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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


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    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


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    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:


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