Up the spiral staircase, again

1) Divergence of the vector field is zero. Therefore, if we make a closed surface, the sum of the fluxes over the constituent surfaces must be zero (according to the Divergence Theorem)
2) There are five parts to the closed surface: Two semi-circular walls (outer / inner), a bottom half-washer, the given surface, and a rectangle of width $1$ and height $\pi$ .
3) Only the fluxes through the given surface and the rectangle are non-zero, so they must be equal and opposite.

On the rectangle:

$\vec{F} = (0, -\frac{1}{r}, 0) \\ \vec{n} = (0,-1,0) \\ \vec{F} \cdot \vec{n} = \frac{1}{r} \\ dS = \pi \, dr \\ \int \vec{F} \cdot \vec{dS} = \pi \int_1^2 \frac{1}{r } \, dr = \pi \, \ell n (2)$

Therefore, the upward flux through the given surface must be $-\pi \, \ell n (2)$ . Multiplying by $1000$ and rounding gives $-2178$

This problem can be solved in at least three interesting ways: Directly from the given parameterisation, by means of Ostrogradsky's theorem (as in Steven's excellent solution), and by Stokes' Theorem.

Let me show the direct approach: The standard normal vector associated with the given parameterisation is $\vec{N}=\frac{\partial\vec{S}}{\partial\theta}\times\frac{\partial\vec{S}}{\partial r}=(\sin \theta, -\cos \theta,r)$ so $\vec{F} \cdot \vec{N}=-\frac{1}{r}$ . The flux is $\Phi =\int\int_S \vec{F} \cdot d\vec{S}=\int \int_D \vec{F} \cdot \vec{N} \ dr \ d \theta=-\int_{0}^{\pi}\int_{1}^{2}\frac{1}{r}\ dr\ d\theta=-\pi \ln(2)\approx -2.17759$ . The required answer is $\boxed{-2178}$ .