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Rotational Volume Using Shell Method

  1. May 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.
    y = 4 - x2
    y = 0

    2. Relevant equations


    3. The attempt at a solution
    Okay I understand that the region is symmetric about the y-axis, however I still don't understand why the integral 2π∫[from -2 to 2] (4x-x3)dx comes out to be 0 when I plug it into my calculator. I know that you can just do the integral from 0 to 2 and then multiply the whole thing by 2 and it comes out correctly but why does doing it from -2 to 2 come out as 0? Aren't both methods trying to calculate the same thing? Why do both get different answers? Wouldn't rotating just the region from 0 to 2 about the y-axis and rotating the region from -2 to 2 about the y-axis give you the same cylindrical solid?
     
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  3. May 24, 2016 #2

    SteamKing

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    You get the wrong answer for the volume using

    $$V=\int_{-2}^2(4-x^2) ⋅ xdx$$

    because you have applied the shell method incorrectly. Since the region y = 4x - x2 is symmetrical about the axis of rotation, in this case the y-axis, you should have made the lower limit of integration a = 0 instead of a = -2.

    http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx
     
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