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Sketching a triple integral

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Here is the question along with the solution and sketch.
    WCyAe4Y.png

    I think the sketch is wrong because the projection in the xy plane shows a rectangular box. I don't think it is a rectangular box because you can solve for an equation relating x and y.

    You know that y = 1-z and x = 1-z^2, so you can solve for some equation with x and y by eliminated z. Which gets you x = -(y-1)^2 + 1 for the relationship in the xy plane and it is clearly not a rectangle...?
     
  2. jcsd
  3. Feb 28, 2013 #2

    HallsofIvy

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    Staff Emeritus
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    That last equation is, of course, x= y(2- y) but it is the projection of the boundary of the y= 1- z and x= 1- z^2, NOT the base itself. The sketch is correct.
     
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