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Boundaries of a triple integral

  1. Jan 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Let D = { (x,y,z) } such that x^2 + y^2 < 1 and -1 < z < 1 } denote the interior of a cylinder. Compute the triple integral of ( xyz dxdydz )


    2. Relevant equations



    3. The attempt at a solution

    Ok so it seems to me that the boundaries should be as follows,

    -1< x < 1
    -Sqrt [1 - y^2] < y < Sqrt [1 - y^2]
    -1 < z < 1

    However all the boundaries above are in the form of the integral between plus or minus (a)

    Now when I perform an integral on xyz i find that the since the a and -a are squared then their respective results are taken from each other you get zero regardless of order.
     
  2. jcsd
  3. Jan 8, 2012 #2
    Here are some comments that I think will help: -Sqrt [1 - y^2] and Sqrt [1 - y^2] are in terms of x, so if you were integrating with these bounds you would be integrating with respect to x, not y. If you had an equation and you integrated it from -Sqrt{1-y^2] to Sqrt[1-y^2], you would end up with a function of y, which is not what you want. I might be unclear, but does that make sense? You should try using polar coordinates when you're integrating along x and y; since you're integrating over a circle.
     
  4. Jan 8, 2012 #3

    SammyS

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    If you want to integrate in that order then notice that if x^2 + y^2 < 1, then [itex]\displaystyle -\sqrt{1-x^2}<y<\sqrt{1-x^2}\,.[/itex]

    The boundaries of D will be easier to determine in cylindrical coordinates. But you would need to express xyz in terms of r, θ, and z , although that's not too difficult.
     
  5. Jan 8, 2012 #4
    yeah sorry that was a typo, so if I integrate through z then y then x, should i get the result of 1/6, I'm new to this and have no way to check the answer.
     
  6. Jan 8, 2012 #5
    That's not what I got. Could you show your work?
     
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