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Triple Integral setup problem

  1. Apr 10, 2005 #1
    Triple Integral setup...

    [tex] \int \int \int_{G} 6x (z+y^3) dx dy dz [/tex] G bounded by [tex] x = 0, \ x = y, \ z = y-y^2, \mbox{and} \ z=y^2 - y^3 [/tex]
    x from 0 to1
    y from 0 to x
    z from z=y-y^2 to y^2 - y^3
    and the integration order becomes dz dy dx
    would this give the right answer?

    what aboiut this one
    [tex] \int \int \int_{G} xy + xz dx dy dz [/tex]

    G bounded by z = x, z=2-x, z = y^2
    z goes from 2-y^2 to y^2
    y goes from 2-x to x
    x goes from 0 to 2
    and the integration order to dz dy dx

    I think the second one is wrong. Please do help!!
  2. jcsd
  3. Apr 10, 2005 #2
    Can you explain your reasoning for 1? Particularly the x and y. What exactly are you trying to do in these problems?
  4. Apr 10, 2005 #3
    the question is evaluate the integral over the given bounded region G

    for the first one z goes like the y function, that s fine
    for the y takes on a min value of 0 and max of x and
    for the x the min value is zero but im not sure about the max value ..
    Last edited: Apr 10, 2005
  5. Apr 10, 2005 #4
    Think of x=0 and x=y as planes, not as lines.

    Also, x takes on a min value of 0, and a max value of y. You don't know much about y.

    Try imagining the region G in your head, its bounded by the yz plane, and the plane y=x and the z function.

    You want to find how far y and x go given your constraints. Solving the functions would not be a bad idea.
    Last edited: Apr 10, 2005
  6. Apr 10, 2005 #5
    if i were to solve what am i solving for???
    which functions would i use?

    x=0, x= y, z = y - y^2 , z = y^2 - y^3
    the intesection of which surfaces??
  7. Apr 10, 2005 #6
    Theres only two equations that could be solved here. You have two functions 'z' in R^3. For a triple integral, you want to find the domain of these curves on the xy plane, where z=0. Solve the z functions to find the range of the y function.

    http://tutorial.math.lamar.edu/AllBrowsers/2415/TripleIntegrals.asp [Broken] Example 2
    Last edited by a moderator: Apr 21, 2017 at 2:30 PM
  8. Apr 10, 2005 #7
    ok i got the y ranges from 0 to 1
    now for the x part since x goes from 0 to y, x goes from 0 to 1
    but y has to be a function of x so y goes from 0 to x?? and then x from 0 to 1?
    Last edited: Apr 10, 2005
  9. Apr 10, 2005 #8
    [tex] \int\int \int_{G} 6x (z+y^3) dx dy dz [/tex] = [tex] \int\int_{D} \int_{y-y^2}^{y^2-y^3} 6x(z+y^3)dz dA[/tex]

    http://www.public.asu.edu/~hyousif/maple.JPG [Broken]
    http://www.public.asu.edu/~hyousif/xzplane.JPG [Broken]

    edit: Its the yz plane, not the xz plane.

    Heres the x-y plane, which is your domain:
    Notice its a triangle with vertices at (0,0) (0,1) (1,1) You can express this as a domain pretty easily.

    http://www.public.asu.edu/~hyousif/xyplane.JPG [Broken]
    Last edited by a moderator: Apr 21, 2017 at 2:31 PM
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