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(repost since typos) Volume of revolution using shell method

  1. Nov 1, 2009 #1
    (EDITED)
    1. Use the shell method to find the volume of the solid generated by revolving about the y-axis. [tex]x=y^2, x=y+2[/tex]

    2. same as #1, except change y and x for the two equations and revolve about x-axis.

    I tried doing [tex]2pi\int_{x=0}^4(x)(\sqrt{x}-x+2)dx[/tex] but the answer is off for #1.

    I tried doing [tex]2pi\int_{y=-1}^4(y)(\sqrt{y}-y+2)dy[/tex], well it's wrong.

    Is there a problem with how I established the shell height?

    the answer for these two questions are supposed to give me 72pi/5.
     
  2. jcsd
  3. Nov 1, 2009 #2

    tiny-tim

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    Hi Call my name! :smile:

    (have a pi: π and a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    Yeees … it's the shell width, isn't it? :wink:
    Are you sure that's the question? If you swap everything, isn't the volume the same?:smile:
     
  4. Nov 1, 2009 #3
    Yeah, I meant that those two questions' answers are the same.

    So is what I have established wrong?
     
  5. Nov 1, 2009 #4

    tiny-tim

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    You've used height instead of width

    (or you've revolved around the wrong axis).
     
  6. Nov 1, 2009 #5
    What do you mean? I'm a newbie so please specify and clarify
     
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