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Volume enclosed by two paraboloids

  1. Apr 2, 2006 #1
    Hello, I have to find the volume enclosed by two paraboloids -
    [tex] z = 9(x^2+y^2) [/tex] and [tex] z = 32-9(x^2+y^2) [/tex]

    I found the limits of integraion by setting them equal to each other. The problem Im having is what function do I integrate?
    The example my teacher gave, he integrated [tex]32-9(x^2+y^2) - 9(x^2+y^2) [/tex], the difference of the two paraboloid equations. Im sure this is right, but I dont understand why. Wouldnt integrating the difference of the paraboloids be the total volume minus the volume enclosed by the two paraboloids?:confused:
     
  2. jcsd
  3. Apr 2, 2006 #2
    not if you take the differance in the right direction
     
  4. Apr 2, 2006 #3
    The way I have it above, it is the volume of the downward opening paraboloid minus the volume of the upward opening paraboloid. This would leave the downward opening paraboloid minus its top.

    Switching the order of the difference would give me the volume of the upward opening paraboloid minus the volume of the downward opening parabolodi. This would leave teh upward opening paraboloid minus its bottom.

    or not?
     
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