Given that $\vec F=\frac{1}{(x^2+y^2+z^2)^{\frac{3}{2}}}(x,y,z)$.
And $S = \{(x,y,z)| z = 1-\frac{(1-x^2-y^2)^{100}}{2}, x^2+y^2 \le 1 \}$.
With a normal that has a positive $z$ component.
And
$L=\{(x,y,z)| x^2+y^2=1, z=0 \}$
with direction against the clock arrows if we look from above.
Find:
$\oint_{\partial S}\vec F \cdot d\vec r$
$\oint_L\vec F \cdot d \vec r$
$\iint_S \vec F\cdot \hat ndS$
The first part of the question was to prove that $\nabla \cdot \vec F=0$ and $\nabla \times \vec F=(0,0,0)$, and that gives me a hint that I need to use gauss and stokes theorems.
But, in order to use gauss theorem, I need to know if $S$ is closed or not, and if not close it. and I know I have a problematic point in $(0,0)$.
In order to use stokes theorem, I need to know that $\vec F \in C^1$ on the curve and on the surface, and on top of that I must "see" the surface so I decide how I'm going to use it and which ways, like for example if I use stokes on a cylinder surface, I need to do it twice, but if I do it on half of a ball, I can just do it on the circle curve and it will give me the flux.
If I could use stokes here the first two integrals will actually give me $0$, (and it's the final answer), and the third one I will need to use gauss, but I have no idea why and how can I use them when I can't visualize the surface $S$ as I've explained.
I would appreciate any help, thanks in advance!
$S = \{(x,y,z)| z = 1-\frac{(1-x^2-y^2)^{100}}{2}, x^2+y^2 \le 1 \}$
Please note that at $r = 0, z = \frac{1}{2}$ and as $r$ increases, $z$ increases with $z = 1$ at $r = 1$.
So it is a surface which is between $\frac{1}{2} \leq z \leq 1$ and is open at $z = 1$ (in absence of a better name for the surface, draw a far resemblance with a paraboloid surface that opens up)
Also as you mentioned, the curl of the vector field is zero over the domain of $S$ and so is divergence.
To apply divergence theorem, close the surface with disk $x^2 + y^2 = 1$ at $z = 1$.
The outward normal vector is $(0, 0, 1)$. So flux through the disk surface is,
$I = \displaystyle \int_0^{2\pi} \int_0^1 \frac{1}{(1 + r^2)^{3/2}} \ r \ dr \ d\theta$
And to find flux through surface $S$, we subtract flux through the disk from flux through closed surface, which is zero.
For line integral over $L$, parametrize the curve as,
$c(t) = (\cos t, \sin t, 0), 0 \leq t \leq 2\pi$
And line integral is,
$\displaystyle \int_0^{2\pi} \vec F(r(t)) \cdot c'(t) \ dt$
which should again be zero.