Understanding the Difference between vector and velocity fields

Question

I'm having difficulty understanding the difference between a vector and velocity field. In particular, there are two questions which I had to complete which ask to find the flux of a velocity field. The two questions are:

1. A hot air balloon has a truncated spherical shape $S$ (shown below). The hot gases escape through the porous surface with a velocity field $\vec F(x,y,z) = \vec \nabla \times (-y,x,0)$. If $R=5$, compute the volume flow rate (flux) of the gases through the surface, taking the normal pointing outwards from the balloon

2. Let $S$ be the surface of the closed region $x^2 \le z \le 1$, with $x\ge 0$. Find the flux of the velocity field $\vec F(x,y,z) = (x-y,y-z,z-x)$ in the direction of the outward normal to $S$

So, for the first question, I used the simplest surface which was the base $D$, and made the connection by Stokes' Theorem $$\iint_S (\vec \nabla \times \vec F)\cdot \mathrm d \vec S = \iint_D (\vec \nabla \times \vec F)\cdot \mathrm d \vec S $$

So I took the curl of $\vec F(x,y,z) = \vec \nabla \times (-y,x,0) = (0,0,2)$, And the curl of $(0,0,2)$ is $0$ but this was very incorrect as the flux became $0$. But then, without knowing why it worked, I set the cross product in the integral equal to $\vec F$ - that is, $$\vec \nabla \times \vec F = \vec \nabla \times (-y,x,0)$$and this somehow worked because a normal to the surface is $\hat n = \hat k$ and the integral

$$\iint_D (\vec \nabla \times \vec F)\cdot \mathrm d \vec S = \iint_D (0,0,2)\cdot(0,0,1) \mathrm dx \mathrm dy = \iint_D 2 \mathrm dx \mathrm dy = 2\times \text{(Area of disk D)}= \frac{25\pi}{8}$$ since the radius of the base is $\frac{R}{4}=\frac{5}{4}$

But then in the second question (a question involving Gauss' Divergence Theorem), I took $\vec F$ as it was, that is, $\vec F(x,y,z) = (x-y,y-z,z-x)$ and got the correct answer $\Big(\frac{3\pi}{4}\Big)$, without any problems. I did the following for (using cylindrical coordinates):

$$\iint_{S=\partial \Omega} \vec F \cdot \mathrm d \vec S = \iiint_{\Omega} \vec \nabla \cdot \vec F \ \mathrm d V = \iiint_{\Omega} 3 \mathrm dV = \int_{\rho=0}^{\rho=1} \int_{\phi=-\frac{\pi}{2}}^{\phi=\frac{\pi}{2}} \int_{z=\rho^2}^{z=1} \rho \mathrm dz \mathrm d\phi \mathrm d\rho = \frac{3\pi}{4}$$

So my question is, why couldn't I use $\vec F$ as it was in $1.$ but I could in $2.$? And what is the difference between a vector and a velocity field and where are both appropriate to be used?

Thanks heaps!