Problem about finding region and function preserving areas This problem is driving me crazy. 
"Let $S$ be a parametrized surface by $\varphi:[a,b]\times[c,d]\rightarrow\mathbb{R^3}$, $\varphi$ is of class $C^1$. Show that exists an open, connected set $D\subset\mathbb{R^2}$ and a bijection $f:D\rightarrow[a,b]\times[c,d]$ of class $C^1$ such that $\varphi\circ f$ preserves areas."
Here is what I did:
I started supposing that $D$ has no holes, doing this it's possible to calculates the area of any region $D\ '\subset D$ by $$\int_{\alpha_1}^{\beta_1}g_1(x)-h_1(x) \ dx +\ldots+\int_{\alpha_n}^{\beta_n}g_n(x)-h_n(x) \ dx$$
where each $g_i(x)$ is the upper function and $h_i(x)$ is the lower function. To make it simpler, but not losing generality, I may consider only one region like this, so we have that $$\int_{\alpha}^{\beta}g(x)-h(x) \ dx$$
is the area of $D\ '$.
*Note that $(\varphi\circ f)(x,y)=\Big(\varphi_1(f(x,y)),\varphi_2(f(x,y)),\varphi_3(f(x,y)\Big)$.
The corresponding area in S is the area of $(\varphi\circ f)(D\ ')$, it's is given by
$$\int_{D\ '} \bigg| \bigg( \frac{\partial\varphi_1(f)}{\partial x},\frac{\partial\varphi_2(f)}{\partial x},\frac{\partial\varphi_3(f)}{\partial x}, \bigg)\times\bigg( \frac{\partial\varphi_1(f)}{\partial y},\frac{\partial\varphi_2(f)}{\partial y},\frac{\partial\varphi_3(f)}{\partial y}, \bigg) \bigg|\cdot |J_f| dA=$$
$$=\int_\alpha^\beta\int_{h(x)}^{g(x)} \bigg| \bigg( \frac{\partial\varphi_1(f)}{\partial x},\frac{\partial\varphi_2(f)}{\partial x},\frac{\partial\varphi_3(f)}{\partial x}, \bigg)\times\bigg( \frac{\partial\varphi_1(f)}{\partial y},\frac{\partial\varphi_2(f)}{\partial y},\frac{\partial\varphi_3(f)}{\partial y}, \bigg) \bigg|\cdot |J_f|\ dy\ dx.$$
This takes me to think that $\bigg| \bigg( \frac{\partial\varphi_1(f)}{\partial x},\frac{\partial\varphi_2(f)}{\partial x},\frac{\partial\varphi_3(f)}{\partial x}, \bigg)\times\bigg( \frac{\partial\varphi_1(f)}{\partial y},\frac{\partial\varphi_2(f)}{\partial y},\frac{\partial\varphi_3(f)}{\partial y}, \bigg) \bigg|\cdot |J_f|=1$, that way I have the equality I'm looking for. This is probably wrong because all I have to do to solve this is equation is to find some $f$, it doesn't depend on $D$.
Please, someone could point out what I'm doing wrong and show me the correct way. 
Thank you very much for your help.
 A: I assume that your representation $\phi$ is regular, i.e., that
$$w(x,y):=|\phi_x(x,y)\times \phi_y(x,y)|\ne 0$$
for all $(x,y)\in R:=[a,b]\times[c,d]$. Instead of $f:\ D\to R\ $ we shall construct its inverse $g:\ R\to D$. 
Note that $w(x,y)$ represents the local area dilatation  of the map $\phi$.  We now take care that $g$ has the same local area dilatation  $w(x,y)$. To this end  let
$$u(x,y):=\int_a^x  w(t,y)\ dt$$
and put
$$g(x,y):=\bigl(u(x,y),y\bigr)\qquad \bigl((x,y)\in R\bigr)\ .$$
Then $g$ maps horizontal lines $y=\eta$ to the same lines $v=\eta$ in the $(u,v)$-plane, and as $u_x(x,y)>0$ for all $(x,y)$ each such horizontal is mapped in a strictly increasing fashion onto its parallel in the $(u,v)$-plane. It follows that the map $g$ is  injective onto some image domain $D$. Furthermore the Jacobian of $g$ computes to 
$$J_g(x,y)=u_x(x,y)=w(x,y)>0\ .$$
It follows that $g$ is regular and thus has a $C^1$-inverse $f$. In addition the area dilatation of $g$ is $w(x,y)$, and therefore the area dilatation of $f:=g^{-1}$  at the point $g(x,y)$ is ${1\over w(x,y)}$.
From this we conclude that $\phi\circ f$ has area dilatation $1$ at all points, in other words: that $\phi\circ f$ preserves areas.
