As you know, the concept of divergence of a vector field $\mathbf{f}\colon\mathbb{R}^n\to\mathbb{R}^n$ is a measure of how closely certain parts of $\mathbb{R}^n$ resemble sinks or sources, so that when a fluid is incompressible and no parts of it are destroyed or created, its divergence is zero everywhere. I thought this would naturally lead to a concept of density—something like sampling the vector field at several locations by placing a number of $n$-dimensional cubes on random spots and computing "how much" of the vector field is inside the cubes, and then dividing each value by the volume of the corresponding cube. If this ratio always stays constant no matter where you put the cubes, the fluid has the same density everywhere and is therefore incompressible (i.e. its divergence is zero everywhere).
But here's the problem: when I tried to define "how much" of the vector field is inside the cubes… it was natural for me to think about integrals, but to my surprise, it looks like vector fields can only be integrated on the boundaries of a region? For vector fields in the plane, only line integrals are ever considered in most courses (e.g. around the contour of a square); for vector fields in space, only surface integrals seem to matter (e.g. through the surface of a solid).
But is it possible to meaningfully define the integral of a two-dimensional field over the interior of a square? Or the integral of a three-dimensional vector field over the interior of a cube, and so on. Even aside from what motivated me to investigate this in the first place, I was shocked to find nothing about these ideas online (or on my textbooks), and at this point I'm just curious to know whether it can be done or not—and if it's impossible, why so.
Best Answer
Let $\mathbf{F}(x,y) = (P(x,y), Q(x,y))$ be a vector field on a region $D \subset \mathbb{R}^2$. As far as I am aware, there is no single concept that mathematicians have designated as "the" integral of $\mathbf{F}$ on $D$. As such, we will have to define it ourselves.
I think the most reasonable definition would be $$\iint_D \mathbf{F}\,dA = \left( \iint_D P\,dA, \iint_D Q\,dA \right)$$ but this results in a vector, and the OP wants a scalar. To that end, the following integrals are all very natural, and could be taken to be the definition of "the integral" of $\mathbf{F}$ on $D$: \begin{align*} \iint_D \text{div}(\mathbf{F})\,dA & = \iint_D (P_x + Q_y)\,dA \tag{1} \\ \iint_D \text{curl}(\mathbf{F}) \cdot \mathbf{k}\,dA & = \iint_D (Q_x - P_y)\,dA \tag{2} \\ \iint_D \Vert \mathbf{F}\Vert\,dA & = \iint_D \sqrt{P(x,y)^2 + Q(x,y)^2}\,dA \tag{3} \\ \left\Vert\iint_D \mathbf{F}\,dA\right\Vert & = \sqrt{ \left( \iint_D P(x,y)\,dA \right)^2 + \left( \iint_D Q(x,y)\,dA \right)^2 } \tag{4} \end{align*} Which one of these we designate as "the" integral of $\mathbf{F}$ on $D$ is a matter of taste, or perhaps a matter of what we want our integral to capture.
As far as I can tell, the OP wants the integral of $\mathbf{F}$ on $D$ to measure how "much" of $\mathbf{F}$ there is on $D$. Here is how to do this: First partition $D$ into sub-rectangles $P_1, \ldots, P_n$ of areas $\Delta A_i$, and choose sample points $(x_i, y_i) \in P_i$. Second, define a scalar quantity called $$\text{Much}(\mathbf{F})(x,y)$$ that measures how "much" of $\mathbf{F}$ there is at $(x,y)$. Finally, form the Riemann sum $$\sum_i \text{Much}(\mathbf{F})(x_i,y_i)\,\Delta A_i$$ so that the limit as the partition gets finer and finer is your integral: $$\iint_D \text{Much}(\mathbf{F})\,dA = \lim \sum_i \text{Much}(\mathbf{F})(x_i,y_i)\,\Delta A_i.$$ In order for this to work, you need to know (at the very least) what you want $\text{Much}(\mathbf{F})$ to mean for, say, a constant vector field like $\mathbf{F}(x,y) = (1,0)$ or $\mathbf{F}(x,y) = (3,4)$.
Heuristically, it follows that $$\text{Much}(\mathbf{F})(x,y) = \lim_{\text{Area}(D) \to 0} \frac{ \iint_D \text{Much}(\mathbf{F})\,dA }{\text{Area}(D)}$$ where the regions $D$ contain the point $(x,y)$, and are shrinking reasonably nicely to the point $(x,y)$ in the limit.
Notice that if we take $\text{Much}(\mathbf{F})$ to mean $\text{div}(\mathbf{F})$, then we get $$\text{div}(\mathbf{F})(x,y) = \lim_{\text{Area}(D) \to 0} \frac{ \iint_D \text{div}(\mathbf{F})\,dA }{\text{Area}(D)} = \lim_{\text{Area}(D) \to 0} \frac{\oint_{\partial D} \mathbf{F} \cdot \mathbf{n}\,ds}{\text{Area}(D)}$$ by using a form of Green's Theorem (or Stokes' Theorem) in the last equality.