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Let $g$ be a continuous $2\pi$-periodic function and $\mathbf{F}=g(\theta ){\mathbf{e}}_{\theta }/r$ on $r>0$. On a circle about the axis, $d\mathbf{r}=r{\mathbf{e}}_{\theta }d\theta$, giving the displayed line integral. A single-valued potential of a conservative vector field on the full punctured space exists if and only if this integral is zero. Necessity is the fundamental theorem for line integrals. For sufficiency, the primitive $P(\theta )={\int }_0^{\theta }g(s)ds$ is $2\pi$-periodic when the mean is zero, and the gradient in cylindrical coordinates gives $\nabla P=\mathbf{F}$.

On a domain with one angular cut removed, the same primitive gives a potential of a conservative vector field even when the mean is nonzero, because the angle has a single-valued continuous branch there. For example $g(\theta )=\sqrt{1+\cos \theta }$ has integral $4\sqrt{2}$ around the full circle but primitive $2\sqrt{2}\sin (\theta /2)$ on $-\pi <\theta <\pi$.

A vector field $X$ is conservative when $X=\nabla \Phi$ for a potential of a conservative vector field $\Phi$. On a simply connected open subset of ${\mathbb{R}}^3$, every continuously differentiable curl-free vector field is conservative. Locally this follows from the Poincaré lemma applied to the differential one-form $X_{i}d{x}^i$; globally, simple connectivity makes its line integral independent of the path. The fundamental theorem for line integrals expresses that path independence as the difference of the potential at the endpoints.

On the punctured plane ${\mathbb{R}}^2\setminus \{(0,0)\}$, consider the vector fieldWith $r^2=x^2+y^2$, its partial derivatives satisfyThus the field has zero scalar curl everywhere it is defined. Nevertheless, on any counterclockwise circle ${\gamma }_R(t)=(R\cos t,R\sin t)$ with $R>0$,The radius cancels exactly.

Applying Green theorem to the enclosed disk would be invalid: $G$ is undefined at the origin, inside that disk. Zero curl on the path alone is insufficient.^[ref]

Now take the annulus $A=\{(x,y):1\le r\le 2\}$. Its positive boundary orientation is counterclockwise on the outer circle and clockwise on the inner circle. The complete boundary line integral is thereforeHere Green theorem applies: an open set containing the closed annulus can avoid the origin. The inner contribution remains $-2\pi$ even for an arbitrarily small positive inner radius. Decreasing the hole's area does not decrease this line integral.

The cancellation has a useful proof picture. Cut the annulus along two radial segments into two half-annuli. Each piece has a single simple closed curve as its boundary; along each cut, the two induced boundary orientations oppose, so those line integrals cancel when the two applications of Green theorem are added. Only the outer and inner circles remain.^[ref]

Locally, a choice of angle $\theta$ in polar coordinates has gradient $\nabla \theta =G$. Globally, following a circle once increases that angle by $2\pi$, so it cannot be a single-valued potential of a conservative vector field on the punctured plane. The nonzero closed-path line integral also rules out such a potential by the fundamental theorem for line integrals.

On a simply connected open region, a continuously differentiable planar vector field with zero scalar curl does have a global potential. The hole is what separates the local derivative condition from the global conclusion.^[ref]

A continuously differentiable vector field on an open subset of three-dimensional Euclidean space is irrotational when its curl is zero. Under the Euclidean identification $\alpha =X_{i}d{x}^i$, this says that the differential one-form $\alpha$ is a closed differential form. The Poincaré lemma gives a potential of a conservative vector field locally. A global potential additionally requires zero line integrals around every closed curve; simply connected domains ensures this. On the punctured plane extended in the third direction, $X=(-y,x,0)/(x^2+y^2)$ has zero curl but integral $2\pi$ around a circle, so it is locally conservative and has no single-valued global potential.

With those constants, integrate ${\varphi }_x=x+3y+3z$ with respect to $x$ to obtain $\varphi =x^2/2+3xy+3xz+g(y,z)$. Matching ${\varphi }_y=3x+y-z$ gives $g=y^2/2-yz+h(z)$. Finally, matching ${\varphi }_z=3x-y-2z$ gives $h=-z^2+C$. Thus a potential of a conservative vector field isThis explicit polynomial establishes a global potential on all of ${\mathbb{R}}^3$; no topological obstruction is present. Its additive constant is arbitrary.

The field is continuously differentiable on all of ${\mathbb{R}}^3$, a simply connected domain. Its vanishing curl therefore implies a global potential of a conservative vector field, so $\mathbf{F}\cdot d\mathbf{r}=d\varphi$ is an exact differential. This global domain condition matters: curl-free fields on domains with holes need not have single-valued potentials.

Integrating ${\varphi }_x=2y{z}^2$ gives $\varphi =2xy{z}^2+g(y,z)$. Matching ${\varphi }_y=2x{z}^2$ forces $g_y=0$, so $g=h(z)$; matching ${\varphi }_z=4xyz$ then forces $h^{\prime }=0$. ConsequentlyThe additive constant does not affect the field or any endpoint difference.

A one-form $A(x,y)dx+B(x,y)dy$ is an exact differential if a single-valued differentiable potential of a conservative vector field $f$ satisfies $A=f_x$ and $B=f_y$ on the domain. Then it equals $df$, and its integral along a path is the endpoint difference of $f$.

Integrate $f_x=10x^4-2x{y}^3$ with respect to $x$:Its $y$ derivative is $-3x^2{y}^2+g^{\prime }(y)$. Matching the required second component forces $g^{\prime }(y)=0$, so the additive constant is arbitrary andThe fundamental theorem for line integrals now givesindependently of the convenient segmented path used before. Merely checking equality of mixed partial derivatives on an unspecified domain would not by itself construct this globally defined potential.

Take $E$ to be continuously differentiable everywhere on ${\mathbb{R}}^3$, as in the question. Its zero curl says ${\partial }_i{E}_j={\partial }_j{E}_i$. Construct a potential of a conservative vector field explicitly byAll segments lie in ${\mathbb{R}}^3$. Differentiating under the integral and using the symmetry of the derivatives,Hence $E=-\nabla \varphi$. This radial integral potential for a curl-free field also works on a star-shaped domain after translating its centre. On an arbitrary domain, zero curl only guarantees local potentials; the everywhere-on-${\mathbb{R}}^3$ interpretation is important for this global assertion.

For the given field, put $q=e^{-x^{2}z}$. The three pairs of derivatives entering its curl areThus the field is irrotational. To find its potential using $E_y=-{\partial }_y\varphi$, integrate ${\partial }_y\varphi =2yq$ to get $\varphi =y^{2}q+\psi (x,z)$. Comparison with $E_x=-{\partial }_x\varphi$ gives ${\psi }_x=0$, and comparison with $E_z=-{\partial }_z\varphi$ gives ${\psi }_z=0$. ThereforeDifferentiating this expression reproduces all three components with the required minus sign; the only freedom on the connected domain is the additive constant.

A continuously differentiable vector field is an irrotational vector field when its curl vanishes:For the global-potential construction, assume that the domain is connected and simply connected, or more generally that all closed-curve line integrals of $\mathbf{F}$ vanish. Then define the potential of a conservative vector field byTwo paths with the same endpoints differ by a closed curve. On a simply connected domain that curve can be spanned, or contracted through piecewise smooth spanning surfaces, inside the domain. Stokes theorem gives zero circulation because the curl is zero. The value of $V$ is therefore path-independent and $V({\mathbf{x}}_0)=0$. To compute a derivative, append a short coordinate segment to a path ending at $\mathbf{x}$. Then $V(\mathbf{x}+h{\mathbf{e}}_i)-V(\mathbf{x})=-{\int }_0^h{F}_i(\mathbf{x}+s{\mathbf{e}}_i)ds$, so ${\partial }_{i}V=-F_i$.

The domain qualification is necessary and is implicit in the potential request. Curl-free alone on an arbitrary domain does not suffice: $\mathbf{F}=(-y,x,0)/(x^2+y^2)$ on the complement of the $z$-axis has zero curl, but its circulation around a unit circle is $2\pi$. It has no single-valued global potential. The intended construction is valid on all of ${\mathbb{R}}^3$ for a smooth curl-free field, and on any other domain satisfying the condition above.

For the given spherical-coordinate field, the angular components have $r{F}_{\theta }=\cos \theta \cos \varphi$ and $r{F}_{\varphi }=m\sin \varphi$, independent of $r$, so the $\theta$ and $\varphi$ curl components vanish. The radial component isIt vanishes identically exactly when $m=-1$. In that case integrate $F_{\theta }=-r^{-1}{V}_{\theta }$ to obtain $V=-\sin \theta \cos \varphi +C(\varphi )$. The azimuthal equation $F_{\varphi }=-(r\sin \theta {)}^{-1}{V}_{\varphi }$ forces $C^{\prime }=0$, and $F_r=0$ removes any radial dependence. At the given north-pole point the first term is zero, so normalization fixes the constant to zero:The Cartesian expression shows that this potential and the field $\mathbf{F}=\nabla (x/r)$ are smooth for $r>0$, including the spherical-coordinate poles. The origin is excluded because the specified field is singular there; that exclusion does not obstruct the displayed global potential.

For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, the necessary and sufficient condition for a conservative vector field is $\nabla \times \mathbf{F}=0$. Necessity follows from equality of mixed partial derivatives of a potential; sufficiency uses the ${\mathbb{R}}^3$ being a simply connected space. On a general domain the topology cannot be omitted.

Here the relevant mixed partial derivatives areThus the curl vanishes. Integrating the first component in $x$ gives $\Phi =x^2\cos y-2x{z}^3+A(y,z)$. Matching the second component gives $A_y=3+2y{e}^z$, hence $A=3y+y^2{e}^z+B(z)$. Matching the last component forces $B^{\prime }=0$. A potential of a conservative vector field with the convention $\mathbf{F}=\nabla \Phi$ is consequentlyIf a physical potential is defined instead through $\mathbf{F}=-\nabla V$, it is $V=-\Phi$.

Parametrize the smooth path by $\gamma :[0,1]\to G$. The chain rule turns the line integral of the gradient into a one-variable derivative:This is the fundamental theorem for line integrals; in particular the integral has path independence.

For a twice continuously differentiable potential of a conservative vector field, the curl of its gradient vanishes because mixed partial derivatives commute:A smooth gradient field is necessarily curl-free.

A sufficient condition is that $G$ be an open simply connected domain, with $\mathbf{F}$ continuously differentiable. Under this condition a curl-free vector field has a global potential of a conservative vector field and hence path independence by the fundamental theorem for line integrals. One way to see the global step is to deform the closed loop formed by two paths into a point inside $G$: the integral is unchanged during the deformation because the curl is zero, as expressed by Stokes theorem. Therefore simple connectivity is a sufficient domain hypothesis. It is not a claim that every particular curl-free field needs such a domain; a given field can have a global potential on a domain with holes.

Away from the origin, the given vector field is the gradient of the globally single-valued radius:The sphere has centre $(5,3,2)$, at distance $\sqrt{38}$ from the origin, and radius one. Its intersection circle is therefore wholly away from the origin, since every point on the sphere has distance at least $\sqrt{38}-1>0$. Along the closed curve the chain rule gives $\mathbf{F}\cdot d\mathbf{x}=dr$. Consequently, independently of orientation,The intersection is nondegenerate: the centre's distance from the plane is $4/\sqrt{35}<1$. No parametrization of the resulting circle is needed because $r$ is a potential of a conservative vector field throughout the punctured space.

Set $s=x^2+y^2>0$. The first two components of the curl vanish since the horizontal components are independent of $z$ and the last component depends only on $z$. For the third,so this is a curl-free vector field everywhere in its domain:Choose the orientation in which the projection onto the $xy$-plane runs counterclockwise, and parametrizeOn this curve the horizontal contribution is $(-\sin t,\cos t)\cdot (-\sin t,\cos t)dt=dt$, while the vertical contribution is $zdz=d(z^2/2)$. Its integral over a closed curve is zero. Thereforewith $-2\pi$ for the opposite orientation. This does not contradict Stokes theorem: the planar disk spanning this curve meets the excluded $z$-axis at $(0,0,200)$, where the vector field is undefined. The horizontal one-form is the angular differential and records one winding about that axis; thus this curl-free vector field has no single-valued global potential of a conservative vector field on its domain.

A conservative vector field on ${\mathbb{R}}^n$ is a vector field $\mathbf{V}=\nabla F$ for a globally defined smooth potential of a conservative vector field $F$. The chain rule givesso its line integral is path independent.

Green theorem states that, for a bounded planar region $D$ with piecewise smooth boundary and continuously differentiable $P,Q$ on a neighbourhood of its closure,The boundary is positively oriented: the region lies on the left while it is traversed, so outer components run counterclockwise and hole boundaries clockwise.

For the proposed potential of a conservative vector field, the fundamental theorem of calculus and differentiation under the integral sign giveSince $Q_x=P_y$, the last integral equals $P(x,y)-P(x,0)$, including negative $y$ with the usual signed-integral convention. Hence the proposed potential is global andThe fact that the vector field is defined on all of ${\mathbb{R}}^2$ ensures both straight integration segments stay inside its domain; a hole in the domain could obstruct a global potential.

Integrating the oscillatory term together with the constant components gives a potential of a conservative vector fieldDirect partial derivatives recover both components of $\mathbf{V}$, proving it is a conservative vector field. Its line integral is therefore path independent. At an integer endpoint $(m,n)$, the sine term vanishes, so for any of the specified pathsAny other potential has the same gradient, so differs from $F$ by a constant on the connected plane. In particular every potential changes by $1$ under $x\mapsto x+1$ and by $2$ under $y\mapsto y+1$. Equivalently the integral from $(0,0)$ to $(1,0)$ is $1$, whereas a unit-periodic potential would make that difference zero. There is no potential periodic with period one in both coordinate directions. This is an example of how periods obstruct a periodic potential: a periodic vector field need not have a periodic potential.

Here a potential of a conservative vector field means a single-valued scalar field $\psi$ with $\nabla \psi =\mathbf{F}$. By the fundamental theorem for line integrals, its line integral around every closed curve must vanish. Here the angular circulation obstruction to a cylindrical potential lets us test this using a circle about the axis.

On $V_1$, the circle from i lies entirely in the region and has line integral $4\sqrt{2}$. Therefore there is no potential of a conservative vector field on $V_1$.

On $V_2$, choose the continuous angular branch $-\pi <\theta <\pi$. By a, the equation for a potential of a conservative vector field reduces to ${\psi }_r={\psi }_z=0$ and ${\psi }_{\theta }=\sqrt{2}\cos (\theta /2)$. ThusThe denominator is positive throughout $V_2$, including when $y=0$ and $x>0$, so this is a well-defined differentiable potential of a conservative vector field on the entire region.

For $V_3$, choose the particular circle $x=2\cos \theta$, $y=2\sin \theta$, $z=0$. Its only point on the removed half-plane's supporting cut is $(-2,0,0)$, where $(x+2{)}^2+z^2=0<1$; this point is retained by $V_3$. Thus the whole circle lies in $V_3$ and again has line integral $4\sqrt{2}$. Consequently there is no potential of a conservative vector field on $V_3$. The opening in the cut allows a closed curve to go around the excluded axis.

Finally $V_4\subset V_2$. Restricting the potential of a conservative vector field already constructed on $V_2$ givesThe extra excluded cylinder does not invalidate an existing potential of a conservative vector field. Thus being simply connected is a sufficient condition in the usual curl-free criterion, not a necessary condition for a particular vector field to have a potential of a conservative vector field.

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.

The sufficiency claim requires a domain qualification. For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, or more generally on a simply connected open domain, the equivalence is valid. On an arbitrary domain, vanishing curl is necessary but may fail to be sufficient.

Necessity. Suppose every closed-curve line integral vanishes. Fix a point $p$ in the open domain and a unit vector $a$. A sufficiently small planar disk $D_{\epsilon }$ of radius $\epsilon$, centred at $p$ and normal to $a$, lies in the domain. Stokes theorem givesDivide by $\pi {\epsilon }^2$ and let $\epsilon \downarrow 0$. By continuity, the average tends to $(\nabla \times \mathbf{F})(p)\cdot a$. Taking the three coordinate directions proves $\boxed{\nabla \times \mathbf{F}=0}$ everywhere.

Sufficiency on ${\mathbb{R}}^3$. A direct proof constructs a potential of a conservative vector field. PutWrite $F_j$ for the components of $\mathbf{F}$. Vanishing curl says ${\partial }_i{F}_j={\partial }_j{F}_i$. Differentiation under the integral, followed by the chain rule, givesThis is the radial integral potential for a curl-free field. The fundamental theorem for line integrals now gives zero on every closed curve. On a simply connected open domain the analogous argument uses path integrals: every loop can be contracted within the domain, and applying Stokes theorem to a piecewise smooth contraction makes its circulation zero. Equivalently, whenever a particular loop bounds an oriented spanning surface entirely in the domain, Stokes theorem immediately gives zero for that loop.

Why the domain matters. On $U={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, the smooth vector fieldhas zero curl, but its line integral around the unit circle in $z=0$ is $2\pi$, as in ii. No spanning surface for this loop can lie entirely in $U$. The literal claim without a condition on the domain therefore has this counterexample.