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Yet another volume

  1. Jan 14, 2006 #1
    Hi,

    I'm having a trouble doing this:

    Compute volume of the solid

    [tex]
    T = \left\{[x,y,z] \in \mathbb{R}^3; x \geq 0, y \geq 0, 0 \leq z \leq 1 - x - y\right\}
    [/tex]

    First I need to express bounds for [itex]x[/itex] and [itex]y[/itex], for [itex]z[/itex] I have it already. So because

    [tex]
    0 \leq z \leq 1 - x - y
    [/tex]

    then

    [tex]
    0 \leq x \leq 1 - z - y
    [/tex]

    and also

    [tex]
    0 \leq y \leq 1 - z - x
    [/tex]

    But that's probably not the right approach, because evaluating integral

    [tex]
    \iiint_{T}\ dx\ dy\ dz = \int_{0}^{1-z-x}\int_{0}^{1-z-y}\int_{0}^{1-x-y}\ dz\ dx\ dy
    [/tex]

    still lefts me with some [itex]z[/itex] and [itex]x[/itex] variables at the end...

    Will somebody point me to the right direction?

    Thank you.
     
  2. jcsd
  3. Jan 14, 2006 #2

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    twoflower, maybe this will help.
    The boundary of this region is given by the plane
    [tex] x+y+z=1 [/tex].
    To find the integral, for some x and y, z varies from 0 to 1-x-y (ie, till the top of the column containing the volume element).
    Now keep x constant and integrate wrt to y. This corresponds to adding the columns with base in the x-y plane (from 0 to the line y=1-x). So, y varies from 0 to 1-x.
    Finally you can integrate wrt to x, which by similar argument varies from 0 to 1.
     
  4. Jan 14, 2006 #3
    Btw what does "wrt" stand for? I would guess "with respect to", but it doesn't make sense to me in the previous post..
     
  5. Jan 14, 2006 #4

    arildno

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    Dearly Missed

    The simplest way to do this is first to recognize that the plane z=1-x-y meets the plane z=0 in the line x+y=1 (the line in z=0, that is)

    Now, regard the part of T lying in the plane z=0.
    This is the triangle bounded by the lines x=0, y=0, x+y=1, that is, we see that 0<=x<=1 and 0<=y<=1-x (alternatively, you could describe the triangle as 0<=y<=1, 0<=x<=1-y)
     
  6. Jan 14, 2006 #5

    Dale

    Staff: Mentor

    In general, if you know that you are going to get a definite volume, then you know that the limits on a given integral cannot depend on any of the inner variables of integration. So in your example, the limits for z can be a function of x and y, the limits for x can be a function of y but not z, and the limits for y cannot be a function of x or z.

    -Dale
     
  7. Jan 14, 2006 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You have
    [tex]T = \left\{[x,y,z] \in \mathbb{R}^3; x \geq 0, y \geq 0, 0 \leq z \leq 1 - x - y\right\}[/tex]
    and you write

    [tex]V= \iiint_{T}dV = \int\int\int dz\ dx\ dy[/tex]
    which indicates that you are going to integrate first with respect to z, then with respect to x, and finally with respect to y.
    (That's an arbitrary choice- doing the integrations in any order must give the same answer- although one way might be easier than another.)
    You are correct, of course, that the answer is a number which means that the "outside" integral, wrt y, must have only numbers as limits.
    How do you find them? Draw a picture! (That should always be the first thing you do with problems like this.) Since you are told that x and y will be larger than or equal to 0, you are looking at the first quadrant in the xy-plane. z= 1- x- y is a plane, of course. It crosses the xy-plane when z= 0= 1-x-y which is the line x+y= 1, through the points (0,1) and (1,0), in the first quadrant. That tells you everything you need to know. Overall, y can be as low as 0 or as high as 1. Your "outside" integral must be [itex]\int_0^1 dy[/itex].
    Now look at the "middle" integral- the one with respect for x. Since you will not yet have integrated with respect to y, the limits of integration may depend upon y. For each y what is the range for x? A fixed y is, of course, a horizontal line running from the vertical line x= 0 to the line x+ y= 1 or x= 1- y. For each y, x ranges between 0 and 1- y. The "middle" integral is [itex]\int_0^{1-y}dx[/itex].
    Finally, when doing the "inner" integral, we have not yet integrated with respect to x or y so the limits of integration may be functions of both x and y. For each x,y, what is the range for z? The plane z= 1- x- y slants up from the line x+y= 1 in the xy-plane to z= 1 at the z-axis (x=y=0). The crucial point is that the xy-plane (z= 0) is the "floor" and the plane z= 1- x- y is the "roof". For a given (x,y) point, z ranges from z= 0 to z= 1- x-y. That "inner" integral is [itex]\int_0^{1-x-y}dz[/itex]

    Personally, I think it helps to write "x= ", "y= ", "z= " in the limits of integration themselves. The volume is given by:
    [tex]\int_{y= 0}^{y=1}\int_{x=0}^{x=1-y}\int_{z=0}^{z=1-x-y}dzdxdy[/tex].

    It would be good practice for you to find the integrals in the other 5 orders, and do the integrations to see that they do indeed give the same answer!
     
    Last edited: Jan 14, 2006
  8. Jan 14, 2006 #7
    Thank you HallsoftIvy very much for comprehensible explanation! I already managed it as soon as I draw it :) First it seemed ugly to me to draw it but it turned out to be pretty straightforward. And then the expressing 'x' for fixed 'y' was easy.
     
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