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Triple Integral to find vol.

  1. Mar 14, 2012 #1
    Can someone help me with this?


    1. The problem statement, all variables and given/known data
    Find the volume V of the solid S bounded by the three coordinate planes, bounded above
    by the plane x+ y+ z = 2, and bounded below by the plane z = x+ y.


    2. Relevant equations
    x + y + z = 2
    z = x + y


    3. The attempt at a solution
    [itex]\int_{0}^{2}\int_{-x}^{2-x}\int_{x+y}^{2-y-x}dzdydx[/itex]
    So I used the above triple integral and got -4(did it twice), wolfram-alpha's calculator gives me 0 and the textbook answer is 1/3

    So obviously I did something wrong in the triple integral and in identifying the limits.
    But I just want to know the right limits for the above question as I have problems identifying them (I drew traces(attached) on the planes but not sure if the region's right)
     

    Attached Files:

  2. jcsd
  3. Mar 14, 2012 #2

    SammyS

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    Hello hqjb. Welcome to PF !

    Where do the planes, x+ y+ z = 2 and z = x+ y intersect ?
     
  4. Mar 14, 2012 #3
    Hey thanks.
    I suppose thats more of a hint than an answer right?

    I think you'll get y=1-x,
    tried and got the right answer with [itex]\int_{0}^{2}\int_{0}^{1-x}\int_{0}^{2-y-x}dzdydx[/itex]
    but Im not sure why
     
    Last edited: Mar 14, 2012
  5. Mar 14, 2012 #4

    SammyS

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    The lower bound for the z integral is wrong --- should be x+y .

    You will often get hints and other guidance, so that you will then understand how to solve a problem.
     
  6. Mar 14, 2012 #5
    Yes, Im hoping to understand this too. That wasn't an attempt to answer but a random integral that got the answer lol.

    I finally got it after plotting the thing in 3d. my x-y trace was wrong it should be the line of intersection between the two planes.

    Why can't it be if I let z=0 for both equations and draw the region?
     
  7. Mar 14, 2012 #6

    SammyS

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    The region only touches the xy-plane (z = 0) at a single point, the origin.
     
  8. Mar 14, 2012 #7
    Alright understood, so i shouldnt use the x-y trace in this case because there isnt one.
    And the bounds are 0 < y < 1-x , 0 < x < 1, x+y < z < 2-x-y ?

    Thanks for your patience and help.
     
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