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Find the Volume (Double Integral)

  1. Apr 14, 2008 #1
    I'm having trouble trying to setup this double integral. The question asks to find the volume of a solid enclosed by the parabolic cylinder [tex]y = x^{2}[/tex] and the planes [tex]z = 3y[/tex], [tex]z = 2+y[/tex]

    I'm not even sure where to start. I have drawn the figure and understand that you have to integrate the two functions [tex]z = 3y[/tex] and [tex]z = 2+y[/tex] and subtract the volumes. However I'm stuck trying to setup the boundaries. Thanks.
     
  2. jcsd
  3. Apr 14, 2008 #2
    To find a volume I suppose you have to set up a triple integral?

    You may, for example, let:
    [tex]3y < z < 2 + y, \quad
    -\sqrt{y} < x < \sqrt{y} \quad,
    0 < y < 1[/tex]

    .. I think, not sure, though.
     
  4. Apr 14, 2008 #3


    If the problem is correct as you stated it the boundaries would be [itex]0\leq y\leq \Gamma[/itex] and [itex]-\sqrt{y}\leq x\leq\sqrt{y}[/itex], where [itex]\Gamma[/itex] ist the y-value for which the two planes intersect.

    So

    [tex]
    \int_0^\Gamma{dy\int_{-\sqrt{y}}^{\sqrt{y}}dx(2y-2)}
    [/tex]
     
    Last edited: Apr 14, 2008
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