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Changing order of integration

  1. Apr 1, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int^{4}_{0}\int^{(4-x)/2}_{0}\int^{(12-3x-6y)/4}_{0}dz*dy*dx[/tex]

    Rewrite using the order dy dx dz.

    2. Relevant equations



    3. The attempt at a solution

    I have it graphed, but from here, I am having a hard time visualizing the bounds of all these dimensions to reorder the integral. My graph is attached.

    The only thing I can determine for sure is that the bounds in the z direction will be 0 to 3. The book's answer is this:

    [tex]\int^{3}_{0}\int^{(12-4z)/3}_{0}\int^{(12-4z-3x)/6}_{0}dy*dx*dz[/tex]

    I can somewhat make sense of the dy bound; they solved the z equation in terms of y, although I can't really see/visualize the reason for this.

    I don't understand the x bound at all, why does the y part disappear? It seems they solved the z equation to get it in terms of only z but I don't see why the y component disappears, or why they would even do that in the first place.
     

    Attached Files:

  2. jcsd
  3. Apr 1, 2010 #2
    x: [0, 4]
    y: [0, (12-3x-4z)/6]
    z: [0, (12-4z)/3]

    12=3x+6y+4z

    Solve for y. (This step is for the y the bounds)

    Then make y=0 and solve for x (This is independent from above so you start with 12=3x+6y+4z when you do this step.)
     
  4. Apr 1, 2010 #3

    vela

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    Ignore the inside integral for the moment. You can think of the problem as an integral of a function over the xz plane.

    [tex]\int\int f(x,z) dx dz[/tex]

    where f(x,z) is your integral over y. If you look at your plot, you see the region on the xz plane is a triangle, so how would you write the limits for the region if your inside integral is wrt x and the outside integral is wrt z?
     
  5. Apr 1, 2010 #4
    Well for x, I think it would go from 0 to something like z = -3/4x+3? And wrt z, 0 to 3?

    But it still doesn't quite explain the dy?
     
    Last edited: Apr 1, 2010
  6. Apr 1, 2010 #5
    when the integration is dydzdx

    dx can't have any variables because if there was a y or z present, how could we solve for it since we already integrated dz?

    From your first integral, you told us (12-3x-6y)/4 as the upper bounds.

    That means z=(12-3x-6y)/4. By putting it in standard form, we obtain 12=4z+3x+6y.

    If you notice in order to obtain the z bounds, we had to solve for z. Now, since we need the first bounds of integration to be in terms of y, we should take the plane and solve for y.
     
  7. Apr 1, 2010 #6
    Back to the x bounds.

    Since we all ready integrated dydz, we need to look at the x axis. What is happening on the x axis? Well x starts at 0 and goes to 4. I know that from the plane. If we set y and z to zero, we get 12=4*0+3*x+6*0 => 12=3x => x=4. The problem giving previously bounded the object in the first octant so we know x=0 is the other point.
     
  8. Apr 1, 2010 #7

    vela

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    You want to solve for x instead of z, but I think you know that.

    For the y limits, you now consider x and z fixed and find what range y covers. It starts at the xz plane, where y=0, and goes up to the plane.
     
  9. Apr 2, 2010 #8
    Once I have solved for x, would I plug this back into the original z bound? That's the only way I get the book answer. Why does that work?

    Also, I'm not quite sure I can "see" that the y goes up to the plane exactly?
     
  10. Apr 2, 2010 #9
    Since the bounds are dy dx dz, once you have solved for x, you would plug the bounds into the corresponding integral. And the integral that is associated with x is the 2nd integral, so I am not 100% I understand what you mean by the original z bounds because the origin z bounds were the first integral which is now the 3rd integral.

    I assuming I know Vela was saying but I am pretty sure Vela doesn't mean going up in the sense of rising. He is saying y goes from the origin and travels down the y axis until it meets the plane 12=3x+6y+4z.
     
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