Understanding the Proof of the Jacobian

[leospyder]

Can someone explain to me this part of the proof of the jacobian?

Idea of the Proof

As usual, we cut S up into tiny rectangles so that the image under T of each rectangle is a parallelogram.



We need to find the area of the parallelogram. Considering differentials, we have

T(u + Du,v) @ T(u,v) + (xuDu,yuDu)

T(u,v + Dv) @ T(u,v) + (xvDv,yvDv)

Thus the two vectors that make the parallelogram are

P = guDu i + huDu j

Q = gvDv i + hvDv j

I don't know what they're talking about...I can follow the rest (the cross product bla bla bla bla bla) but I don't know how they're getting these two vectors...I figured it has something to do with partial differentials but I am still confused. If anyone could provide any insight Id be appreciative.

 

[Galileo]

The notation is a bit screwy, but here's what I think they're doing.

So suppose the surface is parametrised by $\vec T(u,v)$.
Take a small rectangle in the domain with dimensions $\Delta u, \Delta v$, the bottom left corner being the point $(u_0,v_0)$.
The image of this rectangle is a patch of area, which can be approximated by the parallelogram formed by the vectors:

$$\vec T(u_0+\Delta u,v_0)-\vec T(u_0,v_0)$$
and
$$\vec T(u_,v_0+\Delta v)-\vec T(u_0,v_0)$$ (a picture helps here).

These vectors are in turn approximated by
$$\frac{\partial}{\partial u}\vec T(u_0,v_0)\Delta u$$
and
$$\frac{\partial}{\partial v}\vec T(u_0,v_0)\Delta v$$
respectively.

So area patch is about $|\vec T_u \times \vec T_v|\Delta u \Delta v$ and you can figure out the rest.

Hope that helps.

 

[Hurkyl]

It might help (me, anyways) if you would say what you're trying to prove. The Jacobian is a number associated with a matrix; it doesn't make any more sense to ask about a proof of the Jacobian than it does to ask about a proof of the number 2.

 

[leospyder]

The notation is a bit screwy, but here's what I think they're doing.
So suppose the surface is parametrised by $\vec T(u,v)$.
Take a small rectangle in the domain with dimensions $\Delta u, \Delta v$, the bottom left corner being the point $(u_0,v_0)$.
The image of this rectangle is a patch of area, which can be approximated by the parallelogram formed by the vectors:
$$\vec T(u_0+\Delta u,v_0)-\vec T(u_0,v_0)$$
and
$$\vec T(u_,v_0+\Delta v)-\vec T(u_0,v_0)$$ (a picture helps here).
These vectors are in turn approximated by
$$\frac{\partial}{\partial u}\vec T(u_0,v_0)\Delta u$$
and
$$\frac{\partial}{\partial v}\vec T(u_0,v_0)\Delta v$$
respectively.
So area patch is about $|\vec T_u \times \vec T_v|\Delta u \Delta v$ and you can figure out the rest.
Hope that helps.

Oh, I see it better now. THanks a lot, just wanted to say that before I go to bed. If i need further clarification Ill post the fool proof. THakns a lot guys