Gauss theorem applied for curl integral over surface of torus giving wrong answer

Two surfaces are given in cylinder coordinates as respectively $z=r-1$ (cone) and $ z^2 + (r-2)^2=1$ (torus, doughnut). Both surfaces are rotationally symmetric about x-axis.

Define a vector field over $R^3$ : $$ \vec{F}(x,y,z) = (xz-y) \hat{i} + (yz+x) \hat{j} + (x^2 + y^2)\hat{k}$$

Question: A rotationally symmetric body $T$ is bounded by the two surfaces. Sketch the cross section of the body in the $zr$ plane and the projection of the body down in the $xy$ plane. Find the surface integral of the curl of the $F$ over the part $S_2$ which is the surface of $T$ which is common with torus.

I am guessing this common region remarked is the torus itself? At $r=1$ both the cone and torus meet along a boundary line.

I thought of just using the fact that divergence of curl is zero $\nabla \cdot \nabla \times \vec{F}=0$, hence we can apply apply gauss theorem on the surface of torus to find the integral value is zero.. but this is wrong! The true answer is meant to be $6\pi$

The body $T$ is bounded by the surface of the cone as well as the torus. The cone and the torus intersect at $z = 0$ and at $z = 1$ as shown below -

At the intersection of both surfaces -

$(r-1)^2 + (r-2)^2 = 1 \implies (r-1) (r-2) = 0$

So at the intersection $r = 1$ and $r = 2$ which is $z = 0, 1$ resp.

Now as you mentioned, we know that $~\nabla \cdot (\nabla \times \vec F) = 0$.

So the surface integral of the curl of the vector field $\vec F$ through the closed surface $T$ is zero. Now we need to find surface integral over the part of the surface $T$ which is common to the torus ($S_2$). Alternatively we can find the surface integral over the part of the $T$ which is common to the cone ($S_1$) and subtracting from the net flux through $T$ should give us the surface integral over $S_2$.

Parametrize the surface $S_1$ as,

$\phi(r, \theta) = (r\cos \theta, r\sin\theta, r-1), 1 \leq r \leq 2, 0 \leq \theta \leq 2\pi$

$\vec n = \phi_{\theta} \times \phi_r = (r \cos\theta, r\sin\theta, - r)$

$\nabla \times \vec F = (y, -x, 2) = (r\sin\theta, - r\cos\theta, 2)$

Finally the surface integral is,

$ \displaystyle I = \iint_{S_1} (\nabla \times \vec F) \cdot \vec n ~ dS$

and the answer will be $ - I$.

Can you take it from here?