= Solution
{c}
The <curl> is
$$
\nabla\times\mathbf F
=\bigl(\partial_yz^2-\partial_z0,\partial_z0-\partial_xz^2,\partial_x0-\partial_y0\bigr)
=\boxed{\mathbf0}.
$$
To establish that $\mathbf F$ is a <conservative vector field>, construct a <potential of a conservative vector field> directly:
$$
\boxed{\phi(x,y,z)=\frac{z^3}3+C},\qquad
\nabla\phi=(0,0,z^2)=\mathbf F.
$$
This <gradient> calculation is valid on all of $\mathbb R^3$, and hence also on the <cone>. The explicit <potential of a conservative vector field> supplies the required global justification.
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