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Solid region(Triple Integral)

  1. Nov 17, 2007 #1
    [SOLVED] Solid region(Triple Integral)

    1. The problem statement, all variables and given/known data
    Whats the region of the graph in dxdyz, dxdzdy, dydxdz, dydzdx, dzdxdy, dzdydx

    2. Relevant equations

    3. The attempt at a solution
    Attached a picture

    The question asked to evaluate the integrals in 6 different order of integration but I can do that. Just having problem with interpreting the region. Was also only given a picture of the graph and told to evaluate it as I said earlier.

    Attached Files:

  2. jcsd
  3. Nov 18, 2007 #2


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    I don't understand what you are saying. You say you can integrate in the 6 different orders but have a problem with "interpreting the region"? What do you mean by "interpreting"? The figure is obviously a tetrahedron with vertices at (4, 0, 0), (0, 8, 0) and (0, 0, 6). There is, in fact, a fairly simple formula for the volume.

    I suspect you mean that you don't know how to set up the integrals. I'll walk you through one and, hopefully, that will give you the general idea.

    Suppose you choose to integrate with respect to z first, then y, then z. That is:
    [tex]\int\int\int dzdydx[/itex]

    Since the "outside" integral is with respect to x, its limits of integration must be constants. In order to cover the entire figure, x must go from 0 to 4.

    In the next integral, with respect to y, the limits of integration may depend on x. For each x how must y vary? Well, projecting the figure down onto the xy-plane, we have the triangle with vertices (0,0), (4,0) and (0,8). For each x, y must range from 0 up to the line from (4,0) to (0,8). The equation of that line is x/4+ y/8= 1 or y= 8- 2x. For each value of x, y must range from 0 up to 8- 2x.

    Finally, for each (x,y) value, z must range from 0 up to the plane forming the upper boundary. That plane goes through (4,0,0), (0,8,0) and (0,0,6). It has equation x/4+ y/8+ z/6= 1 or z= 6- (3/2)x- (3/4)y.

    The volume is given by
    [tex]\int_{x=0}^4\int_{y=0}^{8-2x}\int_{z= 0}^{6-(3/2)x-(3/4)y} dzdydx[/itex]

    Only five more to go!
  4. Nov 18, 2007 #3
    Thanks alot for responding but I tried it again since I though no one would get to this in time and I got it by myself. The answer turned out to be 32. Only difficulty I had with this was interpreting the region. Everything else was just algebra. Thanks again though for responding.

    Oh just read what you said. What my problem was was that I couldnt see the plane for which I had found the equation for projecting down on the plane but as you could see from my work I did think of it. I just imagined shining a light on the surface and then visualize the resulting orthographic view.
    Last edited: Nov 18, 2007
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