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Triple integral changing order of integration

  1. Nov 9, 2015 #1
    1. The problem statement, all variables and given/known data
    rewrite using the order dx dy dz
    [itex] \int_0^2 \int_{2x}^4\int_0^{sqrt(y^2-4x^2)}dz dy dx [/itex]

    3. The attempt at a solution
    I am having trouble because i don't know what the full 3 dimensional region looks like but the part on the xy plane is a triangle bounded by x = 0 , y = 4 and y = 2x
    In this region z goes from 0 to 4 so I think those will be the limits of integration with respect to z , but I am having trouble finding the others. i know x goes from 0 to y/2 but I don't think that will be the correct limits of integration with respect to x here. I would greatly appreciate if someone can help me get started on this one.
     
  2. jcsd
  3. Nov 9, 2015 #2

    Mark44

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    You can pretty much read off the constraints from the limits of integration.
    Workeing from z to y to x, the region is ##0 \le z \le \sqrt{y^2 - 4x^2}##, ##2x \le y \le 4##, and ##0 \le x \le 2##.

    z does NOT go from 0 to 4 as you say below.
     
  4. Nov 9, 2015 #3

    andrewkirk

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    unless I'm making a mistake, the points (0,0,0) and (0,4,4) are both in the region being integrated.
     
  5. Nov 9, 2015 #4

    andrewkirk

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    Label as ##z_{max}## the upper integration limit of z for any x,y combination, and then write an equation connecting ##x,y,z_{max}##. That equation defines the boundary of the volume being integrated. If you are familiar with the equations of conic sections that should give you an idea of the shape of the volume.
     
  6. Nov 9, 2015 #5

    Mark44

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    That might be, but from the original integration limits, the interval for z is between z = 0 and ##z = \sqrt{y^2 - 4x^2}##, which is an elliptical paraboloid. My point was that the interval along the z-axis is not between the planes z = 0 and z = 4, as the OP implied.
     
    Last edited: Nov 9, 2015
  7. Nov 9, 2015 #6

    andrewkirk

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    I took that reference in the OP to instead mean that, when the order of integration is reversed, the limits of the outer integral will be ##z=0## to ##z=4##, with which I agree. But I suppose either interpretation can be drawn from what was written.
     
  8. Nov 9, 2015 #7

    Mark44

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    I was looking at this from the perspective of the shape of the region, as defined by the original limits of integration.
     
  9. Nov 9, 2015 #8

    LCKurtz

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    @toothpaste666: You aren't likely to solve this problem until you draw at least a first octant sketch. I would suggest you draw the traces in the ##xy## plane and the ##yz## plane for starters. Then draw the traces in the planes ##y=0,~ y=1##, and ##y = 4## in the first octant. That should give you enough to see what the shape of the figure is. It is not a portion of an elliptic paraboloid as has been suggested. Once you have that, come back with your try at the limits for ##dxdydz##.
     
  10. Nov 9, 2015 #9

    Mark44

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    By me, in error. It's elliptic, but not a paraboloid.
     
  11. Nov 10, 2015 #10
    ok I may be completely off but it is starting to look like part of a cone and I think the limits of integration for x will be z/2 to y/2 . then i think the limits for y should be z to 4 because z=y on the zy plane and y = 2x and then I think z will be integrated from 0 to 4
     
  12. Nov 10, 2015 #11

    LCKurtz

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    That is correct. It's part of an elliptic cone, located in the first octant.

    No. Think of a point in the interior of the volume. Move it in the ##x## direction. What is ##x## on the back surface? What is ##x## on the front surface?
     
  13. Nov 10, 2015 #12
    0 to y/2 ?
     
  14. Nov 10, 2015 #13

    LCKurtz

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    No, I think you are just guessing now. You have the equation, and presumably a rough sketch, of the surface. What is ##x## on the front surface?
     
  15. Nov 10, 2015 #14
    I don't want to just guess so first i want to make sure my picture is correct. This is what I have drawn:

    IMG_1109.JPG
     
  16. Nov 10, 2015 #15

    LCKurtz

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    Normally we draw xyz graphs in a right handed coordinate system, unlike your picture. Using your picture, the question would be what is x on the left and x on the right, on the surface, not the xy plane.
     
  17. Nov 10, 2015 #16
    it wouldn't be from 0 to y/2? or wait that would only be on the xy plane? so it would be from 0 to (1/2)sqrt(y^2-z^2)
     
  18. Nov 10, 2015 #17

    LCKurtz

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    Bingo! And your y and z limits you had gotten from the yz plane are correct.
     
  19. Nov 10, 2015 #18
    Thank you all so much!
     
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