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Double iintegral (volume) this is redicuious

  1. Sep 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid bounded by y=2 y=x and z=4-y^2

    2. Relevant equations



    3. The attempt at a solution
    I DO NOT UNDERSTAND WHY I'M WRONG! limits of integration for x=2 to 0 limtis of integration for y=2 to x and 4-y^2 is the integrand I ended up with 4/3 but the answer (according to the answer sheet) is 4! WHY
     
  2. jcsd
  3. Sep 7, 2008 #2
    I think your limits of integration are wrong

    The area in the xy plane is a triangle. I can't really draw it but x doesn't go from 2 to 0 if I read your limits of integration right. Also, y isn't going to go from 2 to x either.
     
    Last edited: Sep 7, 2008
  4. Sep 7, 2008 #3
    Wait I found it. limits for x are y to 0 and for y 2 to 0. If I went from 2 to x for y then that would be like integrating for x again, since the integral lines would be parallel to y.
     
  5. Sep 7, 2008 #4
    I think that would work...however I believe y should go from 0 to 2 otherwise you might get a negative answer.


    Here's what I have:

    [tex]\int^{2}_{0}[/tex][tex]\int^{2}_{x}4 - y^{2}dydx[/tex]

    With that I integrate with respect to y first and then x.
     
    Last edited: Sep 7, 2008
  6. Sep 7, 2008 #5

    tiny-tim

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    Hi evilpostingmong! :smile:
    I don't think well in 3D … :redface:

    but that isn't bounded, is it? :confused:
     
  7. Sep 7, 2008 #6
    Technically you're correct...I assume that they mean it's bounded by the axis planes...
     
  8. Sep 7, 2008 #7

    tiny-tim

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    hmm … I always assume someone copied the question wrong :wink:

    until I'm reassured! :smile:
     
  9. Sep 7, 2008 #8
    If it's not bounded then...the volume is infinite if I am not mistaken.
     
  10. Sep 7, 2008 #9
    It's alright, I did it out. I got 4 so these limits (x=y to x=0 and y=2 to y=0) are correct.
    I appreciate the help, but the limits you put were the same as mine (the ones I put at the first post) which would net you 4/3.
    Yes it is bounded by the xy plane.
     
  11. Sep 7, 2008 #10
    When I did the integral with my limits I got 4.
     
  12. Sep 7, 2008 #11
    Then I made a stupid mistake when I did the integral itself, which I'm known for, lol I just
    did it my original way (the one that we thought was wrong) and yes it is 4. Ur rite.
     
  13. Sep 7, 2008 #12
    One mistake I made was getting to a point where I had 16x/3 and not plugging in the 2 for the x and getting a 16/3 instead of a 32/3 which changes things.
     
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