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Boundary values for integral

  1. Aug 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the volume between z = x2 + y2 and z = 2 - (x2 + y2).


    2. Relevant equations



    3. The attempt at a solution
    if r2 = x2 + y2
    then the lower part of the volume is defined by:
    r2 [itex]\leq[/itex] z [itex]\leq[/itex] 2 - r2
    and: 0 [itex]\leq[/itex] r [itex]\leq[/itex] 1
    the upper part by:
    2 - r2 [itex]\leq[/itex] z [itex]\leq[/itex] r2
    and: 1 [itex]\leq[/itex] r [itex]\leq[/itex] [itex]\sqrt{2}[/itex]

    [itex]\int\int\int[/itex]1 dxdydz, after switching to polar coordinates I get
    [itex]\int\int\int[/itex]r drd[itex]\Theta[/itex]dz

    Theta varies from 0 to 2 pi. That leaves me with taking the integral with respect to r and z.
    I do it for z first, then finally for r. Then add the two volumes. But it's wrong.

    Any ideas?
     
  2. jcsd
  3. Aug 8, 2011 #2
    those 2 curves intersect at x=1 and the plane z=1 , I would break it up into 2 parts and go from z=1 to the top of the paraboloid and then do the second part from the bottom paraboloid to the plane z=1 , because if you don't do that your radius is not always going to the same thing. Actually now that I look at it you kinda did that.
     
  4. Aug 8, 2011 #3

    tiny-tim

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    Hi Inertigratus! :smile:

    (have a square-root: √ and a pi: π and a ≤ :wink:)
    nooo :redface:, that's the whole of the volume :wink:
     
  5. Aug 9, 2011 #4
    Hehe, now you got me confused... :P
    It's the whole volume?
    I split it, because when r = 1, then 1 ≤ z ≤ 1, if you go up to r = √(2) then 2 ≤ z ≤ 0, which looks a bit weird since 0 is not greater than 2...
    So lower part from 0 ≤ r ≤ 1, upper part from 1 ≤ r ≤ √(2)
    Is that wrong? :P
     
  6. Aug 9, 2011 #5

    HallsofIvy

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    I guess the question is: what do you mean by the "lower part"? What you have, [itex]2- r^2\le z\le r^2[/itex], is incorrect because the paraboloid [itex]z= 2- r^2[/itex] is above the parabolid [itex]z= r^2[/itex].
     
  7. Aug 9, 2011 #6
    It's confusing...
    Because, I know that when r = 0, then 2−r2 is obviously larger than r2.
    What's confusing me is that when r is increasing, and reaches r2 = 2, then the inequality doesn't make any sense. 2 ≤ z ≤ 0 makes no sense.
    So I was thinking as I said in my first post, that I would have to split it in to volumes.
    Above z = 1 and below z = 1.
     
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