-hand side

CrossCommonFac

Cross out common factors

N = \frac{3 0 0 g}{1 . 7 \times 1 0 ^{- 4} g}

SimpQuot

Simplify quotient

N = \frac{3 0 0}{1 . 7 \times 1 0 ^{- 4}}

WriteProdFrac

Write as a product of fractions

N = \frac{3 0 0}{1 . 7} (\frac{1}{1 0 ^{- 4}})

FracToNegExponent

$\frac{1}{a ^{m}} = a^{- m}$

N = \frac{3 0 0}{1 . 7} (1 0^{4})

Calculate quotient

N = (176.470588 \dots) (1 0^{4})