When can a double integral be interpreted as a surface area? Let $D$ be a closed, bounded domain in $\mathbb R^2$ and let $\vec r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle$ for $(u, v) \in D$ be a parametrization of a smooth surface $S \subseteq \mathbb R^3$.  Then the area of $S$ is
$$\iint_D \| \vec r_u \times \vec r_v \| \,  dA$$
I am interested in what might be considered a kind of converse of this: Is it always possible to interpret an integral $\iint_D f \, dA$ as the area of some surface?
More precisely, given a (non-negative) function $f:\mathbb R^2 \to \mathbb R$, defined on some domain $D$, under what conditions can we find a surface $S \subseteq \mathbb R^3$ and a parametrization $\vec r(u,v): D \to S$ such that
$$f(u,v) = \| \vec r_u \times \vec r_v \|$$
so that consequently
$$\iint_D f \, dA = \iint_D \| \vec r_u \times \vec r_v \| \,  dA \, ?$$
 A: I assume $f \geq 0$ in $D$ and $f > 0$ almost everywhere in $D$. Otherwise it's not a reasonable representation of a surface area.
Suppose $S$ is actually flat: it can be represented by $z = 0$ on the domain $\{(x(u,v),y(u,v)) \mid (u,v) \in D\}$. Then
$$ \vec r = \langle x, y, 0 \rangle $$
$$ \vec{r}_u = \langle x_u, y_u, 0 \rangle $$
$$ \vec{r}_v = \langle x_v, y_v, 0 \rangle $$
$$ \vec{r}_u \times \vec{r}_v = \langle 0, 0, x_u y_v - x_v y_u \rangle $$
$$ \|\vec{r}_u \times \vec{r}_v\| = |x_u y_v - x_v y_u| $$
(Not surprisingly, the factor in this transformation essentially between subsets of $\mathbb{R}^2$ is a Jacobian determinant.)
And one simple way to get $f(u,v) = \|\vec{r}_u \times \vec{r}_v\| = |x_u y_v - x_v y_u|$ is
$$ \begin{align*} \tilde{f}(u,v) &= \begin{cases} f(u,v) & (u,v) \in D \\
0 & (u,v) \notin D 
\end{cases} \\
 x(u,v) &= \int_0^u \tilde{f}(t,v)\, dt \\
y(u,v) &= v \end{align*} $$
The only real requirement is that $f$ is integrable with respect to $u$ at each fixed $v$. (Similar solutions would exist if $f$ is instead integrable along some other set of level curves.)
