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The space ${\mathbb{R}}^3$ is simply connected, and the electric field has zero curl. Hence it is a conservative vector field. More explicitly, Stokes theorem gives zero circulation around any closed loop, since such a loop bounds a spanning surface in ${\mathbb{R}}^3$. Therefore the line integral between two fixed endpoints is independent of the chosen path.

Fix a reference point ${\mathbf{r}}_0$ and define the electric potentialTaking a short straight terminal segment in each coordinate direction shows ${\partial }_j\phi =-E_j$. ConsequentlyThe electric potential is determined up to an additive constant; the later condition at infinity fixes that constant for the finite charge distribution in part (b).

The curl isTo establish that $\mathbf{F}$ is a conservative vector field, construct a potential of a conservative vector field directly: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.

The displayed vector field is defined on the punctured plane $U={\mathbb{R}}^2\setminus \{(0,0)\}$; it is undefined at the origin. Take the counterclockwise unit circle $\mathbf{r}(t)=(\cos t,\sin t)$, $0\le t\le 2\pi$. Along it,By the fundamental theorem for line integrals, it has no single-valued potential of a conservative vector field on $U$. In particular it cannot be the restriction of the gradient of a smooth function on all of ${\mathbb{R}}^2$.

For $(x,y)\ne (0,0)$, direct differentiation givesThere is no contradiction with Green's theorem: $\mathbf{F}$ is not defined, let alone continuously differentiable, throughout the disk enclosed by $C$. On an annulus the theorem does apply; the counterclockwise outer integral is $2\pi$, and the clockwise inner integral is $-2\pi$, so their sum is zero. The hole prevents this curl-free vector field from being a conservative vector field on its entire domain.

Parametrize the straight segment by $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Then the line integral isThe three different answers are possible because this vector field is not a conservative vector field; for example ${\partial }_y{F}_x=6$ whereas ${\partial }_x{F}_y=0$.

Evaluate the line integral separately on the three oriented line segments. Their parametrizations and contributions areThusThe different value in i demonstrates path dependence. The vector field in this part is different from the conservative vector field in part (a).

On each of the two radial legs, $\mathbf{F}=\mathbf{a}\times \mathbf{r}$ is perpendicular to the tangent. Each contribution is zero, giving $\boxed{\int \mathbf{F}\cdot d\mathbf{r}=0}$. This differs from the straight-path answer, confirming path dependence.

For $\mathbf{G}$, use $\mathbf{r}=(s,0,0)$ with $s$ decreasing from one to zero, followed by $\mathbf{r}=(0,s,0)$ with $s$ increasing from zero to one. The two contributions areThey cancel: $\boxed{\int \mathbf{G}\cdot d\mathbf{r}=0}$. This agreement is a consequence of the conservative vector field property already proved, rather than evidence sufficient by itself to establish it.

In Cartesian components, $\mathbf{F}=(z-y,x,-x)$, soIt cannot be a conservative vector field. For the second field, put $\Phi (\mathbf{r})=-(1+|\mathbf{r}{|}^2{)}^{-1/2}$. Its gradient is $\nabla \Phi =\mathbf{r}/(1+|\mathbf{r}{|}^2{)}^{3/2}=\mathbf{G}$. This smooth global scalar function proves $\mathbf{G}$ is conservative, with $\boxed{\nabla \times \mathbf{G}=0}$. The following path integrals exhibit the different behavior directly.

The field is smooth on the simply connected domain ${\mathbb{R}}^3$. Its curl is $\nabla \times \mathbf{F}=(2\lambda -2,2\mu ,0)$, so a necessary and sufficient condition for a conservative vector field isAt those values it is explicitly the gradient of $\Phi =xy-x^2/2-yz+z^2/2$. Equality of the two previous path integrals would impose only $\lambda -\mu =1$, which is weaker: coinciding path integrals need not imply a conservative field. For example $\lambda =2,\mu =1$ makes those two integrals equal but leaves nonzero curl.

Direct differentiation gives $\boxed{\Phi =x^3{y}^{2}z+C}$ with $\nabla \Phi =\mathbf{G}$, so $\mathbf{G}$ is a conservative vector field and its integral depends only on the endpoints. In contrast,is nonzero, for example at $(1,1,1)$. Thus $\mathbf{F}$ is not conservative. Both earlier paths lie in the surface $y=z$, where the two fields happen to coincide. This illustrates why coinciding path integrals need not imply a conservative field.

On the first segment only $z$ varies and $F_z=x^{2}y=0$. On the second only $y$ varies and $F_y=x^{2}z=0$. On the last segment $y=z=\pi$, so the line integral isThe agreement is the expected path independence of a conservative vector field.

Integrating ${\psi }_x=-(2xy+z)$ with respect to $x$ gives $\psi =-x^{2}y-xz+h(y,z)$. Matching ${\psi }_y=-(x^2+2y)$ requires $h=-y^2+j(z)$, and matching ${\psi }_z=-x$ requires $j$ constant. A scalar potential with the requested sign isThus the force field is a conservative vector field on all of ${\mathbb{R}}^3$.

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).

Direct calculation of the curl givesSince $\mathbf{a}\ne 0$, the first vector field cannot be a gradient field: the curl of a gradient is zero. The second is a smooth polynomial field on all of the simply connected space ${\mathbb{R}}^3$, so its vanishing curl makes it a conservative vector field. A potential is also constructed explicitly in the next part. Thus ${\mathbf{v}}_2$ is a conservative vector field and ${\mathbf{v}}_1$ is not. A vanishing divergence alone, as for ${\mathbf{v}}_1$, is a different property and does not establish conservativity.

On a connected domain, a vector field is a conservative vector field if and only if it is the gradient of a single-valued scalar potential, $\boxed{\mathbf{H}=\nabla \varphi }$. Equivalently, its line integral is path independent, or its integral around every closed curve is zero. For a continuously differentiable field on a simply connected domain, this is equivalent to $\nabla \times \mathbf{H}=0$. Without the domain hypothesis, vanishing curl alone is only a local test and is not the necessary-and-sufficient global statement.

The fundamental theorem for line integrals evaluates this conservative vector field integral by the change in potential. At the initial point $\varphi =0$. At the final point $z=0$, so the exponential factor is one, andThe detailed shape of the path does not affect this line integral.

Parametrize the path by $\mathbf{r}(t)=(x(t),y(t),z(t))$, from $t=a$ to $t=b$. For the conservative vector field $\mathbf{F}=\nabla \Phi$, the chain rule givesIntegrating proves the fundamental theorem for line integrals:Only the endpoints matter, so the integral is independent of the connecting path.

Differentiate the components directly:Thus $\boxed{\nabla \times \mathbf{F}=0}$. The field is smooth on all of ${\mathbb{R}}^3$, a simply connected domain, so the zero-curl condition is sufficient for a conservative vector field. The domain condition matters: zero curl alone need not imply a global potential on a region with holes.

The field $\mathbf{F}$ is not a conservative vector field: either nonzero closed line integral computed below rules out a scalar potential. Equivalently, its planar curl is ${\partial }_x{F}_y-{\partial }_y{F}_x=1-(-1)=2$.

For $\mathbf{G}$, integrating ${\Phi }_x=2x{y}^2$ gives $\Phi =x^2{y}^2+h(y)$. Matching ${\Phi }_y=2y{x}^2$ then gives $h^{\prime }=0$. ThusThis also explains why both closed line integrals of $\mathbf{G}$ vanish.

A continuously differentiable conservative vector field has zero curl. HereThis vanishes everywhere if and only if $\boxed{\beta =0}$. Since the field is smooth on the simply connected domain ${\mathbb{R}}^3$, zero curl is also sufficient; the explicit scalar potential in the next part confirms it.