Docsity
Docsity
Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity

Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips
Guidelines and tips
Sell on Docsity
Sell on Docsity
Docsity AI
Docsity AI
Log inSign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search for your university

Find the specific documents for your university's exams

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents
download point icon20 Points

For each uploaded document

Answer questions
download point icon5 Points

For each given answer (max 1 per day)

All the ways to get free points
Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Calculating the Volume of Solids of Revolution using Shell Method and Disk Method, Study notes of Calculus

Iowa State University (ISU)Calculus

The steps to find the volume of solids of revolution generated by revolving regions about the x-axis and y-axis using both the shell method and disk method. The calculations are based on examples with given regions and boundaries.

Typology: Study notes

Pre 2010

Uploaded on 09/02/2009

koofers-user-zc2
koofers-user-zc2 🇺🇸

9 documents

1 / 9

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Exercises 6.2
1. R is the region bounded by the graphs of y = x2+1, y = 0, x = 0, x = 2. Let S be the solid of
revolution generated by revolving R about the x-axis. Find the volume of S.
.
15
206
15
308096
2
3
16
5
32
3
2
5
1
121x S of Volume
2
0
35
2
0
2
0
24
2
2
xxx
dxxxdx
We now get this volume by the shell method discussed in Section 6.3.
Volume of S =
1
0
5
1
12222 dyyydyy
1
• • • • • • • • • • • • • • • • • •
•
•
•
•
•
•
•
•
•
•
•
x
y
y = x2 + 1
or
1 yx
R
2
pf3
pf4
pf5
pf8
pf9

Partial preview of the text

Download Calculating the Volume of Solids of Revolution using Shell Method and Disk Method and more Study notes Calculus in PDF only on Docsity!

Exercises 6.

1. R is the region bounded by the graphs of y = x

2

+1, y = 0, x = 0, x = 2. Let S be the solid of

revolution generated by revolving R about the x-axis. Find the volume of S.

   

.

15

206

15

96 80 30

2

3

16

5

32

3

2

5

1

Volume ofS x 1 2 1

2

0

5 3

2

0

2

0

4 2 2 2

 

 

   

   

       

x x x

dx x x dx

We now get this volume by the shell method discussed in Section 6.3.

Volume of S = ^   

1

0

5

1

2 y 2 dy 2 y 2 y 1 dy

  • • • • • • • • • • • • • • • • • •
    • • • • • • • • • • •

x

y

y = x

2

or

xy  1

R

  

 

 

 

5

1

2

3 2

5 2

5

1

2

1 2

3

5

1

5

1

1

0

2

y y y

y y y dy

y y y y dy

y y y y dy

3. Let R be the region bounded by the graphs of y = 4 – x

2

, y = 0, x = 0, x = 2.

(a) Let S be the solid of revolution generated by revolving R about the x-axis. Find the volume

of S.

Disk method (Section 6.2)

Volume of S = ^ x ^ dx

2

0

2 2 4

 

.

15

256

5

32

3

64 32

5

1

3

8 16

16 8

2

0

3 5

2

0

2 4

^ 

   

 

 

   

    

x x x

x x dx

Shell method (Section 6.3)

Volume of S = y ydy

4

0

  • • • • • • • • • •
    • • • • • • • • •

x

y

y = 4 – x

2

or

x  4  y

R

 

  

 

2

1

4

1

2

2

1

4

1

4

1

0

2

2

1

4

1

4

1

0

4

2

4

2

4

2 2

2

VolumeofS 2 4 2 2

Shell method

Volumeof S

Disk method

y y

y y dy

dy

y

y dy y

x

dx

x

dx

x

^ 

 

 

9. Let R be the region bounded by the graphs of y = 9 and 0 ,between 2 and 3.

2  x yx  x

Let S be the solid revolution generated by revolving R about the x-axis. Find the volume of S.

2

2

9

9

x y

y x

 

 

  • • • • • • • • • •

x

R

  • • • • • • • • • •
    • • • • • • •

x

  • • • • • • • • • •

x

y

R

2

2

9

9

x y

y x

 

 

.

3

100

3

8

4

3

34

2

8

3

1 0

3

1 27 0 4

3

1 8 5

3

1 2

9

3

1

( 9 ) 4

3

1

2

2 9 2 2 2 9

VolumeofS 2 9 ( 2 ) 2 9 ( 9 )

Shell method

.

3

100

3

8 36

3

8

27 9 18

3

9

9

VolumeofS 9

Disk method

3

5

2

3 2

5

0

2 2

3

5

2

5

0

2

3

5

2 2

5

0

2

3

2

3

3

2

2

3

2

2 2

2

3

 

 

   

 

   

      

   

     

     

     

      

         

  

     

 

 

 

y y y

y y ydy y y dy

y y dy y y y dy

x

x

x dx

x dx

Revolve R about the x-axis.

.

3

512

3

1024 512

12

1 2

4

1

4

2

1

Volume 4

16

0

2 3

16

0

2

16

0

2

^ 

 

   

 

  

 

   

 

     

x x

x x dx x x dx

Revolve R about the y-axis.

.

3

2

3

4 2

8 48

4 4 4 16

Volume

4

0

2 3

4

0

2 4

0

2

^ 

 

   

  

   

 

     

y y

dy

y y dy

y y

R

y

2

=4x

x-2y=

  • (^) x

y

x

y

R

y=4x

2

y=4x

  • • •

The cross section at x is a square with length of a side equal to

2

2  4  x.

The area of the cross section at x is

( )  2 4  4 ^4 .

2

2 2 A x    x    x

The volume of the solid is

.

3

128

3

8 8 8

3

4 4

( ) 4 4

2

2

3

2

2

2

2

2

^ 

    

  

  

 

x x

Axdx x dx

(a) Revolve R about the line x=4.

Volume 4 16 8

8

0

3

7 3

5

8

0

3

4 3

2

2 8

0

3

2

^ 

y y y

y dy y y dy

x

y

x

x

2

+ y

2

length=

2 2  4  x

x

y

R

x=

3

2

2

3

x y

y x