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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.

A scalar potential is $U(x,y)=x^{3}y+y^4$, since $\nabla U=(3x^{2}y,x^3+4y^3)$. The fundamental theorem for line integrals therefore reduces the line integral to its endpoints. The curve starts at $(0,1)$ and ends at $(2,1)$, so

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.

For an oriented piecewise smooth parametric curve $\mathbf{r}:[a,b]\to {\mathbb{R}}^3$ and a continuous vector field $\mathbf{F}$ along it, the vector-field line integral isFor a piecewise smooth curve, sum these integrals over its smooth pieces. The dot product selects the component of the vector field in the direction of motion.

If $\mathbf{F}=\nabla \varphi$ for a smooth function $\varphi$, the multivariable chain rule givesThe fundamental theorem of calculus, with cancellation at intermediate endpoints, therefore proves the fundamental theorem for line integrals:On a closed curve $\mathbf{r}(a)=\mathbf{r}(b)$, so $\boxed{{\oint }_C\nabla \varphi \cdot d\mathbf{r}=0}$.

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.

Integrate $f_x=y{e}^x$ with respect to $x$, obtaining $f=y{e}^x+g(y,z)$. Matching $f_y=e^x+2y{z}^3$ gives $g_y=2y{z}^3$, so $g=y^2{z}^3+h(z)$. Finally, $f_z=3y^2{z}^2+h^{\prime }(z)$ must equal the third component, so $h^{\prime }=0$. Thus a scalar potential isThe fundamental theorem for line integrals now gives ${\int }_{C}v\cdot dr=f(1,1,1)-f(0,0,0)=e+1$ for either path, confirming both direct computations. Since the curl of a gradient is zero for this smooth function, it also confirms the result in part (b). The potential is defined throughout ${\mathbb{R}}^3$, so no domain obstruction to path independence occurs.

Parametrize the segment by $\mathbf{r}(t)=(1-t,t,0)$, $0\le t\le 1$. Its tangent is $(-1,1,0)$ and $\mathbf{F}=(-t,1-t,t-1)$, so $\mathbf{F}\cdot {\mathbf{r}}^{\prime }=1$. Hence its line integral isFor the other line integral, $|\mathbf{r}{|}^2=(1-t{)}^2+t^2$ and $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=2t-1$. ThusThis also equals the change in $\Phi$, by the fundamental theorem for line integrals.

Parametrize the straight path by $\mathbf{r}(t)=(t,t,t)$, $0\le t\le \pi$. Then ${\mathbf{F}}_3\cdot d\mathbf{r}=(4t^3+\sin t)dt$, so direct evaluation of the line integral givesThe fundamental theorem for line integrals supplies an independent check: this is ${\varphi }_3(\pi ,\pi ,\pi )-{\varphi }_3(0,0,0)$.

Parametrize the straight line by $\mathbf{x}(t)=(t,t,t)$, $0\le t\le 1$. The line integral isThe change in scalar potential is $\psi (1,1,1)-\psi (0,0,0)=-3$, so the answer is $-\Delta \psi$, as required by the fundamental theorem for line integrals for $\mathbf{F}=-\nabla \psi$.

On the first segment, $x=z=0$, so both line integrals vanish. On the second, $y=1,z=0$, giving contributions ${\int }_0^{2}3x^{2}dx=8$ and ${\int }_0^{2}2xdx=4$. On the last, $x=2,y=1$, the contributions are ${\int }_0^3-2zdz=-9$ and ${\int }_0^3-6z^{2}dz=-54$. ThereforeThe first also equals $\varphi (B)-\varphi (A)=8-9=-1$, as the fundamental theorem for line integrals predicts.

Parametrize the segment by $\mathbf{r}(t)=(1-t,t,0)$, $0\le t\le 1$, so ${\mathbf{r}}^{\prime }=(-1,1,0)$. For the first field, ${\mathbf{F}}_1=(-t,1-t,-1+t)$ and its dot product with the tangent is one. ThereforeFor the second field, use the fundamental theorem for line integrals. The endpoints both have $r=1$, so their potential values agree andDirectly, its parameter integrand is $(2t-1)/[1+(1-t{)}^2+t^2{]}^{3/2}$, which changes sign under $t\mapsto 1-t$ and integrates to zero.

The gradient part contributes zero around a closed path by the fundamental theorem for line integrals:Along the parametrization, $x^2+z^2=1$, so the first two components of $\mathbf{w}$ vanish wherever the printed expression is defined. The remaining component gives $\mathbf{w}\cdot d\mathbf{x}=xdz={\cos }^2\theta d\theta$. ThusThe printed field needs a qualification: $\mathrm{arctanh}(x^2)$ is undefined at the isolated parameters where $x^2=1$, so $0\mathrm{arctanh}(1)$ cannot literally be evaluated. The result above is the improper line integral, or equivalently the integral of the continuous extension of the restricted field along the curve. That restricted field tends to $(0,0,\cos \theta )$ at each omitted parameter. No extension of the full field to a neighbourhood is being assumed.

Parametrize the directed segment by $\mathbf{r}(t)=(t,t,0)$, $0\le t\le 1$. Its tangent is $(1,1,0)$, while the restricted vector field is $(t^3,t^3,0)$. The line integral is thereforeAlternatively, $\mathbf{G}=\nabla (x^2{y}^2/2)$, so the fundamental theorem for line integrals gives the same endpoint difference.

Parametrize the straight path by $\mathbf{r}(t)=(t,t,2t)$ for $0\le t\le 1$. The conservative field along it is $(6t^5,4t^5,t^5)$ and the tangent is $(1,1,2)$. ThusThe associated potential has ${\varphi }_1(B)-{\varphi }_1(A)=2$, in agreement with the fundamental theorem for line integrals.

On the first straight segment, $y=z=0$ and only $dx$ is nonzero, so both integrals vanish. On the second segment, $x=1,z=0$ and only $dy$ is nonzero; each integrand is one, contributing $1$. On the last segment, $x=y=1$ and only $dz$ is nonzero; each integrand is $\sinh z$, contributing $\cosh 1-1$. ThusThe second result also follows from the fundamental theorem for line integrals, using the potential found in (a). Agreement on this one path does not make $\mathbf{F}$ conservative.

Integrating ${\varphi }_x=2xy$ gives $\varphi =x^{2}y+h(y,z)$. Matching ${\varphi }_y=x^2+3y^2+z^2$ then gives $h=y^3+y{z}^2+k(z)$. Finally ${\varphi }_z=2yz$ forces $k$ to be constant. A scalar potential isThe initial and final points of the helix are $(b,0,0)$ and $(0,b,cL)$, where $L=5\pi /2$. The fundamental theorem for line integrals givesagreeing with the directly evaluated $I_2$.

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.

Evaluate the line integral along the three coordinate segments, treating coordinates not changing along a segment as constants. On the first segment, $y=z=0$, so $F_x=0$ and the contribution is zero. On the second segment,On the final segment,The fundamental theorem for line integrals therefore givesTaking its gradient verifies each component of the field. The constant is $C=\Phi (0,0,0)$; the field specifies only potential differences because the gradient of every constant is zero.

It is helpful first to separate an exact gradient:The fundamental theorem for line integrals makes the gradient contribution $2\alpha +{\alpha }^2$ for either path. On the curved path, $dy=2tdt$, $dz=3t^{2}dt$, $y=t^2$ and $z=t^3$, so the remaining contribution isThus