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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


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    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


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    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.
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