What is the inverse/opposite of a double integral? I am currently taking Calc 3 and we just finished our unit on double/triple integrals. I started thinking about a problem back from AP Physics (where my teacher did an impressive amount of hand waving to somehow avoid directly explaining vector calculus when discussing Maxwell's equations, leading to me spending an entire semester confused about what the heck flux was supposed to be or why it should matter), which basically boils down to $\iiint_D{F(x, y, z)}{dV}=Q(D)$. In Calc 1 and 2, if you have an equation of the form $\int{f(x)}{dx}=g(x)$, then you can take the "inverse integral" (i.e. derivative with respect to x) of both sides, which gives $f(x)=\frac{d}{dx}g(x)$. However, I cannot seem to figure out what the analogue of this would be for a double or triple integral. What is the opposite/inverse operation for a double or triple integral over some domain $D$?
 A: It's convenient but misleading to write $\int f(x) \, dx = g(x)$. The RHS is a function of $x$ but the LHS is not until you write down some bounds of integration, and once you do the inner $x$ has been integrated over so it's a "dummy variable" that has nothing to do with the $x$ on the RHS. Less ambiguous is to write
$$\int_a^x f(t) \, dt = g(x)$$
which makes the dependence on $x$ of the LHS clear and emphasizes that the variable being integrated over is a dummy variable.
This is not just pedantry; the point of doing this is to emphasize what changes when you move from integrals to double integrals. To think of a double integral as a function of two variables, rather than just as a function of a region being integrated over, we need to pick regions to integrate over that are defined by two parameters. An easy choice is a rectangle $[a, x] \times [b, y]$, which lets us write expressions like
$$\int_{t=b}^y \int_{s=a}^x f(s, t) \, ds \, dt = g(x, y)$$
and if we set things up like this we can recover $f$ by differentiating twice, using the fundamental theorem of calculus twice:
$$f(x, y) = \frac{\partial}{\partial x} \frac{\partial}{\partial y} g(x, y).$$
This generalizes to $n$ dimensions too. But it's worth emphasizing that in order to do this we had to choose to integrate over rectangles, and there are lots of other choices of how to parameterize regions to integrate over that we could make instead.
A: To complement  Qiaochu Y. excellent explanation  note that,  same as we write
$$
\eqalign{
  & \int {f(x)dx}  = g(x) = \int_{t = a}^x {f(t)dt}  + c
 = \int\limits_{t \in \left[ {a,x} \right]} {f(t)dt}  + c\quad  \Rightarrow \quad   \cr 
  &  \Rightarrow \quad d\int_{t = a}^x {f(t)dt}  = \int\limits_{t \in \left[ {x,x + dx} \right]} {f(t)dt}
  = f(x)dx = {d \over {dx}}g(x)dx \cr} 
$$
with all cautions as for continuity, and with the algebraic meaning for negative $dx$,
then, specially in physics, we often write
$$
\eqalign{
  & \int\!\!\!\int\limits_A {f(x,y)dxdy}  = \int\!\!\!\int\limits_A {f(x,y)\delta A}  = g(A)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \delta \int\!\!\!\int\limits_A {f(x,y)\delta A}  = \int\!\!\!\int\limits_{\delta A} {f(x,y)dxdy}
  = {d \over {dA}}g(A)\,\delta A \cr} 
$$
where  the $\delta A$ is also taken with the algebraic sign.

From here we reach to Stoke's Theorem.
