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Apply Green theorem with $P=-y/2$ and $Q=x/2$ to a positively oriented regular planar region. Since $Q_x-P_y=1$, its area is the displayed line integral. Clockwise orientation changes the sign; holes have the opposite boundary orientation to the outer curve.

A differential form is a smooth field of alternating multilinear maps that can be integrated over an appropriately oriented manifold of matching dimension. A one-form measures a directed displacement; a two-form measures an oriented area element. The point is to express what is being accumulated along a curve or over a surface without tying the answer to one set of coordinates. This makes line integrals, surface integrals, and the integral theorems of vector calculus parts of one construction.

For a concrete one-form, take $\alpha =xdy$ in the plane. Travel from $(0,0)$ to $(1,1)$ along the bottom and then the right edge of the unit square. The horizontal leg contributes zero because $dy=0$; the vertical leg has $x=1$ and contributes ${\int }_0^{1}dy=1$. Taking the left and then the top edge instead gives zero: the vertical leg has $x=0$, and the horizontal leg again has $dy=0$. The two paths have the same endpoints but different integrals.

Now replace $\alpha$ with the exact differential $\beta =d(xy)=ydx+xdy$. Both paths give $1$, because integrating an exact differential along a curve gives the endpoint difference of its potential. Here that difference is $xy{|}_{(1,1)}-xy{|}_{(0,0)}=1$. This is the fundamental theorem of calculus applied to a function along a parametric curve, and is the basic reason conservative vector fields have path-independent work.

The exterior derivative records the local obstruction in the first example: $d(xdy)=dx\wedge dy$. The Generalized Stokes theorem turns the difference between the two path integrals into the integral of this two-form over the square, whose oriented area is $1$. Interior edges cancel when adjoining pieces are added; only the outer boundary survives. That cancellation connects a local calculation with a global measurement.

More precisely, a degree-$k$ differential form assigns to each point $p$ a multilinear scalar-valued function of $k$ tangent vectors at $p$. Interchanging two arguments changes its sign. A one-form is therefore a covector field, while a two-form vanishes on a pair of parallel vectors. Degree zero simply means a smooth function. The space ${\Omega }^k(M)$ is a vector space; the wedge product of differential forms combines degrees, and the exterior derivative raises degree by one.

The usefulness goes beyond bookkeeping. Pullbacks of a differential form supply the right transformation law automatically under a coordinate change. The Generalized Stokes theorem unifies the fundamental theorem of calculus, Green theorem, Stokes theorem, and divergence theorem. Meanwhile, the distinction between closed differential forms and exact differential forms detects holes through de Rham cohomology. Reyer Sjamaar develops these connections in sections 5.4 and 9.3 of his university notes. ^[ref].

If $X$ is an oriented $n$-manifold with boundary, $F:\partial X\hookrightarrow X$ is the inclusion, and $\omega$ is a compactly supported $(n-1)$-form, thenwhere the boundary has the outward-normal-first boundary orientation. A subordinate partition of unity reduces the theorem to the fundamental theorem of calculus in oriented coordinate half-spaces; local finiteness and compact support ensure that only finitely many terms contribute.

For the counterclockwise unit-square boundary and $\omega =xdy$, the right-hand side is $1$, while the left-hand side is ${\int }_{[0,1{]}^2}dx\wedge dy=1$. The equation says that the boundary integral measures the total local circulation in the region. Reversing the orientation of a smooth manifold reverses both sides together.

In dimension one, the boundary of an interval consists of its endpoint with a plus sign and its starting point with a minus sign; the identity becomes ${\int }_a^{b}df=f(b)-f(a)$. In the plane it gives Green theorem. Applied to appropriate differential forms in three dimensions it gives both Stokes theorem and the divergence theorem. The boundary and the differential being integrated must belong to the stated domain: removing a singular point can make a spanning disc unavailable.

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 a star-shaped planar domain containing zero, let $\eta$ be a smooth closed differential one-form and set $f(x)={\int }_{[0,x]}\eta$. Green theorem on the triangle with vertices $0,x,x+h$ gives $f(x+h)-f(x)={\int }_{[x,x+h]}\eta$. Dividing by a displacement and taking its limit proves $df=\eta$. Thus every such form is an exact differential form, a degree-one version of the Poincaré lemma which uses only the planar integral theorem.

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]

Here a periodic solution must mean a nonconstant periodic orbit: constant equilibria are not excluded by the stated divergence hypothesis. Suppose such an orbit exists. Since the vector field is continuously differentiable, uniqueness for the autonomous differential equation makes its trajectory a simple closed curve $C$, bounding a region $D$ by the Jordan curve theorem. The Green theorem givesAlong the orbit, $dx=fdt$ and $dy=gdt$, so the line integral is identically zero. Reversing orientation if necessary does not change this conclusion. The area integral is strictly positive because its continuous integrand is positive everywhere and $D$ has positive area. This contradiction proves the Bendixson-Dulac criterion in this case.

For the second-order equation set $y=x^{\prime }$, giving $x^{\prime }=y$, $y^{\prime }=-y-\epsilon y^3-\sin x$. Choose the Dulac function $\rho =-1$. ThenThus there is no nonconstant periodic solution for $\epsilon >0$. The equilibria $x(t)=k\pi$ do remain solutions. As a direct check, the mechanical energy $H=y^2/2+1-\cos x$ obeys $H^{\prime }=-y^2-\epsilon y^4$, whose integral over a period could vanish only if $y\equiv 0$.

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 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.

In a manifold chart, write a smooth differential form as $\eta ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$, with $I$ increasing. Define its exterior derivative byTo check that this definition is global, change coordinates to $y$. The chain rule expresses $d{x}^i$ as ${\sum }_j({\partial }_j{x}^i)d{y}^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $d{y}^k\wedge d{y}^j$ is antisymmetric. The coordinate definition satisfies the graded Leibniz rule, so differentiating the transformed expression for $\eta$ leaves exactly ${\sum }_{I}d{a}_I\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. This agrees with the original formula and proves independence of the manifold chart. The same cancellation of mixed derivatives gives $d^2=0$.

The de Rham cohomology is thereforethe vector space of closed differential forms modulo exact differential forms.

Here is a Green-theorem proof of exactness on a disc. Identify the disc with a round disc centred at zero, and let $\eta =Pdx+Qdy$ be a closed differential form. Set $f(x)={\int }_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, Green theorem on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since ${\partial }_{x}Q-{\partial }_{y}P=0$. HenceTaking $h=\epsilon v$ and dividing by $\epsilon$ yields $d{f}_x(v)={\eta }_x(v)$. The radial integral defining $f$ is smooth, so $\eta =df$ andDegenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.

Integration requires an orientation: take $M$ to be an oriented compact $n$-dimensional smooth manifold without boundary. The Generalized Stokes theorem says ${\int }_{M}d\beta =0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an exact differential form leaves its integral unchanged. Thusis a well-defined linear map $H_{\mathrm{dR}}^n(M;\mathbb{R})\to \mathbb{R}$. Compactness alone does not supply the orientation omitted from the printed integration statement.

For an $n$-dimensional Lie group, choose a nonzero $\lambda \in {\Lambda }^n{T}_e^{\ast }G$ and define the invariant volume form on a Lie group byEach $d{L}_{g^{-1}}$ is an isomorphism, so $\omega$ is nowhere zero and defines an orientation. The composition identity $L_{(hg{)}^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^{\ast }\omega =\omega$. Moreover, every left-invariant differential form of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.

Under the supplied invariant-representative assumption, all top-degree de Rham cohomology classes are multiples of $[\omega ]$. For compact $G$, orient it so $\omega$ is positive; then ${\int }_G\omega >0$, whereas the Generalized Stokes theorem makes every exact top form have integral zero. Therefore $[\omega ]\ne 0$ andunder that assumption. For actual compact Lie groups, this argument applies when $G$ is connected. The supplied assumption fails for disconnected groups: on $S^1\times \mathbb{Z}/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the top-degree de Rham cohomology of a compact Lie group is ${\mathbb{R}}^c$, where $c$ is its number of connected components.

For $S^2$, use $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with the plane. The disc argument gives $\eta {|}_U=d{f}_U$ and $\eta {|}_V=d{f}_V$ for any closed differential one-form $\eta$. The overlap $U\cap V$ is connected, so $d(f_U-f_V)=0$ makes the difference a constant. Adjust one potential by that constant; they glue to a global smooth function $f$ with $df=\eta$. ThusFinally, $S^3$ is a compact connected Lie group: the identificationidentifies it smoothly with the special unitary group $SU(2)$. Matrix multiplication and inversion supply its smooth group operations. The preceding invariant-volume argument therefore gives

Green theorem states that for a positively oriented planar region $D$,Here $P=x{y}^2-\lambda y$ and $Q=x^{2}y$, soThe stated orientations make $C_1\cup C_2$ the positive boundary of the region inside the radius-$a$ circle and outside the radius-$b$ circle. Its area is $\pi (a^2-b^2)$ regardless of the inner circle's center. Therefore

The periodic Wirtinger inequality says that if $f$ is continuously differentiable, $L$-periodic, and has mean zero, thenEquality holds exactly for a linear combination of $\cos (2\pi s/L)$ and $\sin (2\pi s/L)$.

First suppose $\partial \Omega$ is one positively oriented simple closed curve, parametrized by arc length as $\gamma (s)=(x(s),y(s))$ for $0\le s\le L$. Translating the origin, which changes neither length nor area, makes both $x$ and $y$ have mean zero. Green theorem and the Cauchy-Schwarz inequality giveApplying Wirtinger's inequality to both coordinate functions and using the unit-speed identity $(x^{\prime }{)}^2+(y^{\prime }{)}^2=1$ yieldsand therefore the planar isoperimetric inequalityFor several boundary components, apply the simple-curve result to the relevant enclosed regions and use ${\sum }_j{L}_j^2\le ({\sum }_j{L}_j{)}^2$; holes only decrease the area.

Equality in both inequalities forcesfor constant vectors $A,B$, with the unit-speed and Cauchy-Schwarz equality conditions making $A$ and $B$ perpendicular and equally long. Thus the boundary is a circle. Conversely, a circular domain has area $\pi R^2$ and perimeter $2\pi R$, so equality holds.

Stokes theorem converts circulation around an oriented boundary into curl flux through a spanning surface. For a sufficiently smooth vector field on an oriented piecewise smooth surface $S$,The boundary orientation follows the right-hand rule relative to $\mathbf{n}$. The field must be smooth on the surface, not merely on its boundary; a singularity crossing the surface cannot be ignored. For a fixed loop, applying the theorem to Faraday's differential law gives the integral flux law of induction.

The planar form is Green theorem. For a region $D$ with positively oriented boundary and sufficiently smooth $P,Q$,The outer boundary is counterclockwise and any holes have clockwise boundaries. For instance $P=-y/2,Q=x/2$ makes the area of $D$ equal to half the line integral of $xdy-ydx$; the unit circle gives area $\pi$.

A closely related use of “Green's theorems” in field theory is Green's identities, obtained by applying the divergence theorem to $\varphi \nabla \psi$:and, after subtracting the expression with $\varphi ,\psi$ exchanged,These relate interior sources to boundary data in electrostatics and other boundary-value problems. For example, a harmonic potential difference vanishing on a Dirichlet boundary condition has zero integral of its squared gradient by the first identity, establishing uniqueness. Both the planar circulation theorem and the field identities are included here to make the convention clear.

Vector calculus studies differentiation and integration of scalar fields and vector fields. Three operators answer different local questions: the gradient describes change of a scalar value, the divergence describes net outward flow, and the curl describes local circulation. Their associated integral theorems turn local derivatives into measurements along paths and across boundaries.

The difference between divergence and curl is especially visible in two smooth vector fields on three-dimensional Euclidean space:At each point they have the same Euclidean norm $\sqrt{x^2+y^2}$, but their directions differ. For $F$, $\nabla \cdot F=2$ and $\nabla \times F=0$. For $G$, $\nabla \cdot G=0$ and $\nabla \times G=(0,0,2)$. One expands radially in each horizontal plane; the other describes rigid rotation.

Around the counterclockwise unit circle in the plane $z=0$, $F$ is normal to the boundary and has zero tangential component. Its outward planar flux integral is $2\pi$, while its line integral around the circle is zero. For $G$ the values are reversed: outward planar flux is zero, and circulation is $2\pi$. The divergence and curl predict exactly which boundary measurement survives.

Why is this useful? A velocity field can move material through a boundary or circulate it within a region. Conservation of mass connects its divergence with density changes through the continuity equation; its curl is vorticity. In electromagnetism, the Maxwell equations use the same operators to connect electric fields, magnetic fields, and their sources. The common mathematical language lets the same integral calculations work across different physical models. ^[ref].

The geometric framework behind these operators is developed in differential forms. The Generalized Stokes theorem unifies the fundamental theorem for line integrals, Green theorem, Stokes theorem, and divergence theorem. In coordinates their formulas look different, but each relates accumulation over a region to a measurement on its boundary.

Wilson's Vector Analysis, published in 1901, was based on J. Willard Gibbs's lectures at Yale. Wilson's preface explains that examples from geometry, mechanics, and physics are interspersed with the mathematical development, so readers encounter applications as soon as they have the tools to understand them. That is a useful way to approach vector calculus: learn each operation through a measurement it makes possible. ^[ref].