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Set $A=10x^4-2x{y}^3$ and $B=-3x^2{y}^2$. Their mixed derivatives agree: $A_y=-6x{y}^2=B_x$. Both components are smooth on all of ${\mathbb{R}}^2$, a simply connected space, so the line integral is path independent.

Choose a horizontal segment from $(0,0)$ to $(2,0)$ followed by a vertical segment to $(2,1)$. On the first, $y=0$ and $dy=0$; on the second, $x=2$ and $dx=0$. ThusThe selected path and its orientation are shown below. No contribution comes from a differential whose coordinate is held constant on a segment.

Construct a global vector-field potential rather than relying only on vanishing curl:Thus $\mathbf{F}$ is a conservative vector field. The endpoints of the given parametric curve are $\mathbf{r}(0)=(3,0,0)$ and $\mathbf{r}(\pi /2)=(0,2,1)$. The fundamental theorem for line integrals givesThe intervening turns of the curve have no effect because this line integral is path independent.

By Stokes theorem, it suffices to calculate the boundary line integral. On the $xz$ plane, $y=0$ and $dy=0$, soThis exact differential has zero integral around a closed curve. ThereforeThe argument applies to any admissible spanning surface, including a curved one; the surface itself need not lie in the $xz$ plane.

Choose the oriented surface parametrization $\mathbf{r}(x,y)=(x,y,1-x)$, with $0\le x,y\le 1$. Then ${\mathbf{r}}_x\times {\mathbf{r}}_y=(1,0,1)$ and $\nabla \times \mathbf{A}=(-3z,x^2,y)$. The surface integral isFor the compatible boundary direction, use successive vertices $(0,0,1)$, $(1,0,0)$, $(1,1,0)$, $(0,1,1)$ and back. The four line integrals are respectivelyTheir sum is $-1$, agreeing with the surface integral. Choosing the opposite normal and reversing the boundary direction gives $+1$ on both sides. Stating an orientation is necessary because the question does not prescribe one.

A vector line integral along an oriented parametrized curve $r(t)$ is $\int a(r(t))\cdot r^{\prime }(t)dt$; for a force field it is the work along that path. On the upper branch, set $y=t$, $x=t^2$, $0\le t\le 1$. ThensoThe stated $x$ endpoints do not distinguish the two branches of $y^2=x$. If the lower branch is intended, the same parametrization runs from $t=0$ to $t=-1$, giving instead $\boxed{4/3}$. The two values correspond to different terminal points, $(1,1)$ and $(1,-1)$; choosing a branch is necessary to make the question unambiguous.

The Stokes theorem states that for an oriented smooth surface $S$ with compatible oriented boundary $C$ and a continuously differentiable vector field,Here $\nabla \times F=(-2,-2,-2)$, whereas a normal vector to the specified plane is $(1,0,-1)$. Their dot product is zero. Thus every closed curve in that plane has zero circulation by the Stokes theorem.

The source says “any curve” and displays an ordinary line integral, without specifying that it is closed or giving endpoints. An open curve need not give zero. On the plane, put $z=x-1$ and $dz=dx$; the pulled-back differential one-form becomesTherefore the complete endpoint-dependent answer iswhich specializes to zero for a closed curve. For example, the path from $(1,0,0)$ to $(2,0,1)$ lies in the plane and has integral $1$. This illustrates a conservative restriction of a vector field to a plane despite nonzero ambient curl.

For an oriented piecewise smooth surface $S$ with boundary $\partial S$, and a vector field $F$ continuously differentiable on a neighborhood of $S$, Stokes theorem statesHere $n$ is the chosen unit normal vector, $dS$ is the unsigned surface area element, and $dr$ is the tangent displacement along the boundary. The right-hand rule pairs the boundary orientation with $n$. If there are several boundary components, the line integral includes all of them with their induced orientations.

A simply connected surface is connected and every closed loop on it can be continuously contracted to a point while remaining on the surface. A disc is an example. An annulus is not simply connected: a loop winding once around its hole cannot contract within it.

Stokes' theorem is valid for both types of surface, under the smoothness and orientation hypotheses above. Simple connectivity is not a requirement of the theorem. For a concrete annular example, take $F=(-y/2,x/2,0)$ and $a\le r\le R$ in the horizontal plane. Its curl is $(0,0,1)$, so the left side is $\pi (R^2-a^2)$. The right side is the counterclockwise outer-circle integral $\pi R^2$ minus the counterclockwise inner-circle integral $\pi a^2$, giving exactly the same answer. Omitting the inner boundary would be an incorrect application of the theorem.

For $n=1$, the azimuthal unit vector is $\hat{\theta }=(-\sin \theta ,\cos \theta ,0)$, soOn a counterclockwise circle $C_r$ at fixed height, $dr=r\hat{\theta }d\theta$. Henceindependent of the positive radius. Clockwise orientation reverses the sign.

There is no contradiction with the zero curl found in solution. The vector field is undefined on the axis, so it is not continuously differentiable on a disc spanning $C_r$; the hypotheses of Stokes theorem fail for that disc. It is smooth on a planar annulus excluding the axis, where Stokes theorem correctly gives zero: the two boundary line integrals cancel, $2\pi -2\pi =0$.

The same check works for general $n$. A counterclockwise circle has circulation $\Gamma (r)=2\pi r^{2-2n}$. Integrating the curl over an annulus $a\le r\le R$ gives, for $n\ne 1$,For $n=1$, both sides are zero. Thus all three preceding results are consistent with Stokes theorem; the necessary distinctions are the regularity of the vector field and inclusion of every oriented boundary component.

A scalar line integral is ${\int }_{C}fds=\int f(r(t))\Vert r^{\prime }(t)\Vert dt$: it weights arc length by a scalar field and is unchanged by reversing orientation. A vector-work line integral is ${\int }_{C}F\cdot dr=\int F(r(t))\cdot r^{\prime }(t)dt$: it measures the tangent component of a vector field and changes sign under reversal. Both are invariant under a regular orientation-preserving reparametrization.

For the prescribed counterclockwise curve, traverse the upper semicircle from $(1,0)$ to $(-1,0)$, then the lower parabola from $(-1,0)$ back to $(1,0)$.

On the upper arc, use $x=\cos t$, $y=\sin t$, $0\le t\le \pi$. Thenso its integral is $-3\pi /2$: the last two terms integrate to zero by symmetry or direct integration. On the parabolic arc, use $x=t$, $y=t^2-1$ for $-1\le t\le 1$. Here $x^2-y=1$ andThereforeAs an independent check, Green's theorem gives the integrand ${\partial }_x(x^2-y)-{\partial }_y(3y)=2x-3$. The region is symmetric in $x$, so its odd part vanishes. Its area is $\pi /2+4/3$, giving $-3(\pi /2+4/3)$, exactly the same value.

A conservative vector field can be written as $F=\nabla \Phi$ for a single-valued scalar potential $\Phi$. Its line integral between two points depends only on the endpoints, and its integral around every closed curve is zero. A necessary local test for a smooth planar field $F=(P,Q)$ is $P_y=Q_x$. On a simply connected domain such as the whole plane, this is also sufficient.

HereAn explicit scalar potential is even more decisive:Indeed ${\Phi }_x=2x{e}^{xy}+x^{2}y{e}^{xy}$ and ${\Phi }_y=x^3{e}^{xy}+2y$. Thus the field is conservative on ${\mathbb{R}}^2$.

An exact differential is a differential one-form equal to $d\Phi$ on its domain; an inexact differential is not the differential of a single-valued potential there. For example, $d(xy)=ydx+xdy$ is exact. For the supplied form $\eta =42ydx+xdy$, the cross derivatives areThey disagree, so $\eta$ is not even closed and hence cannot be exact. Equivalently, its counterclockwise line integral around the unit square is $1-42=-41\ne 0$. ThereforeThe connection is $F\cdot dr=d\Phi$: a vector field is conservative precisely when its work differential one-form is exact. Locally vanishing curl alone need not guarantee global exactness on a domain with holes.

For an oriented piecewise smooth surface $S$, a continuously differentiable vector field $F$ on a neighborhood of it, and its compatibly oriented boundary $C=\partial S$, Stokes theorem statesHere $n$ is the chosen unit normal vector, $dS$ is the unsigned surface area element, and $dr$ is the tangent displacement along $C$. The right-hand rule relates the boundary direction to $n$.

The supplied field has curl $\nabla \times F=(2z,2x,2y)$. Parametrize the plane by $r(x,y)=(x,y,1-x-y)$ over the triangle $D=\{x,y\ge 0:x+y\le 1\}$. Its upward oriented surface area element is $(1,1,1)dxdy$, givingFor the boundary line integral, take the positively oriented vertices $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$. Their projections run counterclockwise in the horizontal plane. With $0\le t\le 1$, the three edges areOn each edge, direct substitution gives $F(r(t))\cdot r^{\prime }(t)=t^2+2t-1$. HenceThe two independent computations agree, including the orientation sign.

Parametrize the curve by $\mathbf{r}(x)=(x,e^x,e^x)$ for $0\le x\le 1$. Then $d\mathbf{r}=(1,e^x,e^x)dx$ and $\mathbf{F}(\mathbf{r}(x))=(e^{2x},x,e^x)$. Their dot product gives the line integralUsing integration by parts for the second term,

Choose the outward normal vector on the side of the elliptic cylinder. Parametrize it bySince ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_z=(\frac{1}{2}\cos \theta ,\sin \theta ,0)$ points outward, the parameter rectangle has its bottom edge traversed with increasing $\theta$, its right cut edge with increasing $z$, its top edge with decreasing $\theta$, and its left cut edge with decreasing $z$. After identification, the two cut edges are the same physical curve with opposite directions. Their line integrals cancel for a single-valued vector field. Viewed from above, the bottom rim is counterclockwise and the top rim clockwise.

To apply Stokes theorem to a flux rather than a curl integral, first find a vector potential:This vector potential is smooth everywhere. On either horizontal rim, $d\mathbf{r}$ is horizontal whereas $\mathbf{A}$ is vertical, giving $\mathbf{A}\cdot d\mathbf{r}=0$. For a cut at ${\theta }_0=\pi /4$, $xy=1/4$, so the upward cut contribution is ${\int }_{-1}^1{z}^2/4dz=1/6$ and the downward one is $-1/6$. ConsequentlyA direct check uses the oriented surface element above: $\mathbf{F}\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)=\frac{1}{2}{z}^2\cos 2\theta$, whose integral over a full turn vanishes. Reversing the chosen orientation still gives zero.

The position vector field is a gradient field:Along any piecewise smooth curve, $\mathbf{r}\cdot d\mathbf{r}=d(|\mathbf{r}{|}^2/2)$. Therefore a closed line integral has identical endpoint values andFor a loop bounding an admissible oriented surface, Stokes theorem gives the same result because $\nabla \times \mathbf{r}=0$. The gradient field proof also applies directly to arbitrary piecewise smooth closed loops without choosing a spanning surface.

Let $S$ be a compact piecewise smooth oriented surface, $\mathbf{n}$ its chosen unit normal vector, and $\partial S$ its entire oriented boundary. Let $\mathbf{A}$ be a continuously differentiable vector field on a neighborhood of $S$. Stokes theorem statesHere $\nabla \times \mathbf{A}$ is the curl, $dS$ is the surface area element, and $d\mathbf{r}$ is the directed line integral element. The boundary follows the right-hand rule relative to $\mathbf{n}$; equivalently the surface lies to the left when walking along its boundary with one's head toward $\mathbf{n}$.

The topological requirement is orientability, together with the regularity and boundary assumptions above. Simple connectivity is unnecessary. An annulus or cylindrical side is allowed, provided every boundary component is included with its induced direction. For a concrete counterexample to a disk-only restriction, take the planar annulus $1\le x^2+y^2\le 4$ and $\mathbf{A}=(-y/2,x/2,0)$. The curl is ${\mathbf{e}}_z$, the surface integral is $3\pi$, and the outer counterclockwise and inner clockwise line integrals sum to $4\pi -\pi =3\pi$.

For $S$ in the $xy$ plane with upward normal vector and $\mathbf{A}=(P,Q,0)$, its curl has vertical component $Q_x-P_y$. Thus Stokes theorem becomes Green's theorem:The outer boundary is counterclockwise and any inner boundaries are clockwise.

For a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of an oriented, piecewise smooth surface $S$, Stokes theorem statesHere $\partial S$ is the boundary curve, $d\mathbf{r}$ is its directed tangent element, $\hat{\mathbf{n}}$ is the chosen unit normal vector, $dS$ is the surface area element, and $\nabla \times \mathbf{F}$ is the curl. The right-hand rule relates the boundary direction to $\hat{\mathbf{n}}$. Thus circulation around the edge equals the flux of curl through the spanning surface. Reversing both orientations changes both signs. One can replace a difficult line integral by an easier surface integral, or replace one spanning surface by another within the region of regularity.

For a bounded volume $V$ with piecewise smooth closed boundary and a continuously differentiable vector field on a neighborhood of its closure, the divergence theorem statesHere ${\hat{\mathbf{n}}}_{\mathrm{out}}$ points out of $V$, including into any holes, $dV$ is the volume element, and $\nabla \cdot \mathbf{F}$ is the divergence. It equates total outward flux to integrated source strength. It converts a closed surface integral into a volume integral, and explains why internal fluxes cancel in a subdivided region. Singularities cannot be included without checking the hypotheses.

For a conservative force, the work around every closed path is zero. Stokes theorem gives zero flux of curl through every sufficiently small patch. Dividing by its area and shrinking the patch shows that every component of the continuous curl vanishes:Here $U$ is potential energy; conversely, zero curl implies a conservative vector field on a simply connected domain. The divergence theorem then gives the compatible relation ${\iint }_{\partial V}\mathbf{F}\cdot \hat{\mathbf{n}}dS=-{\iiint }_V{\nabla }^{2}UdV$. There is no general zero-divergence condition on a conservative force: $\mathbf{F}=-\mathbf{r}=-\nabla (r^2/2)$ is a concrete counterexample, with $\nabla \cdot \mathbf{F}=-3$.

Set $P=10x^4-2x{y}^3$ and $Q=-3x^2{y}^2$. Their partial derivatives satisfy $P_y=-6x{y}^2=Q_x$ throughout ${\mathbb{R}}^2$. The vector field is smooth on the entire simply connected plane, so zero curl proves path independence.

Choose the broken straight path $(0,0)\to (2,0)\to (2,1)$. Along its horizontal segment $y=0$ and $dy=0$, givingAlong its vertical segment $x=2$ and $dx=0$, givingTherefore the line integral is

An exact differential is an expression $Pdx+Qdy$ that equals $df=f_{x}dx+f_{y}dy$ for a single-valued differentiable function $f$ on the domain. Its line integral is an endpoint difference by the fundamental theorem for line integrals.

Integrating $f_x=P$ with respect to $x$ gives $f=2x^5-x^2{y}^3+g(y)$. Matching $f_y=-3x^2{y}^2+g^{\prime }(y)$ to $Q$ requires $g^{\prime }(y)=0$. HenceThe arbitrary additive constant cancels. This constructs the vector-field potential globally and provides a second proof of path independence.

Assume a simple, piecewise smooth closed curve oriented counterclockwise around its enclosed region $D$. Apply Green's theorem with $P=-y$ and $Q=x$:The orientation assumption is essential. A clockwise unit circle, $x=\cos \theta$, $y=-\sin \theta$ for $0\le \theta \le 2\pi$, gives $-2\pi$ although its enclosed area is $\pi$. For a self-intersecting curve the same line integral measures twice the signed area, with multiplicity, rather than necessarily twice an ordinary enclosed area.

For positive semiaxes $a,b$, the ellipse has the counterclockwise parametrization $x=a\cos \theta$, $y=b\sin \theta$. ThenThereforeIf nonzero signed constants are allowed in place of positive semiaxes, the geometric area is $\pi |ab|$.

For a sufficiently smooth vector field on an oriented surface $S$ with compatible boundary orientation, Stokes theorem statesApply it to $\mathbf{F}=(P,Q,0)$ in the plane. The closed line integral is the area integral of $Q_x-P_y$. Thus $P_y=Q_x$ makes it vanish whenever a spanning surface lies within the smooth domain. In a simply connected domain this implies path independence: join one path from $A$ to $B$ to the reverse of another, and their difference is a zero closed line integral.

The domain qualification is essential. On the punctured plane, $P=-y/(x^2+y^2)$, $Q=x/(x^2+y^2)$ satisfy $P_y=Q_x$, but the line integral around the unit circle is $2\pi$. A disk spanning that circle passes through the singularity. Hence the printed implication requires an appropriate domain assumption; zero curl alone is not a global path independence theorem.