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The first field is ${\mathbf{F}}_1=(z-y,x,-x)$, so direct curl calculation givesIt is not conservative because its curl is nonzero. The second is a radial gradient field: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.

Inside the sphere, the electric field has the form $\mathbf{E}=\nabla [-k{r}^2/(2a^3)]$, including at the origin. Thus it is a conservative vector field. Outside, use the gradient of a radial function to obtainThis is also an explicit potential of a conservative vector field, single-valued throughout the exterior region. Equivalently, the curl vanishes in both regions: for a radial coefficient $h(r)$, $\nabla \times (h\mathbf{r})=\nabla h\times \mathbf{r}+h\nabla \times \mathbf{r}=0$ because $\nabla h$ is radial. The interior formula avoids any apparent $1/r$ issue at the origin. The vector-field potentials here have the opposite sign from the electric potential convention used in (b).

Write the differential one-form as $Mdx+Ndy$. An exact differential form must satisfy equality of mixed partial derivatives, $M_y=N_x$. HereEquality at every $(x,y)$ holds precisely when $\boxed{a=2}$. Since these polynomials are smooth on the simply connected plane, the closed-form condition is also sufficient for exactness; the explicit potential of a conservative vector field constructed in (c) verifies that sufficiency directly.

With $a=2$, integrate the $dx$ coefficient with respect to $x$:Its other partial derivative is ${\Phi }_y=x^2/2-2xy+g^{\prime }(y)$. Matching the $dy$ coefficient gives $g^{\prime }=0$. Therefore the differential equation is $d\Phi =0$, and its implicit general solution isThese are level curves of the potential of a conservative vector field. Where $N\ne 0$, they give the usual graph solution $dy/dx=-M/N$. At $N=0$, the implicit description retains possible vertical tangents instead of dividing by zero.

If a periodic vector field has a periodic potential of a conservative vector field, the line integral between points separated by a period vector must vanish, since it is the potential difference. A periodic gradient field need not satisfy this: the constant field $(1,2)$ on the plane has potential $x+2y$, but its integrals over the two unit translations are $1$ and $2$. On the periodic quotient these translation integrals are nonzero periods, obstructing a single-valued potential there.