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Multiple Integral Challenge Question, I just need a hint

  1. May 18, 2015 #1
    1. The problem statement, all variables and given/known data

    I will just post an image of the problem

    JB6FEog.png

    and here's the link if the above is too small: http://i.imgur.com/JB6FEog.png?1


    2. Relevant equations


    3. The attempt at a solution
    I've been playing with it, but I can't figure out a good way to "grip" this problem.

    I can see some things. I can see that the scalar triple product is the volume of the parallelopiped made from the three vector a, b and c. I can see that the [itex]\frac{(\alpha \beta \gamma)^2}{8}[/itex] is analogous to [itex]\int \int \int_E xyz dV = \frac{(xyz)^2}{8}[/itex] if x,y and z were the upper limits and the lower limits were all 0.

    So I've thought that maybe there's a change of variable occuring, and the absolute value of the scalar triple product in the denominator comes from the Jacobian...should I keep heading in that direction? I can't think of a Transform that would be appropriate.
     
    Last edited by a moderator: May 18, 2015
  2. jcsd
  3. May 18, 2015 #2

    Ray Vickson

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    The right-hand-side divides by zero unless the three vectors ##\vec{a}, \vec{b}, \vec{c}## are linearly independent, and when that is the case (that is, when you have linear independence) you can change variables to ## u = \vec{a} \cdot \vec{r}, \:v = \vec{b} \cdot \vec{r}, \:w = \vec{c} \cdot \vec{r}##.
     
    Last edited: May 18, 2015
  4. May 18, 2015 #3
    As a side question, is linearly independent the same as not being coplanar, or is not being coplanar one consequence of linear independence?
     
  5. May 18, 2015 #4

    Mark44

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    Any three vectors that are coplanar are linearly dependent, but you can have two vectors in the same plane being linearly independent, as long as one of them isn't a scalar multiple of the other, so the concepts aren't the same.
     
  6. May 18, 2015 #5

    LCKurtz

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    Just to add to that, for the benefit of the OP, if the three vectors are 3 dimensional then whether or not they are coplanar is equivalent to whether or not they are linearly dependent.
     
  7. May 18, 2015 #6
    I need a bit of further help.

    Should I be trying to get the Jacobian thusly

    [tex] u = \vec{a} \cdot \vec{r} = a_1x + a_2y + a_3z[/tex]

    [tex] v = \vec{b} \cdot \vec{r} = b_1x + b_2y + b_3z[/tex]

    [tex] w = \vec{c} \cdot \vec{r} = c_1x + c_2y + c_3z[/tex]

    then find

    [tex] \left| \frac{\partial(x,y,z)}{\partial(u,v,w)} \right| [/tex]

    Or is there some way that doesn't involve converting the dot product into a function of x,y,z?

    Also, doing it this way, I should get [itex]\frac{\partial x}{\partial u} = \frac{1}{a_1}, \ \frac{\partial y}{\partial u} = \frac{1}{a_2}, \frac{\partial x}{\partial w} = \frac{1}{a_3} ... \frac{\partial x}{\partial v} = \frac{1}{b_1} ... \frac{\partial z}{\partial u} = \frac{1}{c_1} [/itex], yes?
     
  8. May 18, 2015 #7

    Ray Vickson

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    Just use the standard textbook formulas for change-of-variables in multiple integration. If you do not have a textbook, look on-line. (I, personally do not feel I can offer more help, without essentially doing the problem for you.)
     
  9. May 18, 2015 #8

    Dick

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    No, you don't get that. You might find it easier to compute ## \left| \frac{\partial(u,v,w)}{\partial(x,y,z)} \right| ##. What's that and what might that have to do with the problem?
     
  10. May 18, 2015 #9
    I'm going to say it's the multiplicative inverse of the Jacobian that I need to find.
     
  11. May 18, 2015 #10

    Dick

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    Yes, it is that. Now what is it, and what might it have to do with the triple product?
     
  12. May 19, 2015 #11
    It's the reciprocal of the scalar triple product. Thanks for your help.

    Thank you everybody for your help! :D
     
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