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Homework Help: Area betweeen curves

  1. Mar 25, 2007 #1
    1. The problem statement, all variables and given/known data

    This is so basic and yet I'm not getting it.

    Find the volume of the solid when the region enclosed is revolved arond the y axis.

    [tex] x = \sqrt{1 + y}[/tex]
    x=0, y=3

    2. Relevant equations



    3. The attempt at a solution

    So I first graphed the thing, finding that the enclosed area was between y=1 and y=3. I took the volumne of the equation, or A=pi *[tex]\sqrt{1 + y}^2[/tex]. I then took the integral, pi * [tex]\int_{1}^{3} 1 + y dy[/tex], coming out with 6*pi. My book says this isn't right. Did I do anything wrong? Use the wrong formula?
     
    Last edited: Mar 25, 2007
  2. jcsd
  3. Mar 25, 2007 #2

    AKG

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    The enclosed area is not between y=1 and y=3.
     
  4. Mar 25, 2007 #3
    Oh wait...I think I got it.

    I was graphing with the equation set equal to x, when it should have been set equal to y. This would make [tex]\int_{-1}^{3} 1 + y dy[/tex], which, when multiplied by pi, would get 8*pi.

    Thanks.
     
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