Question: Use Green's Theorem to prove: $$\iint_{R}dxdy=\iint_{S}\begin{vmatrix}\frac{\partial(x,y)}{\partial(u,v)}\end{vmatrix}dudv$$
$\Big(\frac{\partial(x,y)}{\partial(u,v)} $is the Jacobian of the transformation$\Big)$
I found this question from the Stewart's Calculus Book. The question only states that $x=g(u,v)$ and $y=h(u,v)$ are the transformation (I suppose it is one-to-one, because in order to apply this, the transformation has to be one-to-one). Here comes the problem.
Let C be a positively-oriented boundary curve of R (simple closed), and $Q=x ,P=0$ so that $\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}=1$ and $S$ be the region of transformed R, then:
\begin{align} \iint_{R}dxdy&=\oint_CP \hspace{0.1cm}dx+Q\hspace{0.1cm}dy\\ &=\oint_Cx\hspace{0.1cm}dy\\ &=\oint_Cg(u,v)\,(h_udu+h_vdv)\\ &=\iint_Sg_uh_v+gh_{vu}-g_vh_u-gh_{uv}\;dA \end{align}
The result can be obtained if $h_{vu}=h_{uv}$. However, it is not obvious to me that these are equal. Also, doing this way provides no information for the 'absolute' of the Jacobian. Can someone please point out my problems? Thank you.
ADD ON:
There has been no answer for three days..., I can't really think out of a solution on my own, I will appreciate a lot if someone answer me, thanks.