Need help computing flux integral of vector field over an unclosed shape

Question

I need help computing this integral: $$ \iint_{S} \frac{x \hat{\imath} + y \hat{\jmath} + z \hat{k} }{ ( x^2 + y^2 + z^2 )^{3/2} } \cdot \hat{n} \, dS$$ over S that's defined as: $$S = \left\lbrace (x,y,z) \middle| z = x^2 + y^2 - 4\; , \quad x^2 + y^2 \leq 1 \: \right\rbrace$$ the normal has a negative Z component:
My thoughts so far
I know that if $0\in S$ then $$ \bigcirc\hspace{-1.4em}\iint_{S} \frac{x \hat{\imath} + y \hat{\jmath} + z \hat{k} }{ ( x^2 + y^2 + z^2 )^{3/2} } \cdot \hat{n} \, dS = 4\pi$$ and otherwise it's $0$.
I can close out S using a disk and then get: $$\bigcirc\hspace{-1.4em}\iint_{S -closed} \frac{x \hat{\imath} + y \hat{\jmath} + z \hat{k} }{ ( x^2 + y^2 + z^2 )^{3/2} } \cdot \hat{n} \, dS - \iint_{x^2+y^2 \leq 1} \frac{x \hat{\imath} + y \hat{\jmath} + z \hat{k} }{ ( x^2 + y^2 + z^2 )^{3/2} } \cdot \hat{n} \, dS = \iint_{S} \frac{x \hat{\imath} + y \hat{\jmath} + z \hat{k} }{ ( x^2 + y^2 + z^2 )^{3/2} } \cdot \hat{n} \, dS$$ the first integral should be $4\pi$ since the normal's negative Z component means it's pointing upwards, I don't really know how to calculate the middle expression or if my logic so far is correct. I think I might need to apply stoke's theorem but I don't really know how over such an expression.

Answer 1

We can close the surface with a disk $S_1: x^2 + y^2 \leq 1$ at $z = - 3$ and apply Divergence theorem to find flux through the closed surface $S \cup S_1$. We can then subtract the flux through the disk surface $S_1$ at $z = -3$.

The given vector field is $~ \displaystyle \vec F = \frac{x \hat{\imath} + y \hat{\jmath} + z \hat{k} }{ ( x^2 + y^2 + z^2 )^{3/2} }$

Note $\nabla \cdot \vec F = 0$. So the flux through the closed surface is zero. Now to find flux through the disk at $z = -3$, we do direct surface integral. The outward normal vector to the disk $z = -3~$ is $~\hat n = (0, 0, 1)$.

$ \displaystyle \vec F \cdot \hat n = \frac{z}{ ( x^2 + y^2 + z^2 )^{3/2} }$

So the surface integral is,

$ \displaystyle I_{S_1} = \int_0^{2\pi} \int_0^1 \frac{-3 r}{(9 + r^2)^{3/2}} ~ dr ~ d\theta$

And then the flux through the paraboloid surface $S$ is,

$I_{S} = - I_{S_1}$