= Solution
The first field is $\mathbf F_1=(z-y,x,-x)$, so direct <curl> calculation gives
$$
\boxed{\nabla\times\mathbf F_1=(0,2,2)=2\mathbf a}.
$$
It is \b[not conservative] because its <curl> is nonzero. The second is a radial <gradient> field:
$$
\mathbf F_2=\nabla\left[-\frac1{\sqrt{1+r^2}}\right],
\qquad\boxed{\nabla\times\mathbf F_2=0}.
$$
The <scalar> expression is a <smooth function> and single-valued on all of space, so $\mathbf F_2$ is a <conservative vector field>. Its <potential of a conservative vector field> is $\Phi=-(1+r^2)^{-1/2}$. Parts (a) and (b) check path dependence directly.
Back to article page
