Jacobians of 2-space to 3-space Transformation

[yungman]

From more than one textbook, they only talking about change of variables from 2-space to 2-space or from 3-space to 3-spare. Eg:

$$\frac{\partial (x,y,z)}{\partial (u,v,w)} \hbox { for 3-space and }\frac{\partial (x,y)}{\partial (u,v)} \hbox{ for 2-space }$$

But for surface area of 3-space in the following example where the vector value function:

$$\vec {r} = u\hat{x} + u cos(v) \hat{y} + u sin(v) \hat{z}$$

You can see this is like:

$$\frac{\partial (x,y,z)}{\partial (u,v)}$$

Which I don’t see this from the book. My question is whether the Jacobian is still:

$$| \frac{\pratial \vec{r}}{\partial u} X \frac{\pratial \vec{r}}{\partial v}|$$ ?

This is the standard way of finding surface area of a 3-space object. But this is like transform from 2 space [tex](u,v)[/itex] to 3-space$(x,y,z)$.[/tex]

 

[yungman]

Anyone?

I know the Jacobian is defined as:

$$|\frac{\partial \vec{r} }{\partial u} X \frac{\partial \vec{r} }{\partial v}|$$

So it is the Jacobian even if $\vec{r} = x(u,v)\hat{x} +y(u,v)\hat{y} + z(u,v)\hat{z}$

It works for finding surface intergrals. I just want to verify this here.

 

[fluxions]

http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant"

The domain and codomain can be of different (finite) dimensions. So in the case you're curious about, you will have
$$J = \frac{\partial(x,y,z)}{\partial(u,v)}$$

 

[yungman]

http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant"

The domain and codomain can be of different (finite) dimensions. So in the case you're curious about, you will have
$$J = \frac{\partial(x,y,z)}{\partial(u,v)}$$

Thanks for the reply.

I am surprised Wikipedia has the answer! I gone through a lot of books and online stuff, they all only talked about square matrix where either it is 2X2 or 3X3!

That's what I suspect, because this 3X2 Jacobian work just as well in every single case.

Thanks

Alan