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Using disk method with respect to y-axis

  1. Jan 30, 2009 #1
    1. The problem statement, all variables and given/known data
    y=[tex]\sqrt{x}[/tex], x=4, y=0


    2. Relevant equations
    [tex]\pi[/tex][tex]\int^{c}_{d}{R(y)}^{2}[/tex]dy



    3. The attempt at a solution
    I solve for x then pluged it into the formula

    [tex]\int[/tex][tex]^{2}_{0}[/tex]{(y[tex]^{2}[/tex])[tex]^{2}[/tex]}

    The answer I got was [tex]\frac{32\pi}{5}[/tex]

    but my text book says the answer should be [tex]\frac{128\pi}{5}[/tex]
     
    Last edited: Jan 30, 2009
  2. jcsd
  3. Jan 30, 2009 #2

    Dick

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    It looks like what they want is the area between y=sqrt(x) and the vertical line x=4 rotated around the y-axis. You'll want to use washers. Or you could subtract your answer from the volume of the cylinder with radius 4 and height 2.
     
    Last edited: Jan 30, 2009
  4. Jan 30, 2009 #3
    Yeah sorry, I thought I put that I was supposed to find the the volume as it rotates around the y-axis. which is what i tried to do by solving y=sqrt(x) for x=y^2 then I used that boundries 0-2 and integrated. How do you suppose I use washers? I thought the problem had to have two functions of x to do that.

    Please help this problem has been bothering me since yesterday!
    Thanks!
     
  5. Jan 30, 2009 #4

    Dick

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    If you cut the volume at a fixed value of y then you get a washer shaped region (a circle with a circular hole in the middle). The inner radius is where it hits y=sqrt(x) and, as you said, that's y^2. The outer radius is where it hits the line x=4. So the outer radius is 4. To use the method of washers, you integrate pi*(outer radius)^2-pi*(inner radius)^2.
     
  6. Jan 30, 2009 #5
    I got it! thank you so much!! I can always depend on the PF members!
     
  7. Jan 30, 2009 #6
    Oh, I was wondering can the disk method ever be used to find the volume around the y-axis? If so how can I tell? thanks!
     
  8. Jan 30, 2009 #7

    Dick

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    Draw a sketch of the region. Do you see what you were doing wrong the first time? You were rotating the region between x=0 (the y-axis) and y=sqrt(x). Disks work fine for that. There's no hole in the middle. But what they wanted was the region between y=sqrt(x) and x=4. Now there's a hole in it and you should use washers. You were rotating the wrong region.
     
  9. Jan 30, 2009 #8
    Ok, so for the volume of the same function about the x-axis the graph essentialliy swings down and around the axis with no gap in between the values of integation 0-4. However, by trying to find the volume of the same function as it swings around the y-axis there is a gap between the values of integration 0-2 which is accounted for by the funtion x=y^2. Right?
     
  10. Jan 30, 2009 #9

    Dick

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    That sounds right. Because in your problem they wanted you to rotate the region OUTSIDE of the function, not inside.
     
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