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Homework Help: Volume of a tetrahedron

  1. May 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Let points P1: (1, 3, -1), P2: (2, 1, 4), P3: (1, 3, 7), P4: (5, 0, 2)...form the vertices of a tetrahedron. Find the volume of the tetrahedron.

    2. Relevant equations
    V = 1/3 ah
    A = area of base
    h = height of tetrahedron

    3. The attempt at a solution
    I wanted to solve this using the fact that | u x v | = area of a parallelogram formed by the vectors u and v. So I designated P1 as the "top vertex" and the other three as the vertices forming the "base" of the tetrahedron.

    Then I figured I would divide the tetrahedron into "triangle slices" whose area was | u x v| / 2 while u and v where the two vectors forming this slice.


    u = (P3 +t(P1 - P3)) - (P2 +t(P1 - P2))
    v = (P4 +t(P1 - P3)) - (P2 +t(P1 - P2))

    And I integrated | u x v | / 2 from t = 0 to t = 1. After simplifying | u x v | / 2 into an expression in terms of t, I get:

    1/2 sqrt(75t^4 - 290t^3 +450t^2-290t + 75)

    So when I integrated 1/2 sqrt(75t^4 - 290t^3 +450t^2-290t + 75) dt from 0 to 1, I got approximately 2.25426. Note: I had to integrate numerically as Mathematica couldn't integrate analytically.
    The real answer is 20/3 or 6.66667. I looked at this and saw that if you multiply my answer by 3 then you get 6.76278, which is within 0.0961158 of the real answer.

    I'm not sure if that means I am missing a constant term in my integral, but that 1/3 in the tetrahedron volume equation caught my eye as I didn't understand the reason for it when I read a proof on a website I can't remember.

    Anyways, now I am thinking that I overlooked the differential and that dt should have been dy in the integral, and I would have to then somehow express dt in terms of dy. But I haven't looked at that part yet. I think I will take a look at it tomorrow, after my AP Macroecon test.

    Thanks a lot,
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 13, 2010 #2


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    do you have to integrate? how about finding the area of the base then a vector perp to the base, then find the height by the projection onto that vector...

    also as another idea, you could parameterise in terms of the vetical height, then the area of a slice will vary by a factor of 1 to 0, as the height goes form the base to the peak, that should be pretty easy to integrate as it will just be t...
    Last edited: May 13, 2010
  4. May 13, 2010 #3
    No, I don't have to integrate. I just wanted to solve it this way because its more challenging than just plugging in height and base area to the tetrahedron volume equation.

    As far as parameterising in terms of vertical height, that is exactly the approach that I am taking. I just didn't know how to put into words what the t variable represents. Like I said though, I think the problem I'm having is what to put for the differential. Because I don't think it makes sense for volume of a "triangular slice" to just be (| u x v | / 2) dt.

    So I think I just have to ponder on exactly what should go after the cross product term.

  5. May 13, 2010 #4


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    why not?
    | u x v | represents the area of the parollelogram given by u & v, and dt represents an infinitesimal slice

    in this you should parmetrise t along a line form base to peak of the tetrahedron. Depending on how you do this you will need to include a scale factor:
    - if you paramterise the line, and t varies from 0 to 1, you need to scale by the height of the tetrahedron
    - if you paramterise the line, and t varies from 0 to h, the scale factor will be 1 as t is direct representation of length (probably teh easiest way)

    the way to do that explicitly, is to first write the volume element, dV = | u x v |ds, where ds is an intesimal length element, then find ds in terms of dt in you line paramterisation & integrate over t, by substituting into your equation
  6. May 13, 2010 #5


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    also note that as t varies from 0 to 1, along t lengths are proportional to t, so the area of a slice A(t) is proportional to t^2. Hence when you integrate you get the factor of 1/3.
    Last edited: May 13, 2010
  7. May 13, 2010 #6


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    i've had a look at you original post and think you may have made a mistake early on. Note the way you paramterise u & v, the direction of the vectors should not change, only the length. This means you should be able to write it as
    [tex] \textbf{u}(t) = \textbf{u}_0 c t[/tex]
    [tex] \textbf{v}(t) = \textbf{v}_0 c t[/tex]
    where c is some scale factor... and [itex]\textbf{u}_0, \textbf{v}_0 [/itex] are the orginal base vectors

    then when you take the cross product, you shold get
    [tex] |\textbf{u}(t) \times \textbf{v}(t)| = |\textbf{u}_0 \times \textbf{v}_0| | c^2 t^2|[/tex]

    so it should just be an integral of t^2
  8. May 14, 2010 #7
    Sorry it has taken me so long to respond, I was merely thinking over your responses.

    I don't understand here how you say "along t", as t isn't a vector but rather a scalar.

    As you can see from the picture I attached, I don't think that u and v can be described by any base vectors. Because as you can see, any two combinations of u and v for two different values of t are parallel. And if they are parallel then they are said to be linearly independent and thus they can't be described by each other.


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  9. May 15, 2010 #8


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    correct that t is a scalar, but changing the paramater t to t+dt will move you along the line and there will be a corresponding length in space ds correspodning to the change dt

    if you parametrise along a line perpindicular to the base, then a volume slice wil be
    dV = A(t) ds = |u(t) X v(t) |/2 ds

    in you diagram the direction of u does not change, only its length. Same for v.

    i'm not sure what you're trying to say here.... parallel vetctors are linearly dependent
  10. May 15, 2010 #9
    damn, i just realized what a stupid statement that was :(. ok ya now i see, since they're parallel then the direction is always the same by definition :D
  11. May 15, 2010 #10
    I found that [tex] |\textbf{u}(t) \times \textbf{v}(t)| = 5\sqrt{3}t^2 - 10\sqrt{3}t + 5\sqrt{3} [/tex]:

    [tex]\textbf{P}_1: (1, 3, -1), \textbf{P}_2: (2, 1, 4), \textbf{P}_3: (1, 3, 7), \textbf{P}_4: (5, 0, 2)[/tex]

    [tex]\textbf{u}_0 = \textbf{P}_3 - \textbf{P}_2[/tex]

    [tex]\textbf{u}(t) = (\textbf{P}_3 + t(\textbf{P}_1 - \textbf{P}_3)) - (\textbf{P}_2 + t(\textbf{P}_1 - \textbf{P}_2))[/tex]
    [tex]\textbf{u}(t) = (1 - t)\textbf{P}_3 - (1 - t)\textbf{P}_2[/tex]
    [tex]\textbf{u}(t) = (1 - t)\textbf{u}_0[/tex]

    [tex]\textbf{v}_0 = \textbf{P}_4 - \textbf{P}_2[/tex]

    [tex]\textbf{v}(t) = (\textbf{P}_4 + t(\textbf{P}_1 - \textbf{P}_4)) - (\textbf{P}_2 + t(\textbf{P}_1 - \textbf{P}_2))[/tex]
    [tex]\textbf{v}(t) = (1 - t)\textbf{P}_4 - (1 - t)\textbf{P}_2[/tex]
    [tex]\textbf{v}(t) = (1 - t)\textbf{v}_0[/tex]

    [tex] |\textbf{u}(t) \times \textbf{v}(t)| = |(1 - t)\textbf{u}_0 \times (1 - t)\textbf{v}_0| [/tex]
    [tex] |\textbf{u}(t) \times \textbf{v}(t)| = (1 - t)^2|\textbf{u}_0 \times \textbf{v}_0| [/tex]
    [tex] |\textbf{u}(t) \times \textbf{v}(t)| = 5\sqrt{3}(1 - t)^2 [/tex]
    [tex] |\textbf{u}(t) \times \textbf{v}(t)| = 5\sqrt{3}t^2 - 10\sqrt{3}t + 5\sqrt{3} [/tex]

    Now I just have to figure out what how to express ds in terms of dt.

  12. May 16, 2010 #11


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    yeah sorry, i think i may have confused things as well - the "line" doesn't enter specfically into the integral, but is how i always think of parameterising the problem, then slices are perp to the line

    say you let t vary from 0 to 1, then just let s = ht, where h is the height of the tetrahedron - this will give the correct physical dimensions & so volume
  13. May 19, 2010 #12
    Cool, I think that should work perfectly.

    Thanks for all the help :D,
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