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Simple volume question

  1. Apr 7, 2005 #1
    You have three continuous parametric functions (in the 3D coordinate system),
    one in the xy-plane, one in the xz-plane, and the other in the yz-plane.

    Then, you connect coordinates (x,t) with (y,t) with (z,t),..forming a triangle "slice" with points from these three planar functions for any given "t" (same for each point). One specific triangle for each t-value.

    Then, I integrate these triangles over a t-interval from t=a to t=b. How do I find the value of this integral ? Next, how do I find the volume integrated by these triangles in the t-interval from t=a to t=b ?
  2. jcsd
  3. Apr 8, 2005 #2


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    It seems you're saying we have functions f, g, h : R -> R³ such that f(t) is in the x-y plane, etc. Then, for each time t, we have a triangle T(t) whose edges are the segments f(t) to g(t), g(t) to h(t), and h(t) to f(t). (You wrote some stuff about (x,t), (y,t), (z,t) but that didn't make a whole lot of sense). So, to find the area of the triangle T(t), compute:

    |1/2{[f(t) - g(t)] x [f(t) - h(t)]}|

    This will give you half the area of the parallelogram with 1 arm going from f(t) to g(t), and the other going from f(t) to h(t), which is precisely the area of the triangle in question. You should know this property of the cross product and its relation to the area of a parallelogram.

    Next, you compute:

    [tex]\int _a ^b \frac{1}{2}\left |\left [f(t) - g(t)\right ] \times \left [f(t) - h(t)\right ]\right |\, dt[/tex]
  4. Apr 8, 2005 #3
    H-that's right (i should have see that);
    but how would one find the volume
    integrated by these triangles?
    (the Total volume, no overlaps)
  5. Apr 8, 2005 #4
    If these are parametric funtions wouldn't we have to consier an x1=f(t) x2=g(t) y1=h(t) y2=s(t) z1=u(t) z2=v(t)?
  6. Apr 8, 2005 #5


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    Uhh... the first equation I gave you (in plain text) was the equation for the area of a given triangle, and the second one (in LaTeX) was the equation for the volume. Let A(t) represent the area of the triangle at time t. Then you should be able to see that the second equation given is:

    [tex]\int _a ^b \frac{1}{2}\left |\left [f(t) - g(t)\right ] \times \left [f(t) - h(t)\right ]\right |\, dt = \int _a ^b A(t)\, dt[/tex]

    You should be able to see easily why this should give you the correct volume. The truth is, actually, that this need not give you the correct volume, as it assumes that there is no overlap between "consecutive" triangles (you should note that the above formula essentially tells you to add the volumes of infinitely many infinitessimally thin triangular prims with cross-sectional area A(t) and thickness dt, you should be used to this type of interpretation for finding volumes of solids, and if we replace A(t) with height h(t), you should be used to this interpretation for finding areas in the plane). Actually, it assumes there is no overlap between any pair of triangles, but the condition that the functions are continuous are not enough to claim that no two triangles intersect. I can't think of a general way to do this, perhaps you are to assume that they don't intersect, or that the intersections make negligible difference to the volume, and thus the volume can be computed straightforwardly as per the formula above.
  7. Apr 8, 2005 #6
    That's what I was trying to figure out--the volume generated in case I do consider possible overlaps in volume between ANY multiple triangle "slices." (well, those whose overlap can affect exact volume calculation)
    That would :frown: involve more complex integral(s), not as nice as the one you have presented here :blushing:
  8. Apr 8, 2005 #7


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    Well, one thing to consider is that any two triangles will intersect on a line, and the area of the line is zero, so counting it twice has no effect, I would imagine. Then again, imagine the lines were defined by (cos(T(t-a)), cos(T(t-a)), 0) for the line in the x-y plane, where T = 200*pi/(b - a), and similarly for the other lines. These will go back and forth over and over again, and so if you computed the volume as I've given it, you'll get 100 times the actual volume. I really don't know how you could deal with the general case. Just assume that the functions are "nice".
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