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Finding Bounds for a Triple Integral

  1. May 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Find [tex]\int \int \int_{D} xydV[/tex], where D is the solid bounded by the coordinate planes, the plane x = 1 and the surface [tex]z = 16 - 4x^2 - y^2[/tex].

    2. Relevant equations



    3. The attempt at a solution

    I have no problem with actually performing the integration, but I'm lost on figuring out the bounds. Any help/suggestions/tips?

    Thanks so much!
     
  2. jcsd
  3. May 8, 2009 #2

    Mark44

    Staff: Mentor

    Have you sketched a graph of the region? The surface z = 16 - 4x2 - y2 is a paraboloid with elliptic horizontal cross sections, and with vertex at (0, 0, 16).

    Many times the most difficult part of these double and triple integrals is determining the limits of integration, and that is usually much more difficult if you don't have a good idea of exactly what the region you're integrating looks like.
     
  4. May 9, 2009 #3
    I looked at the region. Would the bounds for dx be from 0 to 1, dy from 0 to 16, and dz from 0 to 1 as well?
     
  5. May 9, 2009 #4

    Cyosis

    User Avatar
    Homework Helper

    Are you sure you looked at the region? The region you describe is a box with sides 1, 16 and 1. Where as the region D is, as said before, a paraboloid with an elliptical cross section with a slice taken off parallel to the zy-plane. Lets simplify the problem a little and ignore the plane x=1. Then we're left with a paraboloid with an elliptical cross section.

    Now lets replace the function xy by 1 (once we've found the volume of the paraboloid we can just replace 1 by xy and add the x=1 plane to get the correct answer). Now the integral over region D' is the volume of the paraboloid z=16-x^2-y^2 bounded by the coordinate planes.

    A few questions about the properties of this region.
    1)Does the parabola have a maximum or minimum?
    2)What are the coordinates of its maximum/minimum?
    3)What kind of shape does the surface of the base of the parabola have? (circle/ellipse)
    4)What is the radius or semi major/semi minor axis of this circle/ellipse?
    5)Looking at the paraboloid from z=0 and up, the coordinate planes cut the paraboloid in how many pieces? Is it a whole, half or quarter paraboloid?


    If you can answer these questions (hopefully they are clear), then you have properly looked at the region. Write down the volume integral for the paraboloid described above, don't bother calculating it. Now add the x=1 plane, how does this change the boundaries? And finally integrate xy over the region D to get the answer you are looking for.
     
  6. May 10, 2009 #5
    I don't know what I was thinking with those bounds I guessed.

    After looking at the graph more, I see that the base is an ellipse and that the major axis is 4 and minor is 2. This leaves me to believe the bounds for dx would be from -2 to 1 (because of the line x=1 if we do not ignore it), and the bounds for dy would be from 0 to 4 (taking two times the integration for that)? I don't know about dz, however, I really have trouble imagining the third dimension.
     
  7. May 11, 2009 #6

    Mark44

    Staff: Mentor

    No, you're still thinking boxes. If you slice the paraboloid into horizontal cross sections, each slice is an ellipse, with the ellipses getting smaller as you go up (as z increases). None of your slices is an ellipse with major axis 4 and minor axis 2, since your solid is bounded below by the plane y = 1. Actually, each slice is a quarter of an ellipse, since the solid is bounded by the x-z plane (the plane y = 0) and the y-z plane (the plane x = 0).

    This means that you are going to need to pick a typical elliptic slice (at a given z-value), and solve for expressions that give you the x value on one side of the ellipse, and for the y value on the other side of the ellipse. The z values are easy: they go from 1 to 16.
     
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