How much fluid passes through a net per second? Only the part of the velocity that crosses the net counts. The flux integral adds up the normal component $\mathbf{F} \cdot \mathbf{n}$ over an oriented surface. It is the surface version of flux across a curve, and it is the integral that Stokes' theorem and the divergence theorem are about.
Key idea
For a continuous field $\mathbf{F}$ and an oriented surface $S$ with unit normal $\mathbf{n}$:
$$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S \mathbf{F} \cdot \mathbf{n}\,dS = \iint_D \mathbf{F}(\mathbf{r}(u, v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v)\,dA,$$
where $\mathbf{r}_u \times \mathbf{r}_v$ points along the chosen orientation; $d\mathbf{S} = \mathbf{n}\,dS = (\mathbf{r}_u \times \mathbf{r}_v)\,dA$. For a graph $z = g(x, y)$ oriented upward:
$$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \left(-P g_x - Q g_y + R\right) dA.$$
Why it works
In time $\Delta t$, fluid with velocity $\mathbf{F}$ crossing a small patch of area $\Delta S$ fills a slanted box with base $\Delta S$ and height $(\mathbf{F} \cdot \mathbf{n})\,\Delta t$: only the normal component moves fluid across. So the rate is $\mathbf{F} \cdot \mathbf{n}\,\Delta S$.
Then $$\mathbf{n}\,dS = \frac{\mathbf{r}_u \times \mathbf{r}_v}{|\mathbf{r}_u \times \mathbf{r}_v|} \cdot |\mathbf{r}_u \times \mathbf{r}_v|\,dA = (\mathbf{r}_u \times \mathbf{r}_v)\,dA\text{:}$$ the lengths cancel. For a graph, $\mathbf{r}_x \times \mathbf{r}_y = \langle -g_x, -g_y, 1 \rangle$, which gives the second formula.
Then $$\mathbf{n}\,dS = \frac{\mathbf{r}_u \times \mathbf{r}_v}{|\mathbf{r}_u \times \mathbf{r}_v|} \cdot |\mathbf{r}_u \times \mathbf{r}_v|\,dA = (\mathbf{r}_u \times \mathbf{r}_v)\,dA\text{:}$$ the lengths cancel. For a graph, $\mathbf{r}_x \times \mathbf{r}_y = \langle -g_x, -g_y, 1 \rangle$, which gives the second formula.
Worked examples
Worked example: Through a paraboloid
Find the upward flux of $\mathbf{F} = \langle 0, 0, z \rangle$ through $z = 4 - x^2 - y^2$, $z \ge 0$.
- With $g = 4 - x^2 - y^2$: $-P g_x - Q g_y + R = z = 4 - x^2 - y^2$.
- Over the disk $x^2 + y^2 \le 4$, in polar coordinates: $$\int_0^{2\pi} \int_0^2 (4 - r^2)\,r\,dr\,d\theta = 2\pi(8 - 4) = 8\pi.$$
Answer: ${}8\pi$
Worked example: Out of a sphere
Find the outward flux of $\mathbf{F} = \langle x, y, z \rangle$ through the sphere of radius $a$.
- Outward unit normal: $\mathbf{n} = \frac{1}{a}\langle x, y, z \rangle$, so $\mathbf{F} \cdot \mathbf{n} = \frac{x^2 + y^2 + z^2}{a} = a$.
- $\iint_S a\,dS = a \cdot 4\pi a^2 = 4\pi a^3$.
Answer: ${}4\pi a^3$
Worked example: Through a cylinder wall
Find the outward flux of $\mathbf{F} = \langle x, 0, 0 \rangle$ through the side of $x^2 + y^2 = 1$, ${}0 \le z \le 2$.
- $\mathbf{r}_\theta \times \mathbf{r}_z = \langle \cos\theta, \sin\theta, 0 \rangle$ points outward.
- $\mathbf{F} \cdot (\mathbf{r}_\theta \times \mathbf{r}_z) = \cos\theta \cdot \cos\theta = \cos^2\theta$.
- $\int_0^2 \int_0^{2\pi} \cos^2\theta\,d\theta\,dz = 2\pi$.
Answer: ${}2\pi$
Common mistake
Two slips with the normal. Flux needs a dot product with the normal: integrating $|\mathbf{F}|$, or one component of $\mathbf{F}$ chosen by habit, gives a different number. And the scaling must be counted once: either $\mathbf{F} \cdot \mathbf{n}\,dS$ (unit normal, with $dS = |\mathbf{r}_u \times \mathbf{r}_v|\,dA$) or $\mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v)\,dA$ (no unit normal, no extra factor), never a unit normal and the factor $|\mathbf{r}_u \times \mathbf{r}_v|$ together with the unnormalized cross product.