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Consider the globally smooth potential $\Phi (x,y,z)=x{y}^2\sin (xz)$. Direct differentiation givesThus the differential one-form is the exact differential $d\Phi$, proving exactness rather than only checking a necessary condition on its partial derivatives. Along any piecewise differentiable path, the chain rule gives $udx+vdy+wdz=d\Phi$. The fundamental theorem of calculus therefore makes the line integral depend only on its endpoints:

Orient the triangular boundary from $P=(1,0,0)$ to $Q=(0,1,0)$ to $R=(0,0,1)$ and back to $P$. This orientation corresponds to the positive normal vector $(Q-P)\times (R-P)=(1,1,1)$. The cyclic coordinate rotation $(x,y,z)\mapsto (z,x,y)$ sends the three directed edges to one another and transforms the vector field by the same rotation. Since rotations preserve dot products, the three edge contributions to the line integral are equal.

On the first edge use $x=1-t$, $y=t$, $z=0$, $0\le t\le 1$. ThenIts integral is $2/3$, so the required circulation is

For a sufficiently smooth vector field on an oriented piecewise smooth surface $T$, with induced boundary orientation on $C=\partial T$, the Stokes theorem statesTherefore the requested surface integral, with all normal components positive, is also $2$.

To verify it directly, parametrize the triangle by $\mathbf{r}(y,z)=(1-y-z,y,z)$ on $y\ge 0$, $z\ge 0$, $y+z\le 1$. Its two tangent vectors are ${\mathbf{r}}_y=(-1,1,0)$ and ${\mathbf{r}}_z=(-1,0,1)$, giving the vector area elementDifferentiation of the vector field gives $\nabla \times A=2(y+z,z+x,x+y)$. On the triangle $x+y+z=1$, so its dot product with $(1,1,1)$ is $4$. Consequentlyconfirming both the magnitude and the orientation sign.

For a smooth vector field near an oriented piecewise smooth surface $S$, Stokes theorem stateswhere the boundary orientation is induced by the surface unit normal through the right-hand rule. For a constant vector $\mathbf{k}$, the curl of $\varphi \mathbf{k}$ is $\nabla \varphi \times \mathbf{k}$. Apply Stokes theorem and use the scalar triple product:Since this holds for every constant $\mathbf{k}$, equality of all vector components proves the vector-valued Stokes identityFor the specified field, the two axis segments contribute zero because $x^2{y}^4$ vanishes on them. On the arc the direct line integral isIts two terms giveandThereforeOne can also check the orientation and sign by Stokes theorem. For the upward-oriented quarter disk, $(\nabla \times \mathbf{A}{)}_z=2x{y}^4-4x^2{y}^3$. Polar integration givesconsistent with the positively oriented boundary in the question.

For the anticlockwise arc use $x=\cos \theta$, $y=\sin \theta$, $0\le \theta \le \pi /2$. Since $dx=-\sin \theta d\theta$ and $dy=\cos \theta d\theta$, the line integral becomesThe two integrals multiplying $-\alpha$ each equal $1/3$, by substituting $\cos \theta$ and $\sin \theta$ respectively. The remaining integral equals one. Thus

On the first leg $y=0$ and $x$ decreases from one to zero; on the second $x=0$ and $y$ increases from zero to one. Adding their line integrals givesThe negative sign in the first term follows from the direction of traversal.

Parametrize the straight segment by $x=1-t$, $y=t$, $0\le t\le 1$. Then $dx=-dt$, $dy=dt$, and $x^2+xy=1-t$. The line integral is

Assume $P,Q$ are continuously differentiable on a neighbourhood of a bounded region with a positively oriented piecewise smooth simple boundary. To prove Green theorem, first take a region between the graphs $y=f(x)$ and $y=g(x)$, $a\le x\le b$, with $f\le g$. Counterclockwise orientation traverses the lower graph to the right and the upper graph to the left; vertical end pieces contribute nothing to $Pdx$. Thus the fundamental theorem of calculus givesLikewise, for a region between graphs $x=u(y)$ and $x=v(y)$, the right boundary is traversed upwards and the left one downwards. Horizontal end pieces contribute nothing to $Qdy$, soFor a polygonal region, triangulate its interior. Each triangle is of both graph types with piecewise linear bounds, so adding the formulae gives the result: the double integrals add and the integrals on internal edges cancel because the adjacent triangles traverse them in opposite directions. For a regular piecewise smooth simple boundary, choose inscribed simple polygonal approximations to its piecewise smooth parametrization. Their line integrals converge to the original line integral, since the parametrizations converge uniformly and their piecewise constant tangent vectors converge in the integral norm on each smooth arc. Their regions converge in area: the symmetric difference lies in a shrinking neighbourhood of the original boundary, which has planar measure zero. Boundedness of the continuous integrand then gives convergence of the area integrals. Passing to the limit proves the formula for the stated boundary class. Combining the two identities givesPositive orientation is essential for the sign.

Choose $P=-y/2$ and $Q=x/2$, for which $Q_x-P_y=1$. The boundary formula for planar area isFor the ellipse, take $a,b>0$ so the given increasing-angle parametrization is counterclockwise. Then $dx=-a\sin \theta d\theta$ and $dy=b\cos \theta d\theta$, soIf signed nonzero parameters are allowed, the unsigned area is $\pi |ab|$, with orientation corrected accordingly.

A conservative vector field is the gradient of a single-valued differentiable scalar potential. Its line integral is path independent and equals the potential difference. If $\psi \mathbf{A}=\nabla F$ with a twice continuously differentiable potential, equality of mixed partial derivatives givesThe product rule then gives ${\psi }_y{A}_1+\psi A_{1,y}={\psi }_x{A}_2+\psi A_{2,x}$. Rearranging proves the integrating-factor equation for a planar vector field:This argument does not require division by $\psi$, so it remains valid even at its zeros. A nonvanishing integrating factor is normally sought when one wants to recover useful potential curves for the original field.

The equality with part (i) follows from Green's theorem. For the coefficients $P=-1$ and $Q=x^2/(2y)$,They are smooth on a neighborhood of this region, since $y>0$. Thus for the positively oriented boundary,The term $-dx$ itself has zero line integral around the closed curve, although it contributes to the four individual edge integrals.

Take the usual positive, counterclockwise boundary orientation; the printed question does not specify one, and reversing it negates the answer. Write the successive corner points as $P=(1,1)$, $Q=(\sqrt{2},1/\sqrt{2})$, $R=(2,1)$, $S=(\sqrt{2},\sqrt{2})$. Traverse $P\to Q\to R\to S\to P$, keeping the region on the left.

On $PQ$, use $y=1/x$ with $1\le x\le \sqrt{2}$. Here $dy=-dx/x^2$, so the line integral isOn $QR$, use $y=x/2$ with $\sqrt{2}\le x\le 2$:On $RS$, use $y=2/x$ with $x$ decreasing from $2$ to $\sqrt{2}$:Finally, on $SP$, use $y=x$ with $x$ decreasing from $\sqrt{2}$ to $1$:Adding all four line integrals gives ${\oint }_C(x^2/(2y))dy-dx=1/2$ for the positive orientation, or $-1/2$ for the reverse orientation.

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.

Write $r^2=x^2+y^2$. The first two components have no $z$ dependence, and the third component is zero. The only possibly nonzero component of the curl isThus $\nabla \times \mathbf{F}=0$ wherever the field is defined.

The disk spanning ${\gamma }_1$ lies away from the excluded axis: the distance from its center to that axis is $2\sqrt{2}>1$. The field is continuously differentiable on a neighborhood of that disk. The Stokes theorem therefore givesThis value is independent of the choice of orientation.

For ${\gamma }_2$, take the counterclockwise orientation viewed from positive $z$. Put $\mathbf{x}(t)=(\cos t,\sin t,0)$, $0\le t\le 2\pi$. Then $\mathbf{F}(\mathbf{x}(t))=(-\sin t,\cos t,0)={\mathbf{x}}^{\prime }(t)$, so the line integral isClockwise orientation gives $-2\pi$; no orientation is printed for this curve. There is no contradiction with the Stokes theorem: its usual spanning disk intersects the axis where the field is undefined, so the differentiability hypothesis fails. This is a curl-free vector field with nonzero circulation around the excluded axis; local angular potentials cannot be combined into a single-valued global potential.

For an oriented smooth surface $S$ with piecewise smooth, consistently oriented boundary, Stokes theorem givesfor a continuously differentiable vector field defined on a neighborhood of $S$. Choose the upward normal vector here. The surface is an annular band of an elliptic paraboloid, parametrized byUsing ${\partial }_{r}r\times {\partial }_{\theta }r$ as the oriented vector area element givesThe sketch below shows the open band, not a capped solid. Its upper circle has radius one and its lower circle radius one third.

Differentiating the given vector field yieldsThe surface integral is thereforeFor the boundary line integral, the upward normal vector induces counterclockwise traversal of the outer circle and clockwise traversal of the inner circle, as viewed from above. On a circle of radius $R$, parametrized counterclockwise, $z$ is constant andSince each fourth power integrates to $3\pi /4$, the two-circle line integral isconfirming Stokes theorem and the Stokes flux through an annular paraboloid. Reversing the chosen orientation changes both integrals to $-40\pi /27$.

Direct use of partial derivatives givesWith the upward oriented surface element found in part (b), the curl flux integrand over the annulus $A$ becomesEvery term in the numerator is odd in $x$ or $y$, so its integral over the symmetric annulus is zero. Thus the surface integral isEach boundary circle has constant $z$, hence $dz=0$. Its line integral is therefore $\int (-ydx+xdy)$, with no contribution from the third field component. An anticlockwise radius-$r$ circle gives $2\pi r^2$. The outward boundary orientation is anticlockwise at $r=2$ and clockwise at $r=1$, soThe two computed values agree, verifying Stokes theorem with both boundary components. A downward choice would give $-6\pi$ on both sides.

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.

Use $\gamma (t)=(\cos t,\sin t,0)$ for $0\le t\le 2\pi$. On this path $\mathbf{F}=(-\sin t,\cos t,0)={\gamma }^{\prime }(t)$, so its line integral isThe first two components of the curl vanish because the field is independent of $z$ and has zero third component. Writing $q=x^2+y^2$, its third curl component isThus this azimuthal inverse-radius vector field is a curl-free vector field on $G={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, but its line integral is not path independent. This domain is not simply connected: the circle links the removed axis. A spanning disk through the axis is inadmissible for Stokes theorem, since the vector field is undefined there. Zero curl alone does not force global path independence.

The curl is constant:Orient the spanning planar disk by the unit normal $\mathbf{n}=(a,b,c)$ and orient its boundary by the right-hand rule, equivalently counterclockwise when viewed from the side towards which $\mathbf{n}$ points. Stokes theorem converts the line integral to a constant flux integral through a disk of area $\pi R^2$. With this orientationReversing the orientation changes the sign. The centre of the circle does not enter because the curl is constant.

Symmetry of the derivative matrix means ${\partial }_i{F}_j={\partial }_j{F}_i$. Contracting this symmetric pair with the antisymmetric Levi-Civita symbol givesEvery circle bounds its planar disk, and the vector field is smooth on all of ${\mathbb{R}}^3$, including that disk. Stokes theorem therefore gives for either orientationThe global domain is essential to this argument; the same local symmetry condition on a domain with a hole need not make every closed-curve line integral vanish.

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.