= Solution
Both patches project onto the unit <disk> $D$ in the $xy$-<plane>. Their outward <oriented surface elements> are
$$
d\mathbf S_+=\left(\frac{x}{\rho},\frac{y}{\rho},1\right)^Tdx\,dy,
\qquad d\mathbf S_-=(2x,2y,-1)^Tdx\,dy.
$$
The cone's apex has zero <area> and does not affect the <surface integral>. For $k=0$, the <vector field> is $(y,x,1)^T$, so
$$
\Phi_+=\iint_D\left(\frac{2xy}{\rho}+1\right)dx\,dy=\pi,
\qquad
\Phi_-=\iint_D(4xy-1)dx\,dy=-\pi.
$$
The terms involving $xy$ vanish by reflection symmetry of the <disk>. Thus the total outward <flux> is
$$
\boxed{\Phi=0}.
$$
This calculation evaluates the two <surface integrals> directly.
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