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Double Integral Volume

  1. Nov 17, 2017 #1
    1. The problem statement, all variables and given/known data
    z=x^2+xy ,y=3x-x^2,y=x find the volume of the region

    2. Relevant equations


    3. The attempt at a solution
    I graphed y=3x-x^2 and y=x I am confused on which region I use to find the volume. Do I use the upper region or the lower region.
     

    Attached Files:

  2. jcsd
  3. Nov 17, 2017 #2

    berkeman

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    Staff: Mentor

    Doesn't the question give 3 equations as constraints for the volume enclosed? Can you do a 3-D graph of all 3 equations?
     
  4. Nov 17, 2017 #3
    Can't do a 3D graph the two constraints are y=x and y=3x-x^2 , I use the z=x^2+xy to find the volume using the double integrals just having trouble with the set up.
     
  5. Nov 17, 2017 #4

    LCKurtz

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    I would assume you want the upper region. The lower region has a boundary portion of ##y=0## which is not mentioned in the problem.
     
  6. Nov 17, 2017 #5

    Ray Vickson

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    In your (x,y)-space there are four regions between the blue and red curves: (i) the south-west region, in which x and y can both go to -∞; (ii) the southern region, in which x can go to ±∞ but y can just go to -∞; (iii) the north-west region, in which x and y can go to +∞; and (iv) the north region, in which x and y are both bounded. Regions (i)--(iii) have infinite areas, which will lead to infinite volumes when we add a third dimension; only region (iv) gives a finite answer.
     
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