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Write the Integral Equation

  1. Oct 27, 2014 #1
    1. The problem statement, all variables and given/known data
    The volume of the solid obtained by rotating the region bounded by
    x=6y^2 [PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [Broken] [Broken] y=1 [PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [Broken] [Broken] x=0[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [Broken] [Broken] it is rotate about the y- axis.

    2. Relevant equations


    3. The attempt at a solution

    Using the disk method, I figured out by integrating (pi)(6y^2)^2 dy from 0 - 1 and got the answer: 22.619

    However, I cannot write an integral equation using the cylindrical shell method. My attempt is:

    Integrating (2pi)(sqrt(x/6))dx from x = 0 to 6.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 27, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    Check the length of your shells. How long is a shell for some specific x-value? In particular, is it increasing as your approach would suggest?
     
  4. Oct 27, 2014 #3
    What do you mean? Isn't it always x?
     
  5. Oct 27, 2014 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No! It is not. The whole point of the "shell" method is that the shells are parallel to the axis of revolution. Here, that is parallel to the y-axis.

    And if you meant "Isn't it always y?" Again, no it it isn't. x= 6y^2 is the lower boundary, y= 1 is the upper boundary. The length of a shell is the vertical distance between them.
     
  6. Oct 27, 2014 #5
    I got it with this:

    (2pi)(1-sqrt(x/6))(x)dx

    thanks.
     
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