Exercise 31 Page 868

Use the formula for the volume of a cylinder.

About 304.2 cubic centimeters.

Practice makes perfect

A can and a rubberized cylindrical holder can be modeled by the following composite solid.

Since the diameter of the base of the can is 6.5 cm, the radius of the base of the can is 6.52=3.25 cm. We are asked to find the volume of the rubberized cylindrical holder. Its height is h=11.5 cm, including 1 cm for the thickness of the bottom. Now, let's divide the holder into two parts.

THe outer side part.
The bottom small cylindrical part.

Let's find the volumes of these parts.

Outer Side Part

First, let's find the volume of its base.

Notice that the area of the base is equal to the difference between the area of the big and the small circle. c Area of Base = c Area of Big Circle - c Area of Small Circle Let's use the formula for the area of a circle.

Region	Big Circle	Small Circle
Radius	R= 4.25	r= 3.25
Area	A_\text{big}=\pi {\color{#FF0000}{R}}^2	A_\text{small}=\pi {\color{#FF0000}{r}}^2
Area	{\color{#0000FF}{A_\text{big}}}=\pi ({\color{#FF0000}{4.25}})^2={\color{#0000FF}{18.0625\pi}}	{\color{#009600}{A_\text{small}}}=\pi ({\color{#FF0000}{3.25}})^2 = {\color{#009600}{10.5625\pi}}

Now, let's find the area of the base of the outer side part of the holder. c Area of Base = c Area of Big Circle - c Area of Small Circle ⇓ B= 18.0625π- 10.5625π=7.5π This tells us that the area of the base is B=7.5π square centimeters. Finally, let's find the volume of the outer side part.

V_\text{side}=\textcolor{darkorange}{B}\textcolor{darkviolet}{h}

SubstituteII

B= 7.5π, h= 11.5

V_\text{side}=(\textcolor{darkorange}{7.5\pi})(\textcolor{darkviolet}{11.5})

Multiply

V_\text{side}=86.25\pi

Therefore, the volume of the outer side of the holder is 86.25π cubic centimeters.

Bottom Small Part

Now, let's find the volume of the bottom small cylindrical part.

This part is a cylinder with a radius r=3.25 cm and a height h=1 cm. Let's find its volume.

V_\text{bottom}=\pi r^2h

▼

Substitute values and evaluate

SubstituteII

r= 3.25, h= 1

V_\text{bottom}=\pi ({\color{#0000FF}{3.25}})^2({\color{#009600}{1}})

CalcPowProd

Calculate power and product

V_\text{bottom}=10.5625\pi

Therefore, the volume of the bottom small part is 10.5625π cubic centimeters.

Volume of Holder

The volume of the outer side part is 86.25π cubic centimeters, and the volume of the bottom small part is 10.5625π cubic centimeters. Now, let's add these volumes to get the volume of the holder.

V_\text{holder}=V_\text{side}+V_\text{bottom}

▼

Substitute values and evaluate

SubstituteII

V_\text{side}={\color{#0000FF}{86.25\pi}}, V_\text{bottom}={\color{#009600}{10.5625\pi}}

V_\text{holder}={\color{#0000FF}{86.25\pi}}+{\color{#009600}{10.5625\pi}}

AddTerms

Add terms

V_\text{holder}=96.8125\pi

▼

Round to 1 decimal place(s)

π ≈ 3.142

V_\text{holder}\approx 96.8125({\color{#FF0000}{3.142}})

Multiply

V_\text{holder}\approx 304.184875

RoundDec

Round to 1 decimal place(s)

V_\text{holder}\approx 304.2

Tthe volume of the holder is 96.8125π cubic centimeters, which is about 304.2 cubic centimeters.

Subchapter links

Check Your Understanding p.866

Practice and Problem Solving p.866-869

H.O.T. Problems p.869

Standardized Test Practice p.870

Spiral Review p.870

Skills Review p.870

Exercise 31 Page 868

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Outer Side Part

Bottom Small Part

Volume of Holder