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Homework Help: Triple Integral Help

  1. Nov 13, 2007 #1
    Triple Integral Help :(

    Can anyone help me with this triple integral problem? I'm sorry I don't know how to post the script properly; I'm a complete newb.

    It's a surface integral problem- that part is not important- I have to calculate a triple integral where S is the surface of the volume bounded by z=1-x^2 and the planes z=0, y=0, and y+z=2

    I am having difficulty setting this up- if anyone can post any relevant information, or direct me to any sites that are phenomenal in helping with triple integrals, in would be greatly appreciated.
  2. jcsd
  3. Nov 13, 2007 #2


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    What are the variables of integration for the 3 integrals?
  4. Nov 13, 2007 #3
    That's where I'm having the problem, finding the bounds of the three integrals. The computation itself I feel as though I can work out, if I have the right bounds of integration
  5. Nov 13, 2007 #4
    If anyone can calculate the triple integral of 3y for the bounds that I mentioned above, let me know if you get 144/35....if you want to do that for fun? ahaha :(
    The bounds that I used were x ranges from -1 to 1, Z ranges from 0 to 1-x^2, and y ranges from 0 to 2-Z. Does anyone else htink that this looks correct?
  6. Nov 14, 2007 #5


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    I hadn't had time to look at this before. z= 1- x^2 is a "parabolic cylinder" with axis running along the y-axis. It's highest point (z max) comes when x= 0 and is at z= 2. y+ z= 2 is a plane crossing the xz-plane in the line z= 2 (above the parabola) and the xy-plane at y= 2. The plane crosses the parabolic cylinder when z= 2- y = 1- x^2 or y= x^2+ 1. The fact that that does not cross the y= 0 threw me for a moment! What that means is that you have to do the integral in two different parts. As long as we are below the parabola y= x^2+ 1, our upper boundary is z= 1- y^2, the parabolic cylinder. After that the upper boundary is z= 2- y, the plane.
    You will need to do this as two integrals. The first integral will have x from -1 to 1, for each x, y from 0 to x^2+ 1, for each (x,y), z from 0 to 1- y^2. The second integral will have x from -1 to 1, for each x, y from x^2+ 1 to 2, for each (x,y), z from 0 to 2- y.
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