I want to compute $\iint_{\Sigma}(\nabla\times F)\cdot \omega dS$ where $\Sigma:=\{(x,y,z)\in\mathbb{R}^3:\sqrt{x^2+y^2}=z, 2+\frac{x}{4}\leq z\leq 4\}$, $F=(y,z,x)$ and $\omega (-3,0,3)=\frac{\sqrt{2}}{2}(1,0,1)$.
1) Applying the divergence theorem I got $\iint_{\Sigma}(\nabla\times F)\cdot \omega dS=\iiint_{V(\Sigma)}\nabla\cdot (\nabla\times F)dV-\iint_{z=4\cap cone}(\nabla\times F)\cdot \omega dS-\iint_{z=2+x/2\cap cone}(\nabla\times F)\cdot \omega dS=0-0-(-16\pi )=16\pi$
2) To check the result I've also applied Stokes's theorem taking as C the circumference $\sqrt{x^2+y^2}=4$ and I get $\oint_C F\cdot dr=\int_{0}^{2\pi}(4\sin (t),4,4\cos (t))\cdot (-4\sin (t), 4\cos (t),0)dt- \iint_{z=2+x/2\cap cone} (\nabla\times F)\cdot \omega dS =16\pi -0=16\pi$
So, is what I did correct?
HINT:
The contour $x^2+y^2=4$ does bound the open surface given by $z=\sqrt{x^2+y^2}$ with $2+x/4\le z\le 4$ plus the open (ellipse) surface $\sqrt{x^2+y^2}\le 2+x/4=z$. It also bounds the disk $x^2+y^2\le 4$, $z=4$.
But in applying Stokes's Theorem, the line integral either (i) does not include the part of the closed surface $\sqrt{x^2+y^2}\le 4$, $z=4$ or (ii) only accounts for the surface $\sqrt{x^2+y^2} \le 4$, $z=4$.
To apply Stokes's Theorem to a closed surface, simply split the surface into two separate surface that share a common boundary contour with opposing orientations.