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The volume of a solid rotating about a different axis

  1. Oct 5, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]y = {\frac{1}{4+x^2}}[/itex] on the interval [0,2], revolving about y = -1
    Use either the disk/washer or shell method to find the volume.



    2. Relevant equations
    [itex]v = pi\int (outer radius)^2-(inner radius)^2\,dx[/itex]
    [itex]v = 2pi\int (radius)(height)\,dy[/itex]
    [itex]
    x = \sqrt{\frac{1}{y}-4}
    [/itex]



    3. The attempt at a solution
    [itex]v = 2pi\int (y+1)
    \sqrt{\frac{1}{y}-4}\,dy[/itex] from [itex]\frac{1}{4}[/itex] to [itex]\frac{1}{8}[/itex]

    I'm just stuck on setting up the integral. I get confused easily from these washer/shell problems, and it's worse when the axis changes haha. So I don't know if this integral is set up correctly. And, I feel like there's something off about my limits. Do I have to add another integral to integrate from 0? Or would it just be easier to use the washer method?
     
  2. jcsd
  3. Oct 5, 2012 #2

    gabbagabbahey

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    If I were you, I'd start by plotting y(x) for the given interval... can you imagine what it wouldlook like as your rotate it about y=-1? What would be the radius of a disk at x? What would be the area of the disk?
     
  4. Oct 5, 2012 #3
    Yeah, I've tried graphing it but even while I was looking at it, I'm still confused as to how to read the inner radius and the outer radius.
    I figured that the inner radius is (y+1) and the outer is [itex]\frac{1}{4+x^2}[/itex], right?
    I'm not sure about the area though.
     

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  5. Oct 5, 2012 #4

    gabbagabbahey

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    Why have you only shaded in part of the area between y(x) and y=-1? Does the problem tel you to rotate onlythe areabetween y(x) and the x-axis around y=-1? Ifnot, I would say your inner radius is zeroand your outer radius is |y(x)|
     
  6. Oct 5, 2012 #5
    Well, the problem didn't specify anything like that. All I received was the equation, the interval and what axis it's rotating about, so I just presumed that's what the graph would look like for the problem.

    I'll try working it out with inner radius as 0 though, thanks!
     
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