Answer
$0$
Work Step by Step
Since, the surface is the part of the cone $x=\sqrt {y^2+z^2} $ for which $0 \leq x \leq 2$.
The boundary of this surface is a circle parallel to the yz plane and the parameterization of the boundary can be written as: $C: r(t)=2i+2 \cos t j+2 \sin t k \implies dr=0i-2 \sin t j$
Stokes' Theorem states that $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr$
where, $C$ corresponds to the boundary of the surface oriented counter-clockwise.
Now, $$\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr=\int_0^{2 \pi} (arctan (32 \cos t \sin^2 t) i+8 \cos t j+16 \sin^2t k) \times (0 \ i-2 \ \sin t j+2 \cos t \ k) dt \\=\int_{2 \pi}^{0} (-16 \sin t +32 \sin^2 t) \times (\cos t dt)$$
Consider $\sin t =u \implies da =\cos t dt$
Now, $$\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr=\int_0^{0} (-16 ut +32 u^2 t)du=0$$
