The discussion revolves around visualizing a triple integral defined by the function xyz², with the volume V bounded by the planes y=1-x, z=0, and z=y. Participants are exploring how to represent this volume graphically.
The conversation is ongoing, with some participants suggesting that an additional bound is required for z to properly define the volume. Others are clarifying the implications of the existing bounds and how they relate to the graphical representation of the region.
There is an indication that the problem may be constrained by the requirement for a bounded volume, as the current setup allows for z to extend infinitely without a fourth boundary. Participants are also reflecting on the graphical representation of the defined region.
Draw a graph. x is from -1 to 1 so draw two vertical lines at x=-1 and x= 1. y if from 0 to 1- x^2 so draw the the line y= 0 and the parabola y= 1- x^2. The region to be integrated is inside that parabola above y= 0. Finally, the plane z= y crosses the y-axis up to (x, 1, 1) and your three dimensional region comes up to that. As Quinzio said, you need another bound on z or that region is not bounded. As it is, z could go up from that plane to infinity of down to negative infinity. Is there another limit, perhaps 0, on the z-integration?arl146 said:Homework Statement
the function is xyz2
V is bounded by y=1-x, z=0, and z=y.
The Attempt at a Solution
the limits are:
x is from -1 to 1 ?
y is from 0 to (1-x^2) ?
z is from 0 to y ?
the question asks for a picture ... how should that look? there are points on (1,0,0), (-1,0,0), (0,0,1) and (0,1,0) ?