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Triple integral under a wedge

  1. Sep 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the triple integral of z where E is bounded by the planes z=0 y=0 x+y=2 and the
    cylinder z^2+y^2=1 in the first octant.

    2. Relevant equations

    3. The attempt at a solution
    Just want to make sure that my setup is right. The limits of integration of x are 2 to 0,
    for z, sqrt(1-y^2) to 0, and for y, 2-x to 0.
  2. jcsd
  3. Sep 5, 2008 #2


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    Your z limits are not correct. It should go from 0 since it's bounded below by the xy plane to sqrt[1-y^2], limits of y should be reversed just like z.
  4. Sep 6, 2008 #3
    The only reason why I integrated from sqrt(whatever) sorry lol is because the partial circle is "higher" than the xy plane.
    Not trying to argue here, but can you explain why this is wrong? Srill don't quite get it.
    Last edited: Sep 6, 2008
  5. Sep 6, 2008 #4
    Sorry for the double post, but I found that doing it my way gets a negative answer. Is that the reason? Oh and could someone please draw the diagram just so that I know what the graph of this should actually look like.
  6. Sep 6, 2008 #5


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    Upon closer inspection I realised that the limits for y are wrong. The upper limit for the y-integrand is either 2-x or sqrt[1-z^2] depending on where you draw the line parallel to the y-axis through the required volume. Try changing the order of the integration.

    Since y is the problematic variable, let it be the last order of integration you perform. The limits for the other 2 variables are unambiguous.
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