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Volume using intergals

  1. Mar 3, 2007 #1
    If the region bounded by y=9-x^2,y=0,x=0. Is rotated about the x axis the what would the volume be. I'm got 407. For the same region rotate about the y=9, and i got 153. Can someone check these answers i think they are wrong.
     
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  3. Mar 3, 2007 #2

    arildno

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    Set up the equations you used! :smile:
     
  4. Mar 3, 2007 #3

    HallsofIvy

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    You know, I could have sworn that when you rotate something a "[itex]\pi[/itex]" would show up!

    How about showing us exactly what you integrated and how you got those answers?
     
  5. Mar 3, 2007 #4
    The first integral would be, I think: b=3,a=0, on the integral pi*(9-x^2)^2


    The second one, which I'm not sure about, would be same b and a, on the integral pi*(9-(9-x^2))^2
     
  6. Mar 3, 2007 #5

    HallsofIvy

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    The first is set up correctly but that surely is not "407"!

    When you rotate around the y-axis, your "disks" will be moving up the y axis. The radius of each disk will be x as a function of y and their "thickness" is dy.
     
  7. Mar 3, 2007 #6
    Thats what i get when I put it into my ti-83. I'm pretty sure that the first one is right. But I'm having a hard time imaging the shape you get when you rotate the area around y=9.
     
  8. Mar 3, 2007 #7
    [tex] \int_{0}^{3} pi*9^2-(pi*(9-9+x^2)^ dx [/tex] is what the second integral maybe?
     
  9. Mar 3, 2007 #8
    Never mind guys, I just checked my work on maple and the answers I had first were right, I should stop second guessing my self :),
     
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