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Calc2: Solid of revolution about y=2 question

  1. Sep 11, 2011 #1
    Hi All, I had a question about this problem I was doing and was hoping to see if I did it right.

    1. The problem statement, all variables and given/known data
    Using Disk or Washer method Find volume of the solid generated by revolving the region bounded by y=sqrt(x), y=2, x=0 about the line y=2.


    2. Relevant equations
    Disk Method: pi*Integral (R(x))^2dx where R(x) is the distance from the axis of revolution to the regions boundaries.


    3. The attempt at a solution

    Limits of integration are [0,4]

    I was getting caught up with what R(x) is. Since I am revolving up around y=2 I am wondering if the distance is 2-sqrt(x) or sqrt(x)-2

    Doing the problem with R(x)=sqrt(x)-2 I get:

    8pi/3

    Doing the problem with R(x) = 2-sqrt(x) I get:

    8pi/3 again...

    So what I am wondering is, in the future, which one is the correct R(x)? or does it not matter? I feel like it should be 2-sqrt(x) because that more clearly sticks out to me as the radius but im not sure.
     
  2. jcsd
  3. Sep 11, 2011 #2
    Think of it this way: if you simply shifted the graph down by a 2 units, then you could rotate it about the x-axis as you always have. To do so, of course, you would take [itex]\sqrt{x}-2[/itex] and rotate it about the x-axis. I think the typical approach, however, is take the radius to be 2 (like you said) and simply analyze [itex]2-\sqrt{x}[/itex].
     
  4. Sep 11, 2011 #3
    Ok, that makes sense, Thanks!
     
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