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Given the vector field $\mathbf f(x,y,z)$ in the title, I'm asked to find the flux over the paraboloid $x^2+y^2+z=3$ above the plane $z=2$, where the surface's normals point upward. I take the latter condition to mean $z>2$, so that the surface consists of only the paraboloidal segment. (Aside: is "paraboloidal" the correct term?)

I join the paraboloid to the disk $x^2+y^2=1$ at $z=2$ and assume the normals along the disk are also pointing outward so that I can apply the divergence theorem. Then, denoting the paraboloid and disk together by $S$ and the space it bounds by $V$,

$$\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV=\frac\pi2$$

Is this approach valid? I have not considered the details of actually parameterizing the parabolic surface and computing the surface integral over it directly. The vector field seemed to me like a hint to use the divergence theorem.

Next I consider the integral over the disk, denoted by $D$ (hence the paraboloid alone could be denoted by $S-D$). Parameterizing it by $\mathbf r(u,v)=\langle u\cos v,u\sin v,2\rangle$ with $0\le u\le1$, $0\le v\le2\pi$, I end up with

$$\begin{align*} \iint_D\mathbf f(x,y,z)\cdot\mathrm dS&=\iint_D\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm du\,\mathrm dv\\[1ex] &=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}\langle\square,\square,2\rangle\cdot\langle0,0,u\rangle\,\mathrm du\,\mathrm dv\\[1ex] &=4\pi\int_{u=0}^{u=1}u\,\mathrm du \end{align*}$$

But the feeling occurs to me that I should negate the cross product and use $\mathbf r_v\times\mathbf r_u$ because I'm under the impression that this normal vector faces upward when it should be facing downward.

Is my instinct correct? I always get confused by deciding the orientation of the normal vector without a visual aide.

So one of these seems to be the solution:

$$\iint_S\mathbf f\cdot\mathrm dS=\frac\pi2\pm2\pi$$

Which is it?

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