7.2: Volumes by Slicing – Day 2 - Washers

Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2001

Little Rock Central High School,Little Rock, Arkansas

Bellwork– NO CALCULATOR (15 minutes):Read page 461, Example 3. 7.2 – Washer Method Applet

A)Find the area of R.

B)Write, but do not evaluate, an integral expression for the volume of the solid generated when R is rotated about the line x-axis.

C)Write, but do not evaluate, an integral expression for the volume of the solid generated when R is rotated about the line y=1.

Homework: 7.2 Part 2 (15, 21, 29-31, 33, 59, 62, 63, 65, 69)

A)Find the area of R.

Area =

B) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is rotated about the line x-axis.

Volume

¿𝜋∫𝑥1

𝑥2

(𝑅 (𝑥 )2−𝑟 (𝑥 )2 )𝑑𝑥

C) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is rotated about the line y=1.

Volume

¿𝜋∫𝑥1

𝑥2

(𝑅 (𝑥 )2−𝑟 (𝑥 )2 )𝑑𝑥

¿

The region bounded by and is revolved about the y-axis.Find the volume.

2y x 2y x

The “disk” now has a hole in it, making it a “washer”.

If we use a horizontal slice:

The volume of the washer is: 2 2 thicknessR r

2 2R r dy

outerradius

innerradius

2y x

2

yx

2y x

y x

2y x

2y x

2

24

0 2

yV y dy

∫

4 2

0

1

4V y y dy

∫4 2

0

1

4V y y dy ∫

42 3

0

1 1

2 12y y

168

3

8

3

𝐴 (𝑦 )𝑑𝑦=¿

Example 1 – Volume by Washers

This application of the method of slicing is called the washer method. The shape of the slice is a circle with a hole in it, so we subtract the area of the inner circle from the area of the outer circle.

The washer method formulas are:

𝑉=𝜋∫𝑥1

𝑥2

(𝑅 (𝑥 )2−𝑟 (𝑥 )2 )𝑑𝑥

𝑉=𝜋∫𝑦1

𝑦2

(𝑅 ( 𝑦 )2−𝑟 (𝑦 )2 )𝑑𝑦

and

2y xIf the same region is rotated about the line x=2:

2y x

The outer radius is:

22

yR

R

The inner radius is:

2r y

r

2y x

2

yx

2y x

y x

4 2 2

0V R r dy ∫

2

24

02 2

2

yy dy

∫

24

04 2 4 4

4

yy y y dy

∫

24

04 2 4 4

4

yy y y dy ∫

14 2 2

0

13 4

4y y y dy ∫

432 3 2

0

3 1 8

2 12 3y y y

16 64

243 3

8

3

p

Example 2 – Volume by Washers

Third Example of Washers

Problem Find the volume of the solid obtained by rotating about the x-axis the region enclosed by the curves y = x andy = x2

…but about the line y = 2 instead of thex-axis

The solid and a cross-section are illustrated on the next slide

Third Example of Washers (cont’d)

Second Example of Washers (cont’d)

Solution Here

So

2 22 4 22 2 5 4A x x x x x x

1 1 4 2

0 0

15 3 2

0

5 4

85 4

5 3 2 15

V A x dx x x x dx

x x x

∫ ∫

The formula can be

applied to any solid for which the

cross-sectional area A(x) can be

found

This includes solids of revolution, as

shown above…

…but includes many other solids as

well

A Bigger Picture

b

aV A x dx∫

The Method of Cross-Sections

Intersect S with a plane Px

perpendicular to the x-axis

Call the cross-sectional area A(x) A(x) will vary as x increases from a to b

Cross-Sections (cont’d) Divide S into “slabs” of equal width

∆xusing planes at x1, x2,…, xn Like slicing a loaf of bread! To add an infinite number of slices of

bread…..we must integrate