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Green's theorem for a positively oriented simple closed planar curve statesHere $Q_x-P_y=1-(-1)=2$. The trapezium has parallel sides of lengths two and one, separated by height one, so its area is $3/2$. ThereforeEvery congruent trapezium gives the same answer when its boundary is traversed anticlockwise, since the scalar curl is constant and geometric congruence preserves area. Reversing the traversal would instead reverse the sign.

The curl is $\nabla \times \mathbf{F}=(x,y,-2z)$. For the upward orientation of the paraboloid, the vector surface area element is $(2x,2y,1)dxdy$. Therefore Stokes theorem givesThe upward normal gives the prescribed anticlockwise boundary orientation viewed from above.

Write $r=\sqrt{x^2+y^2}>0$, so $\mathbf{F}=(f(z)x/r,f(z)y/r,0)$. The Cartesian curl isThe last component is zero because both differentiated terms are $-f(z)xy/r^3$. ThusThe condition $r>0$ excludes the cylindrical-axis coordinate singularity.

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.

Choose the outward unit normal vector on the upper hemisphere. The curl of $\mathbf{F}=(4y,x,2z)$ isIn spherical polar coordinates, putHere $n_z=\cos \phi$ and $dS=a^2\sin \phi d\phi d\theta$. The surface integral is thereforeThe compatible boundary orientation follows increasing $t$ in $\mathbf{r}(t)=(a\cos t,a\sin t,0)$, $0\le t\le 2\pi$. On this circle,Thus the line integral iswhich agrees with the surface integral as required by the Stokes theorem.

Let $\Sigma$ be a piecewise smooth oriented surface, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of it. Stokes theorem states thatThe boundary orientation is induced by the chosen unit normal vector through the right-hand rule. Smoothness on a neighbourhood makes the spatial partial derivatives in the curl well defined.

Choose the outward orientation of a surface on the lower half of the ellipsoid. At its lowest point the normal points down, so its boundary ellipse is traversed clockwise when viewed from above. UseThe line integral is

For an independent surface integral calculation, first compute $\nabla \times \mathbf{F}=(2y,-2x,1)$. Parametrize the lower half byThe outward vector area element isThusThe two sides agree. Choosing the inward unit normal vector and its corresponding counterclockwise boundary orientation changes both answers to $6\pi$.

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.

Let $a$ be an arbitrary constant vector. By the cyclic identity for the scalar triple product,Apply Stokes theorem to $\mathbf{F}\times a$. Since $a$ is constant, the curl product identity givesConsequentlyFor precision, the operator in the last line has componentsThe partial derivatives act on $\mathbf{F}$, with $\mathbf{n}$ multiplying their values; they do not differentiate $\mathbf{n}$. Since the dot products agree for every constant $a$, the cross-product form of Stokes theorem follows:Here $\wedge$ in the paper denotes the cross product, and the printed surface integral has its area element $dS$ understood.

Compute the curl in Cartesian coordinates:The three components are $6y{z}^2-6y{z}^2$, $0-0$, and $e^x-e^x$. Consequently

Compute the curl component by component from the given vector field:For example, $B_x={\partial }_y{A}_z-{\partial }_z{A}_y=0-(-2xz)$ and $B_z={\partial }_x{A}_y-{\partial }_y{A}_x=(1-z^2)-(z^2-1)$. Its divergence isas expected because the divergence of a curl is zero.

Take $C$ counterclockwise when viewed from positive $z$, the disc $D$ upward oriented, and the hemisphere $H$ outward oriented. These choices give both surfaces the same positive boundary orientation. Parametrize $C$ by $r(t)=(\cos t,\sin t,0)$, $0\le t\le 2\pi$. On it, $A=(-\sin t,\cos t,0)$ and $dr=(-\sin t,\cos t,0)dt$, so the line integral isOn $D$, use polar coordinates $(x,y)=(r\cos \varphi ,r\sin \varphi )$ and $d\mathbf{S}=(0,0,1)rdrd\varphi$. Since $B_z=2$ there, the surface flux isOn the unit hemisphere, the outward unit normal vector is $n=(x,y,z)$. Since $x^2+y^2=1-z^2$, we have $B\cdot n=4z(1-z^2)$. In spherical polar coordinates, $z=\cos \theta$ and $dS=\sin \theta d\theta d\varphi$, giving the explicit surface integralIndeed the inner integral is $[{\sin }^4\theta {]}_0^{\pi /2}=1$.

These equalities illustrate Stokes theorem twice: $B=\nabla \times A$ and either upward $D$ or outward $H$ spans the oriented circle $C$, so each surface flux equals the same line integral. For the divergence theorem, close the upper half-ball by adjoining the downward oriented base, namely $-D$. Its outward surface flux isThe minus sign on the base is required because its outward normal vector points down. Reversing a specified orientation reverses that integral's sign.

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.

Orient the three faces outward from the tetrahedron. The two coordinate-plane faces have normals parallel to $-{\mathbf{e}}_x$ or $-{\mathbf{e}}_y$, so their flux of the curl is zero. On the inclined face $z=1-x-y$, its upward vector area is $(1,1,1)dxdy$, with $x,y\ge 0$, $x+y\le 1$. Thus the direct surface integral isThis surface induces the counterclockwise boundary $O\to X\to Y\to O$ viewed from above, where $X=(1,0,0)$ and $Y=(0,1,0)$. At $z=0$, $\mathbf{A}=(2x-y,0,0)$. On $OX$, take $(x,y)=(t,0)$, giving integral ${\int }_0^{1}2tdt=1$. On $XY$, take $(x,y)=(1-t,t)$, giving ${\int }_0^1(-2+3t)dt=-1/2$. On $YO$, $dx=0$ and the integral vanishes. ThereforeBoth sides have been calculated directly, verifying the Stokes theorem with consistent orientations.

Differentiate the Cartesian components to calculate the curl:

A differentiable gradient field must have equal mixed derivatives. Here $\partial F_y/\partial z=3z^2$ but $\partial F_z/\partial y=-x$, which are not identically equal on any open region. Equivalently,Thus no scalar potential exists for this field on an open region. Vanishing of the curl on a special lower-dimensional set would not supply a three-dimensional potential.

The Cartesian curl is obtained by cross-differentiating the components:

No scalar potential exists on the full three-dimensional domain. A sufficiently smooth gradient field has zero curl, since mixed partial derivatives commute. The calculated curl is not identically zero. The fact that it vanishes at some individual points does not make the whole field a gradient field.

The $z$ component of the curl is ${\partial }_x{F}_y-{\partial }_y{F}_x=e^{x+y}-x{e}^{xy}$, which is not identically zero; at the origin it is one. Hence the field is not conservative on ${\mathbb{R}}^3$ and has no global scalar potential.

Each component depends only on its own coordinate, so all cross partial derivatives vanish. Therefore the curl isThis also follows from the identity that the curl of a gradient vanishes for a twice continuously differentiable potential energy.

Taking the curl component by component gives

The curl is $\nabla \times {\mathbf{F}}_2=(2z,3z^2,0)$, which is not identically zero. A gradient of a smooth scalar potential has zero curl, so ${\mathbf{F}}_2$ is not conservative on ${\mathbb{R}}^3$.