Using jacobian to transform area elements from one system to another.

Question

Please excuse my lack of rigor,I'm just average Physics undergraduate.

I have a transformation from $(u, v)$ to $(x, y)$. So an infinitesimal area element from $ (u, v)$ to $(x, y)$ plane will generally map as shown.

Using Jacobian we relate these area elements as

$d A=J. d u .d v$

where $d A \neq d x d y$.

So we can write for some $f(x, y)$ on a region

$\int f(x, y) d A=\int f(u, v) J d u d v$

But how does this follow

$\int f(x, y) d x d y=\int f(u, v) J d u d v$

Since we know that $d A \neq d x d y$ then why is $dA$ replaced by $dx dy$. Every online source does it, replacing a parallelogram $dA$ by a square $dx.dy$.

I'd be grateful if someone helped. Thank you.

Answer 1

Since $x = x(u,v)$ and $y=y(u,v)$ are functions of $u$ and $v$, you can write their differentials using the standard formula:

$$ dx = x_u du + d_v dv \hspace{0.5cm} \text{ and } \hspace{0.5cm} dy = y_u du + y_v dv $$ I've used the subscript notation for partial derivatives, so $x_u$ means $\frac{\partial x}{\partial u}$, etc...

Now assuming $dA$ means $dx \, dy$ (which it usually does), you get $$ dA = dx \, dy = x_u y_u du^2 + x_v y_v dv^2 + x_uy_v d_u d_v + x_v y_u dv du $$ The symbols "$dx$", "$dy$", "$du$", "$dv$", and "$dA$", are usually explained in low-level undergrad calculus courses as represented a "really small" length or area, as your picture indicates. But this is a convenient lie. They are really algebraic objects called "differential forms". See the Wikipedia page for more info.

The important part for this discussion is two properties of multiplication of differential forms.

1. If you square a differential form, you get $0$. So $du^2 = 0$ and $dv^2=0$.
2. It is not commutate, and $du \, dv \neq dv du$. In fact it is "anti-commutative" which means $dv du = - du dv$.

Applying these two rules to the equation above, you get

$$ dA = dx \, dy = (x_uy_v - x_vy_u) \, du \, dv = J \, du \, dv$$