Volume of Bounded Region: Solving Limits for Multiple Integration

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Homework Help Overview

The discussion revolves around determining the limits for a multiple integration problem related to finding the volume of a region bounded by the equations x + 2z = 4 and 2y + z = 2, with the constraints that x, y, and z are all non-negative.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss potential limits for the variables x, y, and z, with some uncertainty about the upper limit for z and the order of integration. There is a suggestion that for any z >= 0, corresponding x and y can be found, leading to a debate about the feasible range for z.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the boundary conditions. Some guidance has been offered regarding the limits for x and y, but the upper limit for z remains unclear, with conflicting views on whether it can extend to infinity or if it is bounded.

Contextual Notes

Participants note multiple boundary conditions that must be satisfied, and there is a recognition that selecting a value for z beyond a certain limit may not be valid, indicating a need for further clarification on the constraints.

windy906
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I'm having problems finding the limits for the follwing integration, can anyone help?

Find the volume bounded by x+2z=4 and 2y+z=2 with x,y and z>=0
 
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I think the following limits apply but can't find the upper one for z, or in which order they should go.

x: 0 -> 4-2z
y: 0 -> 1-0.5z
z: 0 -> ?
 
It seems like if you pick any z>=0, you can find x,y that satisfy the boundary condiitions. Hence, z should be between 0 and [tex]\infty[/tex]
 
Office_Shredder said:
It seems like if you pick any z>=0, you can find x,y that satisfy the boundary condiitions. Hence, z should be between 0 and [tex]\infty[/tex]

There are five boundary conditions:
  1. x+2z<=4
  2. 2y+z<=2
  3. x>=0
  4. y>=0
  5. z>=0

Picking a value for z beyond some finite limit will not work.
 
So z <= 2!

Thank you so much.
 
Last edited:

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