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The Cartesian curl isEach component measures circulation in the plane perpendicular to its corresponding coordinate axis.

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.

The curl of the constant vector field $\mathbf{a}\times \mathbf{b}$ vanishes. The curl of the position field $\mathbf{r}=(x,y,z)$ also vanishes, since all its cross-coordinate partial derivatives are zero. Thus

Again write the field as $\mathbf{b}(\mathbf{a}\cdot \mathbf{r})-\mathbf{r}(\mathbf{a}\cdot \mathbf{b})$. The second term has zero curl. Applying the product rule to the first gives $\nabla (\mathbf{a}\cdot \mathbf{r})\times \mathbf{b}=\mathbf{a}\times \mathbf{b}$. Thus

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.

The three components of the curl cancel pairwise:The domain is all of ${\mathbb{R}}^3$, which is simply connected, so this is consistent with a conservative vector field. To construct its scalar potential, integrate ${\Phi }_x=y+z$ to obtain $\Phi =x(y+z)+h(y,z)$. Matching ${\Phi }_y=x+z$ gives $h_y=z$, hence $h=yz+k(z)$. Finally ${\Phi }_z=x+y$ forces $k^{\prime }(z)=0$. ConsequentlyEvery scalar potential differs by a constant on this connected domain. The endpoint values differ by one, agreeing with the line integral in part (a).

For the second vector field, the curl isIt is nonzero, for example $(0,-2,2)$ at $(1,0,0)$. A differentiable scalar potential would have zero curl of its gradient, so $\mathbf{G}$ is not conservative. The fact that its curl vanishes on the line $x=y=z$ does not make it vanish throughout the domain.

Split the vector field into $a(-y,x,0)$ and $e^{-r^2}(x,y,z)$. The second part is the gradient of $-\frac{1}{2}{e}^{-r^2}$, so its curl vanishes. The first part has only a vertical curl component, ${\partial }_x(ax)-{\partial }_y(-ay)=2a$. Hence

The curl found in part (a) is constant, so its divergence isThis also follows from the identity divergence of a curl is zero, applicable here because the field is smooth everywhere.

The upward unit vector extracts the vertical curl component:It is constant over the unit disc, soThis agrees with part (d) by Stokes theorem: the boundary circle has counterclockwise orientation when viewed from the upward normal, including for the parametrizations $2\pi t^n$.

For $a=0$, differentiate the proposed scalar potential directly:ThusFor $a\ne 0$, the curl is the nonzero vector $(0,0,2a)$. A gradient of a twice differentiable scalar field has zero curl, so no such scalar field exists. Equivalently the closed line integral in part (d) is nonzero, contradicting the path independence of a conservative vector field. The field is smooth on all of ${\mathbb{R}}^3$, so no excluded singular set complicates this conclusion.

In cylindrical coordinates, $x^2+y^2=r^2$ and $r>0$, soThe potential vector has components $(A_r,A_{\theta },A_z)=(0,0,a\ln r)$. Expanding the prescribed cylindrical curl formula givesThereforeSince ${\hat{\mathbf{e}}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, this agrees with the Cartesian result and its clockwise orientation.

The logarithm is defined away from the $z$ axis, where $x^2+y^2>0$. For the vector potential $\psi {\hat{\mathbf{e}}}_z=(0,0,\psi )$, its curl is $({\psi }_y,-{\psi }_x,0)$. Sincewe obtainFor $a>0$ this is clockwise circulation about the axis, the velocity structure of a point vortex in each transverse plane.

The curl of $\mathbf{G}=\mathbf{a}\times \mathbf{r}$ is $2\mathbf{a}$. Thus for $\mathbf{a}\ne 0$ it cannot be a gradient field, while for $\mathbf{a}=0$ it is zero and all its vector-field potentials are constants. This zero-vector case is allowed by the question.

For the radial vector field $\mathbf{H}$, write $\rho =|\mathbf{r}|$. Choose a radial primitiveFor $\rho >0$, the chain rule givesHence all vector-field potentials on a connected domain where the field is defined areThe assumed existence of the needed derivatives and integrals justifies this construction. If the domain includes the origin and the field is regular there, choose ${\rho }_0=0$ and extend the primitive there. Any two vector-field potentials differ by a constant on each connected component.

For $\mathbf{F}$, all three curl components vanish for every $a$. More constructively,Hence $\mathbf{F}$ is conservative for every real $a$. Its endpoint vector-field potential difference is $a$, checking both line integrals.

For the other vector field, direct partial differentiation givesIt therefore has a vector-field potential exactly when $b=0$, with no restriction on $c$:The value of $a$ is irrelevant to $\mathbf{G}$. Its rotational part for general $b$ is $b(1,1,1{)}^T\times \mathbf{r}$. Both selected paths lie in $x=y$, and this rotational part has zero dot product with every tangent vector to that plane. That explains why these paths cannot detect its nonzero curl when $b\ne 0$.

The curl of a gradient is zero pointwise, so its surface integral over this sphere isFor this particular polynomial, $\nabla \Phi =2(x,y,z{)}^T$, whose curl vanishes directly as well.

For $\mathbf{G}$, the product rule for the curl givesSince $\mathbf{a}\ne 0$, this is not identically zero on ${\mathbb{R}}^3$. A gradient field has zero curl, so no scalar potential exists for $\mathbf{G}$ on the whole space.

For the radial vector field $\mathbf{K}$, observe thatThus all vector-field potentials areThese are smooth everywhere, including the origin. Any two differ by a constant because their gradient difference vanishes on the connected whole space.

On the circular arc, $z=0$ and $dz=0$, so the line integral reduces toThis exact differential evaluation applies to the specified arc; for example its parametric equation is $(x,y,z)=(1-\cos \theta ,\sin \theta ,0)$, $0\le \theta \le \pi /2$. On the vertical segment $x=y=1$, only $dz$ contributes, giving ${\int }_0^1(\lambda +1)dz=\lambda +1$. The whole path therefore givesTo answer the final conservativeness requirement, compute the curl:It vanishes identically exactly whenOn the simply connected whole space this condition is sufficient for a conservative vector field. Explicitly a vector-field potential isBoth path calculations then give $13/6$, equal to $\varphi (1,1,1)-\varphi (0,0,0)$ by the fundamental theorem for line integrals. Equality of just these two path values would only force $\mu =2\lambda +1$; it is the curl calculation that establishes full path independence.

Put $\mathbf{W}=\nabla \times \mathbf{F}$ and $\mathbf{v}=\dot{\mathbf{r}}$. All vector fields and their spatial partial derivatives below are evaluated at $\mathbf{r}(t)$. The product rule for a dot product givesBecause the vector field has no explicit time dependence, the multivariable chain rule gives $d\mathbf{W}/dt=J_{\mathbf{W}}\dot{\mathbf{r}}$, where the Jacobian matrix has entries $(J_{\mathbf{W}}{)}_{ij}={\partial }_j{W}_i$. ConsequentlyThe first term measures the change of the curl sampled as the point moves; the second measures the change in the velocity vector used in the dot product.

To give the result explicitly in $P,Q,R$, define the directional derivative operator $L=\dot{x}{\partial }_x+\dot{y}{\partial }_y+\dot{z}{\partial }_z$. Sincethe same expression isFor example $L(R_y-Q_z)=\dot{x}(R_{yx}-Q_{zx})+\dot{y}(R_{yy}-Q_{zy})+\dot{z}(R_{yz}-Q_{zz})$; the other two terms expand in the same way. Thus only the first and second spatial partial derivatives of $P,Q,R$, together with the velocity and acceleration of the parametric curve, occur.

By the Cartesian formula for the curl,