Why is there no Jacobian in the definition of the surface integral, $\iint_UfdS = \iint_Df(r(s,t))|r'_s \times r'_t|dsdt$? I'm extremely confused by Jacobians and when to use them and I think my lack of understanding can be boiled down to this question: 

Why is there no Jacobian in the definition of the surface integral:
  $$\iint_UfdS = \iint_Df(r(s,t))|r'_s \times r'_t|dsdt$$
  where U is some surface and D is another surface? 

Isn't $U$ a surface like any other? Don't we need to compensate for the area when going from $U$ to $D$? Has it got something to do with that $U$ is a "function surface"(?) and D is a surface in $R^2$?
 A: $U$ is a surface in $3$-space, whereas $D\subset{\mathbb R}^2$ is the parameter domain of $U$ in the $(s,t)$-plane. In this plane we have the natural "surface element" ${\rm d}(s,t)$, that just measures ordinary euclidean area. There is indeed a scaling factor between ${\rm d}(s,t)$ and the corresponding area element ${\rm d}S$ on $U$. This scaling factor appears explicitly in your formula, it is $|{\bf r}_s\times{\bf r}_t|$. In other words: To a tiny rectangle in $D$ of area ${\rm d}(s,t)$ corresponds a tiny parallelogram on $U$ of area $|{\bf r}_s\times{\bf r}_t|\>{\rm d}(s,t)$.
The Jacobian, on the other hand, is the analogous local scaling factor if we  compute a volume integral over a three-dimensional domain $\Omega$ in "geometric" three space by parametrizing $\Omega$ using some nice domain $D\subset{\mathbb R}^3$ in an auxiliary parameter space. In this situation we have an essentially bijective $$f:\quad D\to\Omega,\qquad(u,v,w)\mapsto \bigl(x_1(u,v,w), x_2(u,v,w), x_3(u,v,w)\bigr)\ ,$$
and then have to compute the Jacobian $J_f(u,v,w)={\rm det}\bigl(df(u,v,w)\bigr)$.
