In James Stewart's Calculus: Early Transcendentals ($8$e), problem $31$ of Section $16.4$ asks us to prove the change of variables formula
$$\iint_{R}dx\text{ }dy=\iint_{S}\left|\frac{\partial(x,y)}{\partial(u,v)}\right|du\text{ }dv$$
using Green's theorem. Here, $R$ is the image of $S$ under the transformation $T$ defined by the equations $x=g(u,v)$, $y=h(u,v)$. Here's my reasoning:
Notice that $\iint_{R}dA$ is the area of the region $R$. It can be shown that Green's theorem implies that this area is given by the line integral $\int_{\partial R}x\text{ }dy$. Thus, if $\partial R$ can be parameterized by $\textbf{r}(t)=\left<x(t),y(t)\right>$ ($a\leq t\leq b$), we can write
\begin{align*} \iint_{R}dA &= \int_{\partial R}x\text{ }dy\\ &= \int_{a}^{b}x(t)y'(t)\text{ }dt \end{align*}
It was assumed that the image of $S$ under $T$ is $R$, so if $T$ preserves orientation, it must be the case that $T$ maps $\partial S$ to $\partial R$. It follows that
\begin{align*} \left<x(t),y(t)\right> &= \textbf{r}(t)\\ &= T\left(\textbf{r}_{S}(t)\right)\\ &= \left<g(\textbf{r}_{S}(t)),h(\textbf{r}_{S}(t))\right>\\ &= \left<g(u(t),v(t)),h(u(t),v(t))\right> \end{align*}
for all $t\in[a,b]$, where $\textbf{r}_{S}(t)=\left<u(t),v(t)\right>$ parameterizes $\partial S$ (the parameter domain doesn't change because $\partial S$ is a closed curve, and thus must return to $\textbf{r}_{S}(a)$ once $t$ reaches $b$).
We deduce that
\begin{align*} x(t) &= g(u(t),v(t))\\ y(t) &= h(u(t),v(t))\\ y'(t) &= h_{u}(u(t),v(t))u'(t)+h_{v}(u(t),v(t))v'(t) \end{align*}
and consequently
\begin{align*} \int_{a}^{b}x(t)y'(t)\text{ }dt &= \int_{a}^{b}g(u(t),v(t))\left[h_{u}(u(t),v(t))u'(t)+h_{v}(u(t),v(t))v'(t)\right]dt\\ &= \int_{a}^{b}[g(u(t),v(t))h_{u}(u(t),v(t))u'(t)\\ & +g(u(t),v(t))h_{v}(u(t),v(t))v'(t)]dt\\ &= \int_{a}^{b}g(u(t),v(t))h_{u}(u(t),v(t))u'(t)\text{ }dt\\ & +\int_{a}^{b}g(u(t),v(t))h_{v}(u(t),v(t))v'(t)\text{ }dt\\ &= \int_{\partial S}g(u,v)h_{u}(u,v)du+g(u,v)h_{v}(u,v)dv \end{align*}
Applying Green's theorem gives
\begin{align*} \int_{\partial S}g(u,v)h_{u}(u,v)du+g(u,v)h_{v}(u,v)dv &= \iint_{S}\left(\frac{\partial}{\partial u}\left(g(u,v)h_{v}(u,v)\right)-\frac{\partial}{\partial v}\left(g(u,v)h_{u}(u,v)\right)\right)dA\\ &= \iint_{S}(g_{u}(u,v)h_{v}(u,v)+g(u,v)h_{vu}(u,v)\\ & -g_{v}(u,v)h_{u}(u,v)-g(u,v)h_{uv}(u,v))\text{ }dA\\ &= \iint_{S}\left(g_{u}(u,v)h_{v}(u,v)-g_{v}(u,v)h_{u}(u,v)\right)dA \end{align*}
The last expression is precisely $\left|\frac{\partial (x,y)}{\partial (u,v)}\right|$ (the orientation has not changed), so we may finally write
$$\iint_{R}dA=\iint_{S}\left|\frac{\partial(x,y)}{\partial(u,v)}\right|dA$$
I appreciate any and all feedback. If you identify any mistakes, I kindly ask that you only give me hints so I can correct them myself.
Extra: with a few tweaks, this argument can be used to prove the more general result
$$\iint_{R}f(x,y)\text{ }dA=\iint_{S}f(g(u,v),h(u,v))\left|\frac{\partial (x,y)}{\partial (u,v)}\right|dA$$