Shell method

  • #1

Homework Statement


Find the volume of the solid of revolution:
F(x)=2x+3 on [0,1]
Revolved over the line x=3 and y=5

Homework Equations


Shell Method: 2[tex]\pi[/tex][tex]\int[/tex][tex]^{b}[/tex][tex]_{a}[/tex]x[f(x)-g(x)]dx
obviously just sub y for dy
Disk Method: [tex]/pi[/tex][tex]/int[/tex][tex]^{b}[/tex][tex]_{a}[/tex][F(x)[tex]^{2}[/tex]-G(x)[tex]^{2}[/tex]dx
 

Answers and Replies

  • #2
The last part since I wasn't allowed so many [tex]

Homework Statement


Find the volume of the solid of revolution:
F(x)=2x+3 on [0,1]
Revolved over the line x=3 and y=5

Homework Equations


Shell Method: 2[tex]\pi[/tex][tex]\int[/tex][tex]^{b}[/tex][tex]_{a}[/tex]x[f(x)-g(x)]dx
obviously just sub y for dy
Disk Method: [tex]\pi[/tex][tex]\int[/tex][tex]^{b}[/tex][tex]_{a}[/tex][F(x)[tex]^{2}[/tex]-G(x)[tex]^{2}[/tex]dx


The Attempt at a Solution


line x=3: 2[tex]\pi[/tex][tex]\int[/tex](3-x)(2x+3)dx =115.19

answer key is unfortunately in disk method which I don't like as much:
[tex]\pi[/tex][tex]\int[/tex][tex]^{3}[/tex][tex]_{0}[/tex](9-4)dy + [tex]\pi[/tex][tex]\int[/tex][tex]^{5}[/tex][tex]_{3}[/tex](3-((y-3)/2))[tex]^{2}[/tex]-4dy

=78.91


line y=5: 2[tex]\pi[/tex][tex]\int[/tex][tex]^{5}[/tex][tex]_{0}[/tex](5-y)(1-((y-3)/2)) =130.8996

answer key/ disk method: [tex]\pi[/tex][tex]\int[/tex][tex]^{1}[/tex][tex]_{0}[/tex](25-(5-(2x+3))[tex]^{2}[/tex]dx

=77.206
 
  • #3
The last part since I wasn't allowed so many [tex]

The Attempt at a Solution


line x=3: 2[tex]/pi[/tex][tex]/int[/tex](3-x)(2x+3)dx =115.19

answer key is unfortunately in disk method which I don't like as much:
[tex]\pi[/tex][tex]\int[/tex][tex]^{3}[/tex][tex]_{0}[/tex](9-4)dy + [tex]\pi[/tex][tex]\int[/tex][tex]^{5}[/tex][tex]_{3}[/tex](3-((y-3)/2))[tex]^{2}[/tex]-4dy

=78.91

line y=5: 2[tex]\pi[/tex][tex]\int[/tex][tex]^{5}[/tex][tex]_{0}[/tex](5-y)(1-((y-3)/2)) =130.8996

answer key/ disk method: [tex]\pi[/tex][tex]\int[/tex][tex]^{1}[/tex][tex]_{0}[/tex](25-(5-(2x+3))[tex]^{2}[/tex]dx

=77.206
 
  • #4
the first number in the integral is the upper bound and the second is the lower.... I just can't get it to be formatted the right way for some reason.

EDIT: sry about the repeat... said database error so I thought it didn't go through.

EDIT: OMG there's 2 double posts.... srry
 
Last edited:
  • #5
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement


Find the volume of the solid of revolution:
F(x)=2x+3 on [0,1]
Revolved over the line x=3 and y=5
Is this two separate problems? It doesn't appear to be from your work but what do you mean by "revolved over" two separate lines? In any case, "F(x)= 2x+3 on [0,1]" doesn't define a region. Do you mean the region bounded by y=2x+ 3, y= 0, x= 0, and x= 1?

Homework Equations


Shell Method: 2[tex]\pi[/tex][tex]\int[/tex][tex]^{b}[/tex][tex]_{a}[/tex]x[f(x)-g(x)]dx
obviously just sub y for dy
Disk Method: [tex]\pi[/tex][tex]\int[/tex][tex]^{b}[/tex][tex]_{a}[/tex][F(x)[tex]^{2}[/tex]-G(x)[tex]^{2}[/tex]dx


The Attempt at a Solution


line x=3: 2[tex]\pi[/tex][tex]\int[/tex](3-x)(2x+3)dx =115.19

answer key is unfortunately in disk method which I don't like as much:
[tex]\pi[/tex][tex]\int[/tex][tex]^{3}[/tex][tex]_{0}[/tex](9-4)dy + [tex]\pi[/tex][tex]\int[/tex][tex]^{5}[/tex][tex]_{3}[/tex](3-((y-3)/2))[tex]^{2}[/tex]-4dy

=78.91


line y=5: 2[tex]\pi[/tex][tex]\int[/tex][tex]^{5}[/tex][tex]_{0}[/tex](5-y)(1-((y-3)/2)) =130.8996

answer key/ disk method: [tex]\pi[/tex][tex]\int[/tex][tex]^{1}[/tex][tex]_{0}[/tex](25-(5-(2x+3))[tex]^{2}[/tex]dx

=77.206
 

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