1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Shell method

  1. Mar 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid of revolution:
    F(x)=2x+3 on [0,1]
    Revolved over the line x=3 and y=5

    2. Relevant 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
     
  2. jcsd
  3. Mar 26, 2008 #2
    The last part since I wasn't allowed so many [tex]

    1. The problem statement, all variables and given/known data
    Find the volume of the solid of revolution:
    F(x)=2x+3 on [0,1]
    Revolved over the line x=3 and y=5

    2. Relevant 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


    3. 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. Mar 26, 2008 #3
    The last part since I wasn't allowed so many [tex]

    3. 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
     
  5. Mar 26, 2008 #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: Mar 26, 2008
  6. Mar 27, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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?

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Shell method
  1. The shell method (Replies: 2)

  2. Shell method (Replies: 0)

  3. Shell Method (Replies: 1)

  4. Shell method (Replies: 3)

  5. Shell Method (Replies: 7)

Loading...