Let S be the surface $\{(x,y,z)\in R^3:x^2+y^2+2z=2,z\geq 0\}$ and let $\hat{n}$ be the outwards unit normal to S. If $\vec{F}=\langle y,xz,x^2+y^2\rangle$ then find the value of $\int_{S} \vec{F}.\hat{n}\,dS$.
My main concern here is that, I believe $$S:\{(x,y,z)\in R^3:x^2+y^2+2z=2,z\geq 0\}$$
must be a closed surface? (If not then why?)
If it is indeed closed, then by divergence theorem, since $\nabla.\vec{F} = 0$, the final answer is zero.
However, if it is not closed, then,
Let $S'$ be such that it includes $S$ aswell as the $P:x-y$ plane.
$S'$ is closed and
$\int_{S'}\vec{F}.\hat{n}\,dS = \int_{S}\vec{F}.\hat{n}\,dS + \int_{P}\vec{F}.\hat{n}\,dS = \int_{S}\vec{F}.\hat{n}\,dS - \int\int_{R} (x^2+y^2)dydx$ where $R:x^2+y^2=2$
$\Rightarrow 0= \int_{S}\vec{F}.\hat{n}\,dS - 2\pi $ $\; \; ($Since $S'$ is closed, and using divergence theorem)
$\Rightarrow \int_{S}\vec{F}.\hat{n}\,dS = 2\pi$ which agrees with the answer given to me.
However, I don't know why the first approach is not valid since I don't see how $S$ is not closed.
$S$ is not a closed surface. It is an inverted (circular) paraboloid with vertex at $(0, 0, 1)$. Since you restrict $z\geq 0$, so it is bounded but "open" at the bottom i.e., the boundary at the bottom (a circular disk $x^2+y^2\leq 2$ on the $xy$-plane) is not part of the surface $S$.