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Volume by Integration

  1. Oct 21, 2012 #1
    Find the volume of the solid obtained by rotating the region bounded by the curves:

    x = 2 sqrt(y), x = 0, y = 9; about the y-axis

    I have this graphed and everything so I'm not sure there's a need to worry about that.

    Setup (integrating with respect to y):
    Outer radius = x = 2 sqrt(y). Inner radius = x = 0. Bounded by y = 9 and 0.

    9
    ∫π[2 sqrt(y)]^2 dy
    0

    Which is equivalent to:
    _______9
    (2πy^2)| = 36π
    _______0

    However, my book says the answer is 162π. What am I doing wrong?
     
  2. jcsd
  3. Oct 21, 2012 #2

    Zondrina

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    Homework Helper

    Ahhh now you might've missed this :

    x = 2[itex]\sqrt{y}[/itex]

    So your volume integral becomes :

    [itex]\pi \int_{0}^{9} 4y \space dy[/itex]

    Since [itex]f^2(y) = 4y[/itex]
     
    Last edited: Oct 21, 2012
  4. Oct 21, 2012 #3
    That doesn't change what I get at the end. I'll still factor out the 4 and the π and what's left in the integral will still come to be (y^2) / 2. Multiply that by what's outside the integral and I still have 2πy^2.
     
  5. Oct 21, 2012 #4

    Zondrina

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    Homework Helper

    Oh whoops, in my last post i meant your integral is really :

    [itex]π \int_{0}^{9} 4y \space dy[/itex]

    So, integrating it you get :

    [itex]π \int_{0}^{9} 4y \space dy = 4π \int_{0}^{9} y \space dy = 2π[y^2]_{0}^{9} \space dy[/itex]

    Do you see it now?
     
  6. Oct 21, 2012 #5
    Aah. I know what I did. I just thought that came out to 36π for some stupid reason. Alright thanks. Sorry about that. I'm a bit of a dope : D
     
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