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Homework Help: Triple Integral-Calculus

  1. May 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Hey there, I'll be delighted to get some help in the following question:
    Let A be the region in space bounded by the next planes:
    [tex] x=1[/tex], [tex]x=2[/tex], [tex]x-y+1=0[/tex],
    [tex]x-2y=2[/tex], [tex]x+y-z=0[/tex] , [tex]z=0[/tex]...

    Write the integral [tex] \int \int \int_{A} f(x,y,z) dxdydz [/tex] as shown in the theorem above.

    The problem is I can't figure out how the region A looks like...
    Hope you'll be able to help me dealing with this question...

    Thanks in advance

    2. Relevant equations
    Let E be a closed region with a surface in R^2 and let [tex] g^1, g^2[/tex] be two real functions, continous in E. Let's look at A:
    [tex] A=( (x,y,z)|(x,y) \in E, g^1(x,y)\leq z \leq g^2(x,y) [/tex]. Then if f is a continous function with 3 variables, continous in A, then:
    [tex] \int \int \int_{A} f(x,y,z) dxdydz = \int \int_{E} (\int_{g^1(x,y)}^{g^2(x,y)} f(x,y,z)dz) dxdy [/tex]...

    3. The attempt at a solution
  2. jcsd
  3. May 17, 2010 #2


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    Here's a picture to help you get started. It isn't to scale; the x axis is stretched. You will need to be careful with your limits, especially if you are required to do x first. The top plane is in red, the bottom in brown, and the others are "see through".

  4. May 17, 2010 #3

    Perhaps it would help if you wrote the equations as:

    [tex] x=1[/tex]



    Those are all actually surfaces in 3D space. Can you just make rough sketches of those using the standard "isometric" axes: horizontal x-axis, diagonal y-axis into the page, and vertical z-axis. You know what I mean. For example, using that axis setup, for the first two, I would just draw a straight line at x=1 parallel to the y-axis, then another one parallel but at x=2. Now, using that same setup, draw a line y=2x+1. That going to start at the point (0,1), then another point at say (2,3) but draw it using that isometric axes. If you don't know what that is, try and find out. Do the same for y=1/2x-1. Keep in mind these are all surfaces and we're just drawing the cross-section at z=0. Try and understand that, then try and learn to draw it in Mathematica or other software.

    Also, any chance you typed the differentials incorrectly? That looks way difficult that way but is much easier as dzdydx. Looks to me anyway. :)
    Last edited: May 17, 2010
  5. May 18, 2010 #4
    Hey there,

    I'm not sure if I'm right, but the answer might be:
    E=( (x,y) | 1 \le x \le 2 , 1+\frac{x}{2} \le y \le 1+x ) [/tex]
    [tex] g^1(x,y)=0,g^2(x,y)=x+y [/tex] ??? (In the wording of the theorem I've quoted)

    Am I right?

    Thanks !
  6. May 18, 2010 #5
    I drew it and the way it looks to me, to solve it using a single triple integral, you have to solve it using dzdydx. Now you have the limits for x and z correct for doing it that way but not for y. Integrate first in the z direction from zero to the surface z=x+y, then integrate in the y-direction between the lines y=1/2x-1 to the line y=2x+1, then finally integrate in the x-direction between the points 1 and 2. For example, if f(x,y,z) is one, then I get for the volume of the region 117/8.
  7. May 18, 2010 #6
    Yes but I don't need to solve or calculate anything-just to write the triple integral in the way the theorem I've quoted does...
    This specific integral is of course :
    [tex] \int_{0}^{x+y} dz \int_{0.5x-1}^{x+1}dy /int_{1}^{2} f(x,y,z)dx [/tex] ...But in the theorem I've quoted we need to find a region E and two function g1&g2...

    Hope you'll be able to verify my calculation above

  8. May 18, 2010 #7
    Nvm...I understand it now

    Thanks a lot!
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