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Volume of Object

  1. Nov 8, 2006 #1
    The top of an object is bounded by [tex] z=4-4(x^2+y^2)[/tex] and the bottom is bounded by [tex]z=(x^2+y^2)^2 -1 [/tex], find the volume.

    So I converted to polar coordinates such that
    top: [tex]z=4-4r^2[/tex]
    bot: [tex]z=r^2 - 1[/tex]

    Then I integrated the first quadrant and multiplied by 4, with the bounds of dzdrdø such that the volume was:
    [tex]V= 4[ \int_{\theta=0}^{\pi/2} \int_{r=0}^{r=1} \int_{z=0}^{z=4-4r^2} r dz dr d \theta + \int_{\theta=0}^{\pi/2} \int_{r=0}^{r=1} \int_{z=r^2-1}^{z=0} r dz dr d\theta ][/tex]

    The answer I got was 7pi/2, but the answer from the book is 8pi/3. I think I have done all the arithematic right, but did I set it up right?
     
    Last edited: Nov 8, 2006
  2. jcsd
  3. Nov 8, 2006 #2

    quasar987

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    what is the region of integration in terms of x&y?
     
  4. Nov 8, 2006 #3
    It isn't given, the problem just gave the object in terms of its top and bottom. I sketched it out, and the object is essentially one of those plastic easter eggs (where the top is longer and narrower than the bottom).

    But, just going back then in terms of x&y the integration will be the very sticky:
    [tex]V= 4[ \int_{y=0}^{y=1} \int_{x=0}^{x=\sqrt{1-y^2}} \int_{z=0}^{z=4-4(x^2+y^2)} dz dx dy + \int_{y=0}^{y=1} \int_{x=0}^{x=\sqrt{1-y^2}} \int_{z=(x^2+y^2)^2-1}^{z=0}dz dx dy][/tex]
    Which does not look appealing to calculate.
     
    Last edited: Nov 8, 2006
  5. Nov 8, 2006 #4

    quasar987

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    Ah, but [tex]z=(x^2+y^2)^2 -1 [/tex] transforms into [tex]z=r^4 -1 [/tex]! power 4 not 2.
     
  6. Nov 8, 2006 #5
    Oh no, you are right! Oops.

    Thanks.
     
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