Codex @codex  0

A connected component of a boundary. A surface with a hole can have several boundary components, each with its own induced boundary orientation.

Every conformal automorphism of the upper half-plane is a Möbius transformation with real coefficients and positive determinant, and conversely every such map is a conformal bijection of the upper half-plane to itself:The real-coefficient and sign conditions matter. The complex Möbius transformation $z\mapsto iz$ sends $i$ to $-1$ on the boundary, while the real Möbius transformation $z\mapsto 1/z$ has negative determinant and sends $i$ to $-i$ in the lower half-plane. ^[ref]

For two concrete upper half-plane automorphisms, takeBoth matrices have determinant one. Here $T(i)=1+i$, $S(i)=i$, and $S(1+i)=(-1+i)/2$. On the extended real line, $T$ fixes infinity and $S$ exchanges zero and infinity. On the unit disc, the same maps become $C\circ T\circ C^{-1}$ and $C\circ S\circ C^{-1}$; the second is simply $w\mapsto -w$.

Multiplying all four coefficients by a nonzero real scalar leaves $f$ unchanged. Dividing them by $\sqrt{\Delta }$ normalizes the determinant to one; the two remaining representatives $A$ and $-A$ give the same map. Thus the group under function composition is the projective special linear groupwhere $S{L}_2(\mathbb{R})$ is the real special linear group of degree two. Composition of the Möbius transformations corresponds to multiplication of their matrices. ^[ref]

On the boundary,So $f$ preserves the cyclic boundary orientation of the extended real line; when $c\ne 0$, its pole at $-d/c$ prevents treating it as a single increasing function across that point on the ordinary real line. Three distinct boundary images determine a Möbius transformation uniquely, but an upper half-plane automorphism can realize only triples with matching cyclic boundary orientation.

A brief proof outline starts with the identitiesThey show that $f$ and its inverse function preserve the upper half-plane. For completeness of the classification, conjugate an arbitrary conformal bijection by the Cayley transform. Normalize its image of zero with an automorphism of the unit disk, then apply the Schwarz lemma to the resulting map and its inverse function; both inequalities force a rotation. Conjugating back gives the real Möbius transformation above.

Finally, $|f^{\prime }(z)|=\Delta /|cz+d{|}^2$ and the displayed imaginary part identity show that $|df(z)|/\mathrm{Im}f(z)=|dz|/\mathrm{Im}z$. Every upper half-plane automorphism is therefore also an isometry of the Poincaré half-plane model. The same formulas describe both holomorphic coordinate symmetries and motions of the hyperbolic plane.

Let $D_n$ act on Euclidean space by rotations through $2\pi /n$ about a vertical axis and half-turns about $n$ equally spaced horizontal axes. A round solid torus about the vertical axis, drilled through radially along the $2n$ places where the horizontal axes cross its tube, has a smooth boundary of genus $2n+1$ after equivariant rounding of its rims. All nonidentity rotation axes lie in the complement of this boundary, so the induced group action is a free action of a group. Each bore replaces two closed discs by an annulus, lowering Euler characteristic by two and increasing genus by one. The boundary orientation is preserved; its induced Riemann surface has an unramified quotient space of genus two by the Riemann-Hurwitz formula.

Green’s theorem converts circulation around a positively oriented planar boundary into the area integral of scalar curl.

Let $D$ be a bounded region in the Euclidean plane whose boundary is a finite union of disjoint, piecewise smooth simple closed curves. Suppose the vector field $F=(P,Q)$ is continuously differentiable on an open set containing the whole closed region. With positive boundary orientation,The partial derivatives $Q_x-P_y$ give the scalar curl: the third component of the curl of $(P,Q,0)$. For a parametric curve $\gamma (t)=(x(t),y(t))$, the boundary line integral has integrand $P(\gamma (t))x^{\prime }(t)+Q(\gamma (t))y^{\prime }(t)$.^[ref]

Positive boundary orientation keeps $D$ on the left as one walks along each boundary component: the outer curve is counterclockwise, and curves around holes are clockwise. Every component belongs in the line integral. The vector field must be regular throughout $D$, but need not extend into a hole.^[ref]

The useful tradeoff is between boundary and interior: integrate tangential components along a curve, or integrate a partial derivative over a region, whichever is simpler. The boundary formula for planar area uses this freedom in the opposite direction, recovering area from a line integral. As a planar case of Stokes theorem, Green theorem also explains why a local curl can determine a complete boundary measurement while leaving room for the global obstruction in Green theorem with holes.

A short proof starts with a rectangle $D=[a,b]\times [c,d]$. Add its four positively oriented edge line integrals:The fundamental theorem of calculus turns the differences into integrals of $-P_y$ and $Q_x$; Fubini's theorem then gives the required double integral. On subdividing a region, adjacent edge line integrals cancel because their directions oppose. For curved boundaries, the proof uses pieces bounded by graphs and double integrals; cancellation works the same way. This is the geometric reason only the region's complete boundary survives.^[ref]

Take the vector field $F=(-2y,3x)$. Its scalar curl is the constant $3-(-2)=5$, so Green theorem predicts that its positively oriented boundary line integral is five times the enclosed area. This one field makes both a round and a rectangular calculation transparent.

For the circle $x^2+y^2=4$, choose the counterclockwise parametric equations $\gamma (t)=(2\cos t,2\sin t)$, $0\le t\le 2\pi$. Direct substitution in the line integral givesThe interior double integral is shorter: the disk has area $4\pi$, hence ${\iint }_{D}5dA=20\pi$. Reversing the boundary orientation gives $-20\pi$.

For the rectangle $D=[0,3]\times [0,2]$, traverse its bottom, right, top, and left edges in that order. The four line integrals are respectivelyTheir sum is $30$, agreeing with ${\iint }_{D}5dA=5\cdot 3\cdot 2$. In both examples the path geometry changes the separate contributions, while constant curl makes only area matter to the final answer.

On the punctured plane ${\mathbb{R}}^2\setminus \{(0,0)\}$, consider the vector fieldWith $r^2=x^2+y^2$, its partial derivatives satisfyThus the field has zero scalar curl everywhere it is defined. Nevertheless, on any counterclockwise circle ${\gamma }_R(t)=(R\cos t,R\sin t)$ with $R>0$,The radius cancels exactly.

Applying Green theorem to the enclosed disk would be invalid: $G$ is undefined at the origin, inside that disk. Zero curl on the path alone is insufficient.^[ref]

Now take the annulus $A=\{(x,y):1\le r\le 2\}$. Its positive boundary orientation is counterclockwise on the outer circle and clockwise on the inner circle. The complete boundary line integral is thereforeHere Green theorem applies: an open set containing the closed annulus can avoid the origin. The inner contribution remains $-2\pi$ even for an arbitrarily small positive inner radius. Decreasing the hole's area does not decrease this line integral.

The cancellation has a useful proof picture. Cut the annulus along two radial segments into two half-annuli. Each piece has a single simple closed curve as its boundary; along each cut, the two induced boundary orientations oppose, so those line integrals cancel when the two applications of Green theorem are added. Only the outer and inner circles remain.^[ref]

Locally, a choice of angle $\theta$ in polar coordinates has gradient $\nabla \theta =G$. Globally, following a circle once increases that angle by $2\pi$, so it cannot be a single-valued potential of a conservative vector field on the punctured plane. The nonzero closed-path line integral also rules out such a potential by the fundamental theorem for line integrals.

On a simply connected open region, a continuously differentiable planar vector field with zero scalar curl does have a global potential. The hole is what separates the local derivative condition from the global conclusion.^[ref]

The dihedral group here has order $2n$. Substituting $h=2$ into part (d) givesIn particular, $n=3$ requires genus seven. A geometric construction is to start with a round solid torus about the vertical axis, then drill $2n$ thin radial tunnels through its tube in the horizontal midplane, equally spaced at azimuths $k\pi /n$, $0\le k<2n$. The tunnels run from the central opening to the exterior; their directions are horizontal and radial. Round all their rims in the same way, preserving the dihedral symmetry, and let $X$ be the boundary of the resulting solid.

Here is why the construction has exactly the required properties. Every nonidentity power of $\beta$ is a rotation in three dimensions fixing the vertical axis, which lies in the central opening and is disjoint from $X$. Each element $\alpha {\beta }^j$ is a half-turn about a horizontal line at azimuth $j\pi /n$, modulo $\pi$. The $n$ such lines meet the original solid torus in $2n$ short radial segments. Each segment lies strictly inside one drilled tunnel, so these axes are also disjoint from $X$. A nonidentity rotation in three dimensions fixes only its axis. Thus no nonidentity element fixes a point of $X$, proving a free action of a group.

Rotation by $2\pi /n$ permutes the tunnels, while $\alpha :(x,y,z)\mapsto (x,-y,-z)$ preserves their arrangement and the original solid torus. Hence $X$ is preserved by the two generators and therefore by $G$. The tunnels can be chosen sufficiently thin to be pairwise disjoint, and the rounding can be made group-invariant in small disjoint neighbourhoods of their rims. It neither changes the genus nor introduces an intersection with any axis.

Each tunnel removes two discs from the existing boundary and inserts the cylindrical wall of the bore. Its contribution to Euler characteristic is $-2$: the two removed discs contribute $-2$, and the inserted annulus and its boundary circles have zero Euler characteristic. The boundary remains connected and orientable, so each tunnel increases genus by one. Starting from a torus and drilling $2n$ tunnels therefore gives $g=1+2n$. The construction also supplies a Riemann surface: use the complex structure associated with the induced oriented Riemannian metric. The ambient generators are isometries preserving the solid and ambient orientation of a vector space, so they preserve the induced boundary orientation and act by biholomorphisms. Part (d) then confirms that the quotient space has genus two. This is a free dihedral action on a drilled torus.

For $0<b-a<2\pi$, let $D=\mathbf{r}([a,b]\times [c,d])$ and use the orientation of a surface determined by $N={\mathbf{r}}_u\times {\mathbf{r}}_v/|{\mathbf{r}}_u\times {\mathbf{r}}_v|$. Its positive boundary orientation follows the bottom side in increasing $u$, the right side in increasing $v$, the top side in decreasing $u$, and the left side in decreasing $v$. The Gauss-Bonnet theorem for this disc with four corners readswhere the ${\epsilon }_j$ are the signed exterior angles.

The bottom and top sides have cancelling geodesic curvature integrals: a screw motion takes one to the other with opposite boundary orientation. Explicitly, along increasing $u$ at fixed $v$, the speed is one and the signed geodesic curvature is $-\sin u/\sqrt{1+{\sin }^{2}u}$ on both sides. The right and left sides contribute, respectively, the integral along ${\gamma }_b$ and minus the integral along ${\gamma }_a$.

Although the coordinate sides are generally not perpendicular, the corner contributions still sum to $2\pi$. Indeed, let $\theta (u)$ be the angle between ${\mathbf{r}}_u$ and ${\mathbf{r}}_v$, soStarting at the bottom-left corner, the exterior angles are $\pi -\theta (a)$, $\theta (b)$, $\pi -\theta (b)$ and $\theta (a)$. Their sum is $2\pi$. Cancelling this from the Gauss-Bonnet theorem leaveswhich is the identity in part (e). For a rectangle covering more than one turn, apply the same argument on the parameter rectangle with its induced Riemannian metric, or partition it into narrower rectangles and cancel the artificial internal boundary contributions. This gives the version counting multiplicity, as required by the correction in part (e).

For an oriented piecewise smooth surface $S$ and a continuously differentiable vector field, Stokes theorem stateswhere the boundary orientation is induced by $n$ through the right-hand rule. Choose the upward orientation of the elliptic paraboloid. Its band has cylindrical radius $1/2\le r\le 1$, and a parametrization isIts oriented surface element and scalar area element areThe requested sketch shows the open band, not a capped solid.

For the specified vector field, direct differentiation gives $\nabla \times B=(0,0,3x^2+3y^2)$. HenceTo check this by Stokes theorem, parametrize a counterclockwise circle of radius $\rho$ at height ${\rho }^2$ by $r_{\rho }(\phi )=(\rho \cos \phi ,\rho \sin \phi ,{\rho }^2)$. Along it $dz=0$, andUsing ${\int }_0^{2\pi }{\sin }^4\phi d\phi ={\int }_0^{2\pi }{\cos }^4\phi d\phi =3\pi /4$ gives $3\pi {\rho }^4/2$. The induced outer circle is counterclockwise as viewed from above, and the inner circle is clockwise. ThereforeThis verifies the direct surface integral and the Stokes flux through an annular paraboloid. Reversing the chosen surface orientation reverses the final sign.

Throughout the hyperbolic discussion take curvature $-1$ and $g\ge 2$. A Y-piece is a pair of pants with a hyperbolic metric and three geodesic boundary components. Three labelled positive boundary lengths ${\ell }_1,{\ell }_2,{\ell }_3$ determine it up to isometry. To see existence and uniqueness, cut along the three shortest perpendicular seams. The two resulting right-angled hyperbolic hexagons have alternating side lengths ${\ell }_i/2$, which determine each hexagon. Conversely, any three positive alternating lengths produce such a hexagon, and doubling along the other sides produces the desired Y-piece. In particular the seam opposite ${\ell }_i$ satisfies

An X-piece is a four-holed hyperbolic sphere with geodesic boundaries, obtained by joining two Y-pieces along one pair of equal-length boundaries. For a chosen separating curve, it is determined bywhere $s$ is the common gluing length and $\tau$ is the relative arclength twist, measured between chosen seam endpoints. The identification reverses boundary orientation so that the surface is oriented. For an unmarked gluing, $\tau$ is periodic modulo $s$; a marking retains $\tau \in \mathbb{R}$, because a full Dehn twist changes the marking. These six parameters describe the X-piece with its chosen separating curve; different curve choices or boundary symmetries can describe the same unmarked surface.

Choose a pants decomposition of a closed oriented genus-$g$ surface. Each Y-piece has Euler characteristic $-1$, and gluing boundary circles does not change the total Euler characteristic. Thus there are $2g-2$ pieces. Their $3(2g-2)$ boundary circles are paired, giving $3g-3$ gluing curves. Choose their disjoint geodesic representatives, then prescribe each cuff length and its relative twist. The Fenchel–Nielsen coordinates areThey determine the marked hyperbolic surface and hence its unmarked isometry class, with $6g-6$ real parameters. One must also fix the combinatorial gluing pattern of the chosen pants decomposition. Full twists and other changes of marking identify some parameter sets in unmarked moduli space of Riemann surfaces. The genus restriction matters: a sphere and a torus do not admit a closed curvature-$-1$ metric or such a decomposition into geodesic Y-pieces.

For a fixed oriented topological surface $S_g$, Teichmüller space ${\mathcal{T}}_g$ consists of marked Riemann surfaces $(X,f:S_g\to X)$. Two markings represent the same point when a conformal isomorphism between the surfaces carries one marking to the other up to homotopy. By the uniformization theorem, for $g\ge 2$ these are equivalently marked curvature-$-1$ metrics. The displayed Fenchel–Nielsen coordinates identify ${\mathcal{T}}_g$ with a cell of real dimension $6g-6$; the mapping class group quotient gives the unmarked moduli space of Riemann surfaces.

The spectral version of the Wolpert generic spectral rigidity theorem says that outside a closed proper real-analytic exceptional subset of ${\mathcal{T}}_g$, the Laplace spectrum determines the hyperbolic surface up to isometry. The exceptional subset has positive codimension and is invariant under changes of marking. Orientation reversal must be allowed, since the Laplacian cannot detect orientation. Equivalently, a generic surface is determined by its unmarked length spectrum, counted with multiplicities. This is a generic uniqueness statement, not a claim that every isospectral pair is isometric; the construction in Question 4 lies in the exceptional locus.

Three major ingredients and their roles are as follows.

For completeness, a related finiteness statement sometimes grouped with this theorem is that a fixed closed hyperbolic surface has only finitely many isospectral isometry classes. Its short proof uses three particularly concrete tools: a common discrete length spectrum; Mumford's compactness theorem, applied to the common positive systole; and finitely many determining marked lengths. Compactness would give an accumulation point if an isospectral family were infinite. Nearby markings make each determining length converge, while discreteness of the fixed spectrum makes it eventually constant. The determining lengths then force eventual equality of the marked surfaces, a contradiction. This finiteness of isospectral hyperbolic surfaces statement does not replace the stronger generic uniqueness assertion above.

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.

Stokes theorem equates the circulation around the induced boundary orientation of an oriented surface to the normal flux of curl:The usual version requires a piecewise smooth oriented surface and a vector field continuously differentiable on a neighbourhood of it. We use the upward orientation wherever the surface is a graph; each outer circle is counterclockwise viewed from above, and an inner circle is clockwise.

Write $\rho =(x^2+y^2{)}^{1/2}$. The surface of revolution has $z=\sqrt{{\rho }^2+1-\lambda }$ with $0\le z\le 1$. For $0<\lambda <1$, it is a disk-shaped cap of a two-sheeted hyperboloid, running from $\rho =0,z=\sqrt{1-\lambda }$ to $\rho =\sqrt{\lambda },z=1$. For $\lambda =1$, it is a right circular cone, $z=\rho$, including its apex. For $\lambda >1$, it is an annular strip of a one-sheet hyperboloid, between the circles $\rho =\sqrt{\lambda -1},z=0$ and $\rho =\sqrt{\lambda },z=1$.

For $z>0$, differentiation of the graph gives $z_x=x/z$ and $z_y=y/z$, so the upward vector surface element of a graph and its scalar area areFor $\lambda >1$ this graph expression is singular at the lower circle because the surface has a vertical tangent there, rather than a singular geometric surface. A regular parametrization isIt proves that the graph surface integral is interpreted by its integrable endpoint limit. The same parametrization works away from the pole or apex in the other regimes.

Direct differentiation of the first vector field givesIts surface integral is twice the projected area, namely $2\pi \lambda$ for $0<\lambda \le 1$ and $2\pi [\lambda -(\lambda -1)]=2\pi$ for $\lambda >1$. On an oriented circle of radius $R$, $\mathbf{F}\cdot d\mathbf{x}=R^{2}d\varphi$. The outer circulation is $2\pi \lambda$; an inner circle, when present, contributes $-2\pi (\lambda -1)$. ThusAt $\lambda =1$, excise a small apex circle of radius $\epsilon$, apply Stokes theorem to the smooth truncated cone, and let $\epsilon \to 0$. The removed flux and its inner circulation are both $2\pi {\epsilon }^2$, so the limiting equality holds. This avoids silently applying a smooth-surface theorem at the cone's nonsmooth apex.

The second vector field is the azimuthal inverse-radius vector field, $\mathbf{G}={\mathbf{e}}_{\varphi }/\rho$. Away from the axis,The other two curl components vanish because the field is independent of $z$ and has zero $z$ component. On any positively oriented circle around the axis, $\mathbf{G}\cdot d\mathbf{x}=d\varphi$, so its circulation is $2\pi$, independent of radius. ThereforeFor $\lambda >1$ the surface avoids the axis, so Stokes theorem applies directly and returns zero curl flux. For $0<\lambda <1$ the disk cap meets the axis, and for $\lambda =1$ the apex lies on it: $\mathbf{G}$ is undefined there, so the neighbourhood hypothesis of Stokes theorem fails. Removing a small circle gives cancelling inner and outer circulations; the inner circulation remains $-2\pi$ as its radius tends to zero. Hence it cannot be discarded as it could for $\mathbf{F}$. A nonzero circulation here is compatible with zero curl away from the excluded axis.

On the part with $1\le z\le 2$, writeThis is the upper portion of a hyperboloid of one sheet. Choosing the outward orientation, whose normal points radially away from the $z$-axis, the vector area element isIts radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below.

Forthe curl isOn the surface, $z^2={\rho }^2-1$, and thereforeDirect integration gives

To verify this with the Stokes theorem, note that on a circle of fixed $\rho ,z$,The induced boundary orientation runs in the negative $\varphi$ direction on the top circle and the positive $\varphi$ direction on the bottom circle. Hence

For $S^{\prime }$, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive $\varphi$ direction at $z=-1$, where ${\rho }^2=2$, and negative $\varphi$ direction at $z=\sqrt{2}$, where ${\rho }^2=3$. Thus

Use the spherical coordinates parametrizationThen ${\mathbf{R}}_{\varphi }\times {\mathbf{R}}_{\theta }=\sin \varphi \mathbf{R}$, so the outward unit normal vector is $\mathbf{n}=\mathbf{R}=\sin \varphi {\mathbf{e}}_r+\cos \varphi {\mathbf{e}}_z$ and the surface area element is $\sin \varphi d\varphi d\theta$. Since $\sin \varphi >0$ throughout this band, the vector field is smooth near it. The surface integral of the curl isFor the induced boundary orientation, the lower circle $\varphi =\beta$ is traversed with increasing $\theta$, and the upper circle $\varphi =\alpha$ with decreasing $\theta$. One way to check the signs is that the outward conormal is ${\mathbf{e}}_{\varphi }$ at the lower circle and $-{\mathbf{e}}_{\varphi }$ at the upper circle; the positive unit tangent vector is $\mathbf{n}\times \text{conormal}$.

Along a circle of fixed $\varphi$, $d\mathbf{r}=\sin \varphi {\mathbf{e}}_{\theta }d\theta$. Therefore the circulation iswhich equals the boxed surface integral and verifies Stokes theorem with the required orientation.

Let $S$ be a piecewise smooth oriented surface with piecewise smooth boundary $\partial S$, and let $\mathbf{F}$ be a continuously differentiable vector field on a neighbourhood of $S$. The Stokes theorem statesThe boundary orientation is induced by the chosen unit normal vector $\mathbf{n}$ through the right-hand rule: with the right thumb along $\mathbf{n}$, the curled fingers indicate positive boundary traversal. Equivalently, when walking along the boundary with one's head pointing along $\mathbf{n}$, the surface lies to the left. An upward oriented disk therefore has counterclockwise boundary traversal when viewed from above. Reversing the orientation reverses both the line integral and the surface integral.

On the unit circle, the positively oriented unit tangent vector is $\mathbf{u}=(-y,x,0)$, and $|\mathbf{u}|=1$. Hence $d\mathbf{r}=\mathbf{u}ds$ andSince the scalar field $\psi$ has no $z$ dependence, the curl product rule for a scalar field givesOn the unit upper hemisphere $S_1$, the outward unit normal vector is $(x,y,z)$, so its vertical component is $z=\cos \theta$. Applying the Stokes theorem to the vector field $\psi \mathbf{u}$ therefore provesThe smoothness needed here is that $\psi$ extend as a continuously differentiable scalar field to a neighbourhood of the projected unit disk. The left side is a scalar line integral with respect to arc length; the positive boundary orientation is used when identifying it with the vector line integral.

Choose the outward orientation on the upper half of the ellipsoid, giving counterclockwise boundary orientation on $C$ as viewed from $z>0$. Parametrize the boundary circle byThen $\mathbf{u}(\mathbf{c}(t))=(-\sin t,\cos t,0)={\mathbf{c}}^{\prime }(t)$, so the line integral isFor a direct surface integral, use the parametrized surfaceIts outward vector surface element is supplied by the cross productUsing the curl from a,The two integrals agree for every $\alpha >0$, verifying the Stokes theorem. Only the vertical component of the vector surface element enters, which explains the absence of $\alpha$ in the answer.

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 compact, piecewise smooth oriented surface with piecewise smooth boundary, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of $\Sigma$. The Stokes theorem statesThe boundary orientation must agree with the chosen unit normal vector by the right-hand rule. For an upward orientation on a horizontal disk, its boundary is traversed counterclockwise when viewed from above. Reversing the orientation reverses both the line integral and the surface integral.

Use the outward orientation on the unit hemisphere and counterclockwise boundary orientation on the unit circle. With $\mathbf{r}(t)=(\cos t,\sin t,0)$ and $f=x=\cos t$, the vector line integral isFor the surface integral, $\nabla f=\mathbf{i}$ and the outward unit normal vector is $\mathbf{n}=(x,y,z)$, since the radius is one. The cross product isThe integral of $y$ is zero by reflection symmetry. Using the same spherical polar coordinates as in ii, with $a=1$, the other component isHenceverifying the vector-valued Stokes identity directly, component by component. The wedge symbol in the paper denotes the cross product here.