1. Limited time only! Sign up for a free 30min personal 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 Integration (need help)

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Use the shell method to calculate the volume of rotation about the x-axis for the region underneath the graph.

    y=[itex]\sqrt[3]{x}[/itex]-2, 8[itex]\leq[/itex]x[itex]\leq[/itex]27


    2. Relevant equations

    I was taught that I would set up the integral by using the area which would look something like [itex]\int[/itex]2pi(r)(h).

    3. The attempt at a solution

    I saw that the maximum height was 27 and so to get the height of the cyclinder I thought that I would have to subtract the height of the equation [itex]\sqrt[3]{x}[/itex]-2 from 27 to get the height I need to set up the integral. Since I was rotating around the x-axis I first converted the equation so that it would be in terms of y which came out to be x=[itex](y+2)^{3}[/itex]. I then started plugging it into the integral. I knew that since 8[itex]\leq[/itex]x[itex]\leq[/itex]27 I would be integrating from (0,1) because those are the possible y values for this equation within that range. My integral came out to be [itex]\int^{1}_{0}[/itex]2pi(y)(27-[itex](y+2)^{3}[/itex]).

    From that integral my answer came out to be [itex]\frac{54pi}{5}[/itex] which I know is wrong because when I use the disk method the answer comes out to [itex]\frac{38pi}{5}[/itex] which also happens to be the answer in the textbook. Could someone please check my integral to see if I have it correct? Thanks a lot.
     
  2. jcsd
  3. Sep 18, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your integral is set up correctly and will give 38pi/5 if you don't make any arithmetic errors.
     
  4. Sep 18, 2011 #3
    I was wondering why every time I do the integral I got a different number.

    2pi[itex]\int[/itex]y(27-[itex](y+2)^{3}[/itex])dx=2pi[itex]\int[/itex](27y-y[itex](y+2)^{3}[/itex])dx=2pi[itex]\int[/itex](27y-y([itex]y^{3}[/itex]+[itex]6y^{2}[/itex]+12y+8))dx=2pi[itex]\int[/itex](27y-([itex]y^{4}[/itex]+[itex]6y^{3}[/itex]+[itex]12y^{2}[/itex]+8y))dx =2pi[itex]\int[/itex](27y-[itex]y^{4}[/itex]-[itex]6y^{3}[/itex]-[itex]12y^{2}[/itex]-8y)dx=2pi([itex]\frac{27y^{2}}{2}-\frac{y^{5}}{5}-\frac{6y^{4}}{4}-\frac{12y^{3}}{3}-\frac{8y^{2}}{2})^{1}_{0}[/itex]=2pi[([itex]\frac{27}{2}[/itex]-[itex]\frac{1}{5}[/itex]-[itex]\frac{3}{2}[/itex]-4-4)-(0)]=2pi(12-[itex]\frac{1}{5}[/itex]-8)=2pi(4-[itex]\frac{1}{5}[/itex]) =2pi([itex]\frac{20}{5}[/itex]-[itex]\frac{1}{5}[/itex])=2pi([itex]\frac{19}{5}[/itex])=[itex]\frac{38pi}{5}[/itex]

    YESSSSSSS. Thanks so much for your help LCKurtz. I was doing it over and over changing the integral but by knowing that it was an arithmetic error, I realized when I did (y+2)(y+2)(y+2) I multiplied incorrectly leading to the whole thing being wrong. Thanks.
     
  5. Sep 18, 2011 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's a dy integral, by the way. Instead of multiplying the cubic out like that, which, as you know, is very error prone, a much easier way to do the integral is to substitute u = y+2, du = dy. :cool:
     
  6. Sep 18, 2011 #5
    Woops. I totally messed up with the dx's haha. Hmm. I will try with the substitution. Thanks so much for your help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Shell Method Integration (need help)
  1. Shell method help. (Replies: 1)

Loading...