Answer
$\int_{R} dx dy=\iint_{S} [\dfrac{\partial (x,y)}{\partial (u,v) }]$ (Proved)
Work Step by Step
We have: $x = g(u,v); y= h(u,v)$
The line integral can be written as follows:
$A=\oint_{C} x dy$
So, $\int_{R} dx dy=\oint_{\partial R} x dy$
and $dy= \dfrac{\partial h}{\partial u} du+ \dfrac{\partial h}{\partial v} dv$
Now, $\oint_{\partial R} x dy=\oint_{\partial S} (g(u,v) \dfrac{\partial h}{\partial u}) du+ (g(u,v)\dfrac{\partial h}{\partial v}) dv$
Green's Theorem states that:
$\oint_CP\,dx+Q\,dy=\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dA \\=\iint_{\partial S} ( \dfrac{\partial g}{\partial u} \dfrac{\partial h}{\partial v}-\dfrac{\partial g}{\partial v} \dfrac{\partial h}{\partial u})+ g[\dfrac{\partial }{\partial u}(\dfrac{\partial h}{\partial v})-\dfrac{\partial }{\partial v}(\dfrac{\partial h}{\partial u})] dA$
and $\int_{R} dx dy=\iint_{S} [\dfrac{\partial (x,y)}{\partial (u,v) }]$
Therefore, the result has been proved.
