# Multiple Integral Challenge Question, I just need a hint

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1. May 18, 2015

### kostoglotov

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

I will just post an image of the problem

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 $\frac{(\alpha \beta \gamma)^2}{8}$ is analogous to $\int \int \int_E xyz dV = \frac{(xyz)^2}{8}$ 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. May 18, 2015

### Ray Vickson

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
3. May 18, 2015

### kostoglotov

As a side question, is linearly independent the same as not being coplanar, or is not being coplanar one consequence of linear independence?

4. May 18, 2015

### Staff: Mentor

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.

5. May 18, 2015

### LCKurtz

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.

6. May 18, 2015

### kostoglotov

I need a bit of further help.

Should I be trying to get the Jacobian thusly

$$u = \vec{a} \cdot \vec{r} = a_1x + a_2y + a_3z$$

$$v = \vec{b} \cdot \vec{r} = b_1x + b_2y + b_3z$$

$$w = \vec{c} \cdot \vec{r} = c_1x + c_2y + c_3z$$

then find

$$\left| \frac{\partial(x,y,z)}{\partial(u,v,w)} \right|$$

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 $\frac{\partial x}{\partial u} = \frac{1}{a_1}, \ \frac{\partial y}{\partial u} = \frac{1}{a_2}, \frac{\partial x}{\partial w} = \frac{1}{a_3} ... \frac{\partial x}{\partial v} = \frac{1}{b_1} ... \frac{\partial z}{\partial u} = \frac{1}{c_1}$, yes?

7. May 18, 2015

### Ray Vickson

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.)

8. May 18, 2015

### Dick

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?

9. May 18, 2015

### kostoglotov

I'm going to say it's the multiplicative inverse of the Jacobian that I need to find.

10. May 18, 2015

### Dick

Yes, it is that. Now what is it, and what might it have to do with the triple product?

11. May 19, 2015

### kostoglotov

It's the reciprocal of the scalar triple product. Thanks for your help.

Thank you everybody for your help! :D