Multiple Integral Challenge Question, I just need a hint

[kostoglotov]

Homework Statement

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

Homework Equations

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 occurring, 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.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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 occurring, 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.

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 $u = \vec{a} \cdot \vec{r}, \:v = \vec{b} \cdot \vec{r}, \:w = \vec{c} \cdot \vec{r}$.

 

[kostoglotov]

the three vectors $\vec{a}, \vec{b}, \vec{c}$ are linearly independent.

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

 

[Mark44]

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

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.

 

[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.

 

[kostoglotov]

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 $u = \vec{a} \cdot \vec{r}, \:v = \vec{b} \cdot \vec{r}, \:w = \vec{c} \cdot \vec{r}$.

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?

 

[Ray Vickson]

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?

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.)

 

[Dick]

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?

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?

 

[kostoglotov]

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?

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

 

[Dick]

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

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

 

[kostoglotov]

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

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

Thank you everybody for your help! :D