The problem involves finding the bounds for a triple integral of the function xy over the solid region D, which is defined by the coordinate planes, the plane x = 1, and the surface z = 16 - 4x^2 - y^2. The context is centered around understanding the geometric interpretation of the region involved in the integration.
The discussion is ongoing, with participants exploring different interpretations of the region and questioning the initial assumptions about the bounds. Some guidance has been offered regarding the need to consider elliptical cross-sections of the paraboloid and how to approach the integration limits based on these slices.
There is a noted difficulty in visualizing the three-dimensional aspect of the problem, particularly in determining the correct bounds for z and how the solid is shaped by the coordinate planes. Participants are also considering the implications of the plane x = 1 on the overall bounds.
Mark44 said: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.
Cyosis said: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. Let's simplify the problem a little and ignore the plane x=1. Then we're left with a paraboloid with an elliptical cross section.
Now let's 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.
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).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.