Stokes's Theorem

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Why a Single Theorem Governs the Boundary The Theorem and Its Proof Consequences and Classical Theorems

Why a Single Theorem Governs the Boundary

The fundamental theorem of calculus says that integrating a derivative over an interval returns the original function evaluated at the two endpoints. The interior integral of \(f'\) is determined entirely by data on the boundary \(\{a, b\}\). This is an economy of information. To recover the total change of \(f\) across \([a, b]\) one need not examine every point inside. The endpoints already encode it.

Stokes's theorem is the assertion that this economy is not special to intervals. On an oriented manifold of any positive dimension, the integral of the derivative of a compactly supported form over the interior equals an integral over the boundary alone.

We are now in a position to make that statement precise, because the two halves of it are already in hand. On an oriented smooth manifold we know how to form the integral of a top-degree form, and we know how to differentiate a form of any degree by the exterior derivative \(d\). The remaining ingredient is the boundary. An oriented manifold with boundary carries a canonical orientation on its boundary, the Stokes orientation, fixed precisely so that the signs come out right. Three pieces are now in place: an integral, a derivative, and an oriented boundary. With them the relationship can be stated in one line, and the bulk of this page is spent proving it and reading off its consequences.

What that one line earns is unification. The fundamental theorem itself is its lowest-dimensional instance, with the endpoints \(\{a, b\}\) playing the role of the boundary. Specialized to the plane and to three-dimensional space it reproduces the classical integral theorems of vector calculus: Green's theorem, the divergence theorem, and the classical Stokes theorem for the curl. Each of the three was historically a separate result with a separate proof, and here they become a single statement under different choices of form and domain. Green's theorem is derived at the end of this page. The other two need a metric to translate forms into vector fields, and they are derived when integration on Riemannian manifolds is developed. The point now is that the unification is genuine, not a slogan.

Why the Result Is Worth the Machinery

It is fair to ask whether the apparatus of forms, orientations, and exterior derivatives earns its keep, when the classical theorems can each be proved by hand. The answer is that the apparatus is what makes the theorems one theorem. A computation tailored to surfaces in space says nothing about a region in seven dimensions, or about a manifold not given as a subset of any Euclidean space. The form-theoretic statement holds in every positive dimension and on every oriented manifold, with a single proof. This generality is not idle. The theorem is needed in exactly this form wherever integration runs over abstract manifolds. Those manifolds include configuration spaces and the curved domains on which geometric models of data are built.

The boundary relationship is also the hinge of de Rham cohomology, the structure in which integration alone can certify that a closed form fails to be exact. Integration can likewise certify that a compact oriented submanifold without boundary, with its given orientation, is not the oriented boundary of any compact oriented embedded submanifold with boundary of one dimension higher. Several of the corollaries below are the first visible signs of that structure.

The Theorem and Its Proof

We state the theorem in full generality and then prove it. The strategy of the proof is the one that has served the development so far. We verify the claim first in a single coordinate model, where it reduces to the fundamental theorem of calculus applied one variable at a time. The general case is then assembled from coordinate pieces using a partition of unity. The model is the upper half-space, and the boundary signs that emerged when its boundary orientation was computed are exactly the signs that will make the two sides agree.

Theorem: Stokes's Theorem

Let \(M\) be an oriented smooth \(n\)-manifold with boundary, \(n \geq 1\), and let \(\omega\) be a compactly supported smooth \((n-1)\)-form on \(M\). Then \[ \int_M d\omega = \int_{\partial M} \omega, \] where \(\partial M\) carries the Stokes orientation and the right-hand side is read as the integral of the pullback \(\iota_{\partial M}^*\omega\) along the inclusion \(\iota_{\partial M} : \partial M \hookrightarrow M\). If \(\partial M = \varnothing\), the right-hand side is interpreted as zero.

The degrees match by design. The derivative \(d\omega\) is an \(n\)-form, of exactly the degree integrable over the \(n\)-manifold \(M\), while \(\omega\) has degree \(n-1\), exactly the degree integrable over the \((n-1)\)-dimensional boundary. Differentiation raises degree by one and passing to the boundary lowers dimension by one, so the two operations meet at a single equality. Compact support makes both sides defined even when \(M\) is noncompact. The support of \(d\omega\) is a closed subset of the compact set \(\operatorname{supp}\omega\), hence compact. The support of the pullback \(\iota_{\partial M}^*\omega\) is closed in \(\partial M\) and contained in \(\operatorname{supp}\omega \cap \partial M\), which is compact because \(\partial M\) is closed in \(M\), so \(\iota_{\partial M}^*\omega\) is compactly supported on \(\partial M\). When \(n = 1\) the boundary is a \(0\)-manifold, a discrete space whose compact subsets are finite, and the right-hand side is the finite signed sum fixed by the zero-dimensional convention. When \(\partial M = \varnothing\) the right-hand side is zero, so the theorem then asserts that the exterior derivative of a compactly supported \((n-1)\)-form integrates to zero over a boundaryless manifold. That degenerate case will turn out to carry real content.

Proof.

The proof proceeds in three steps of increasing generality: the half-space \(\mathbb{H}^n\) itself, then a form supported in a single chart, then an arbitrary form patched together from charts. The first step carries all the analytic content. The later steps are bookkeeping.

Step 1: The upper half-space.
Suppose first that \(M\) is the upper half-space \(\mathbb{H}^n = \{x^n \geq 0\}\) itself, with its standard orientation. Because \(\omega\) is compactly supported, there is a number \(R \gt 0\) for which the support of \(\omega\) lies in \((-R, R) \times \cdots \times (-R, R) \times [0, R)\), inside the box \(A = [-R, R] \times \cdots \times [-R, R] \times [0, R]\). In particular \(\omega\) vanishes on the faces \(x^i = \pm R\) for \(i \lt n\) and on the face \(x^n = R\). Write \(\omega\) in standard coordinates, using the hat to mark the omitted factor: \[ \omega = \sum_{i=1}^{n} \omega_i\, dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^n. \] Each \(\omega_i\) is a smooth function vanishing outside \(A\). We compute \(d\omega\) term by term. The exterior derivative of \(\omega_i\) is \(\sum_j (\partial \omega_i / \partial x^j)\, dx^j\), so \[ d\omega = \sum_{i, j} \frac{\partial \omega_i}{\partial x^j}\, dx^j \wedge dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^n. \] In the wedge \(dx^j \wedge dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^n\), every coordinate factor except \(dx^i\) already appears, so the term vanishes unless \(j = i\). When \(j = i\), the factor \(dx^i\) sits at the front and must be moved into its natural position, passing the \(i - 1\) factors \(dx^1, \dots, dx^{i-1}\) ahead of it. Each transposition contributes a sign, for a total of \((-1)^{i-1}\). Hence \[ d\omega = \sum_{i=1}^{n} (-1)^{i-1}\, \frac{\partial \omega_i}{\partial x^i}\, dx^1 \wedge \cdots \wedge dx^n. \]

Integrating over \(\mathbb{H}^n\) means erasing the wedges and integrating over the box \(A\). The integrand is continuous on \(A\), so its integral may be computed as an iterated integral in any order, and for each term we carry out the \(x^i\)-integration first. The fundamental theorem of calculus turns the inner integral of \(\partial \omega_i / \partial x^i\) into a difference of boundary values of \(\omega_i\): \[ \int_{\mathbb{H}^n} d\omega = \sum_{i=1}^{n} (-1)^{i-1} \int \cdots \int \Bigl[\, \omega_i\, \Bigr]_{x^i = (\text{lower})}^{x^i = (\text{upper})} \, dx^1 \cdots \widehat{dx^i} \cdots dx^n. \] Now the support condition does the pruning. For an index \(i \neq n\), the variable \(x^i\) ranges over the full interval \([-R, R]\), and \(\omega_i\) vanishes at both endpoints \(x^i = \pm R\). The bracketed difference is zero, and the entire term drops out. The only surviving term is \(i = n\), where \(x^n\) ranges over \([0, R]\). At the upper limit \(x^n = R\) the coefficient \(\omega_n\) vanishes, but at the lower limit \(x^n = 0\) it need not, so the difference of boundary values reduces to \(-\,\omega_n(x^1, \dots, x^{n-1}, 0)\). Therefore \[ \int_{\mathbb{H}^n} d\omega = (-1)^{n-1} \cdot (-1) \int_{\mathbb{R}^{n-1}} \omega_n(x^1, \dots, x^{n-1}, 0)\, dx^1 \cdots dx^{n-1} = (-1)^{n} \int_{\mathbb{R}^{n-1}} \omega_n(\cdot, 0)\, dx^1 \cdots dx^{n-1}, \] the factor \((-1)\) coming from the lower limit of integration.

It remains to evaluate the other side, \(\int_{\partial \mathbb{H}^n} \omega\). The boundary is \(\{x^n = 0\}\), identified with \(\mathbb{R}^{n-1}\) through \((x^1, \dots, x^{n-1})\). Pulling \(\omega\) back to the boundary kills every term but one. For \(i \neq n\), the factor \(dx^n\) survives in the wedge, and the pullback of \(dx^n\) to the slice \(x^n = 0\) is zero, since \(x^n\) is constant there. Only the \(i = n\) term survives, since its wedge \(dx^1 \wedge \cdots \wedge dx^{n-1}\) contains no \(dx^n\). So the pullback is \(\omega_n(x^1, \dots, x^{n-1}, 0)\, dx^1 \wedge \cdots \wedge dx^{n-1}\), and the boundary integral is the integral of \(\omega_n(\cdot, 0)\) over \(\mathbb{R}^{n-1}\). That integral is taken with the Stokes orientation of the half-space boundary, which differs from the standard orientation of \(\mathbb{R}^{n-1}\) by precisely the factor \((-1)^n\). Inserting it, we obtain \[ \int_{\partial \mathbb{H}^n} \omega = (-1)^{n} \int_{\mathbb{R}^{n-1}} \omega_n(\cdot, 0)\, dx^1 \cdots dx^{n-1}. \] The two computations agree, and the theorem holds on \(\mathbb{H}^n\). For \(n = 1\) the integrals over \(\mathbb{R}^{n-1}\) degenerate, and the computation reads as follows. The form \(\omega = \omega_1\) is a function, \(\int_{\mathbb{H}^1} d\omega = -\omega_1(0)\), and \(\partial\mathbb{H}^1 = \{0\}\) is a single point, which the Stokes orientation marks with the sign \((-1)^1 = -1\), so the signed sum on the right is also \(-\omega_1(0)\).

Replacing \(\mathbb{H}^n\) by \(\mathbb{R}^n\) repeats the computation with one change. The variable \(x^n\) now also ranges over the full interval \([-R, R]\), so the \(i = n\) term vanishes along with the rest. Both sides are then zero. The left side vanishes because every boundary difference does, and the right side because \(\mathbb{R}^n\) has no boundary, so the equality holds trivially.

Step 2: A form supported in one chart.
Let \(M\) now be an arbitrary oriented smooth manifold with boundary, and suppose \(\omega\) is compactly supported in the domain of a single smooth chart \((U, \varphi)\) that is positively or negatively oriented. Write \(\varepsilon = \pm 1\) for its sign, with \(\varepsilon = 1\) for a positively oriented chart. We cannot simply assume \(\varepsilon = 1\), because when \(n = 1\) the right endpoint of \([0, 1]\), with its standard orientation, admits no positively oriented boundary chart. Suppose first that \((U, \varphi)\) is a boundary chart. The coordinate form \((\varphi^{-1})^*\omega\) is compactly supported in the open subset \(\varphi(U)\) of \(\mathbb{H}^n\), so extending it by zero outside \(\varphi(U)\) gives a smooth compactly supported \((n-1)\)-form on \(\mathbb{H}^n\), to which Step 1 applies. Pushing the form to coordinates, the definition of the integral over a chart gives \[ \begin{align*} \int_M d\omega &= \varepsilon \int_{\varphi(U)} \bigl(\varphi^{-1}\bigr)^* d\omega\\\\ &= \varepsilon \int_{\mathbb{H}^n} d\Bigl( \bigl(\varphi^{-1}\bigr)^*\omega \Bigr), \end{align*} \] where the second equality uses the commutation of pullback with the exterior derivative, together with the locality of \(d\): the exterior derivative of the extended form agrees with \(d\bigl((\varphi^{-1})^*\omega\bigr)\) on \(\varphi(U)\) and vanishes off its compact support. Step 1 equates the last integral with the integral of \((\varphi^{-1})^*\omega\) over \(\partial \mathbb{H}^n\), taken with its Stokes orientation.

It remains to match that boundary integral with \(\int_{\partial M}\omega\). By invariance of the boundary, \[ \partial M \cap U = \varphi^{-1}(\partial\mathbb{H}^n), \] and \(\varphi\) restricts to a diffeomorphism \(\varphi_\partial\) of \(\partial M \cap U\) onto the open subset \(\varphi(U) \cap \partial\mathbb{H}^n\) of \(\partial\mathbb{H}^n\). We claim that, when both boundaries carry their Stokes orientations, \(\varphi_\partial\) is orientation-preserving if \(\varepsilon = 1\) and orientation-reversing if \(\varepsilon = -1\). Fix \(p \in \partial M \cap U\). The differential \(d\varphi_p\) carries positively oriented bases of \(T_pM\) to bases of \(T_{\varphi(p)}\mathbb{H}^n = \mathbb{R}^n\) that are positively oriented if \(\varepsilon = 1\) and negatively oriented if \(\varepsilon = -1\). Tangency to the boundary and pointing outward are both read off from the last component in the coordinates \((x^i)\), on \(M\) as on \(\mathbb{H}^n\), and \(d\varphi_p\) preserves components. Hence it carries \(T_p\partial M\) onto \(T_{\varphi(p)}\partial\mathbb{H}^n\) and outward-pointing vectors to outward-pointing vectors. Now a basis \((E_1, \dots, E_{n-1})\) of \(T_p\partial M\) is positively oriented for the Stokes orientation exactly when \((N_p, E_1, \dots, E_{n-1})\) is a positively oriented basis of \(T_pM\) for an outward-pointing vector \(N_p\). Any such vector serves equally. Two outward-pointing vectors at \(p\) have negative last components, so they differ by a positive multiple plus a vector tangent to \(\partial M\), and this change multiplies the determinant of the enlarged basis by that positive multiple. Applying \(d\varphi_p\) to this enlarged basis, the condition holds exactly when \((d\varphi_p N_p, d\varphi_p E_1, \dots, d\varphi_p E_{n-1})\) is positively oriented in \(\mathbb{R}^n\) if \(\varepsilon = 1\), and negatively oriented if \(\varepsilon = -1\). Since \(d\varphi_p N_p\) is outward-pointing for \(\mathbb{H}^n\), this is exactly the statement that \((d\varphi_p E_1, \dots, d\varphi_p E_{n-1})\) is positively, respectively negatively, oriented for the Stokes orientation of \(\partial\mathbb{H}^n\). When \(n = 1\) the bases \((E_i)\) are empty and the same comparison is made through signs. If \(\mu\) is an orientation form of \(M\), the orientation form \(\iota^*(N \lrcorner\, \mu)\) given by the proposition on the induced orientation is then the function \(\mu(N)\), so the sign of a boundary point is \(+\) exactly when an outward vector there is positively oriented, and \(d\varphi_p\) matches these signs if \(\varepsilon = 1\) and exchanges them if \(\varepsilon = -1\). This proves the claim.

The form \(\iota_{\partial M}^*\omega\) is supported in \(\partial M \cap U\), and its pullback through \(\varphi_\partial^{-1}\) is the restriction of \((\varphi^{-1})^*\omega\) to \(\partial\mathbb{H}^n\), because \[ \varphi^{-1} \circ \iota_{\partial\mathbb{H}^n} = \iota_{\partial M} \circ \varphi_\partial^{-1} \] on \(\varphi(U) \cap \partial\mathbb{H}^n\). Both forms are compactly supported in these open subsets, so their integrals over the open subsets and over the whole boundaries agree, since charts inside the open subsets compute both. By part (d) of the properties of the integral, diffeomorphism invariance, applied to \(\varphi_\partial^{-1}\), the boundary integral computed over \(\partial\mathbb{H}^n\) in Step 1 therefore equals \(\varepsilon \int_{\partial M}\omega\). That compatibility is exactly what the Stokes orientation was defined to secure. Tracing the equalities back, and using \(\varepsilon^2 = 1\), \[ \int_M d\omega = \varepsilon \cdot \varepsilon \int_{\partial M} \omega = \int_{\partial M} \omega \] for any form supported in a single boundary chart. When \((U, \varphi)\) is an interior chart, \(U\) lies in \(\operatorname{Int} M\), so \(\operatorname{supp}\omega\) does not meet \(\partial M\) and the right-hand side is zero. The left-hand side is \(\varepsilon\) times the integral of \(d\bigl((\varphi^{-1})^*\omega\bigr)\), the coordinate form extended by zero to \(\mathbb{R}^n\), and it vanishes by the \(\mathbb{R}^n\) version of Step 1.

Step 3: Patching with a partition of unity.
Finally let \(\omega\) be an arbitrary compactly supported \((n-1)\)-form. As in the definition of the integral, cover its support by finitely many domains \(U_1, \dots, U_m\) of positively or negatively oriented charts, and choose a smooth partition of unity \(\{\psi_0, \psi_1, \dots, \psi_m\}\) subordinate to the open cover \(\{M \setminus \operatorname{supp}\omega, U_1, \dots, U_m\}\) of \(M\). Since \(\psi_0\) vanishes on \(\operatorname{supp}\omega\), we have \(\omega = \sum_{k=1}^{m} \psi_k\,\omega\), and the sums below run over \(k \geq 1\). Each \(\psi_k \omega\) is a compactly supported \((n-1)\)-form lying inside a single chart, so Step 2 applies to it. Summing over \(k\), \[ \begin{align*} \int_M d\omega &= \int_M d\!\left( \sum_k \psi_k\, \omega \right) = \sum_k \int_M d(\psi_k\, \omega)\\\\ &= \sum_k \int_{\partial M} \psi_k\, \omega = \int_{\partial M} \left( \sum_k \psi_k \right) \omega = \int_{\partial M} \omega. \end{align*} \] The interchange of \(d\) with the finite sum is the linearity of the exterior derivative, and the interchange of each integral with the sum is the linearity of the integral. Since the sums are finite, no convergence question arises. This proves the theorem in general.

Consequences and Classical Theorems

With the theorem in hand, a sequence of consequences follows almost mechanically. Several of them are nevertheless far from trivial in content, and one is the first place we meet the idea that integration can detect the global shape of a manifold. We take them in order of increasing depth, then close with the classical theorem of Green as a direct specialization.

The Fundamental Theorem for Line Integrals

The first consequence is a sanity check that also recovers a classical fact. Let \(\gamma : [a, b] \to M\) be a smooth embedding, so that its image \(S = \gamma([a,b])\) is an embedded \(1\)-dimensional submanifold with boundary, oriented so that \(\gamma\) is orientation-preserving. Its boundary consists of the two endpoints \(\gamma(a)\) and \(\gamma(b)\), and the Stokes orientation assigns \(\gamma(b)\) the sign \(+\) and \(\gamma(a)\) the sign \(-\), as the outward direction at each end dictates. A smooth function \(f \in C^\infty(M)\) is a \(0\)-form, so the theorem applies to its restriction \(\iota_S^* f\) along the inclusion \(\iota_S : S \hookrightarrow M\). Pullback commutes with the exterior derivative, so \(\iota_S^*(df) = d(\iota_S^* f)\), and the theorem gives \[ \int_S df = \int_{\partial S} f = f(\gamma(b)) - f(\gamma(a)), \] since integrating a \(0\)-form over an oriented finite set of points is, by definition, the signed sum of its values.

The integral on the left is the integral of the \(1\)-form \(df\) over the curve, so this is the statement that the integral of the differential \(df\) along an embedded curve depends only on the endpoints. In particular, taking \(\gamma\) to be the inclusion of \([a,b]\) into \(\mathbb{R}\), it is the ordinary fundamental theorem of calculus. No machinery beyond the theorem itself is used. The entire content is that the boundary of a segment is its two signed endpoints.

Two Vanishing Theorems

Two special cases occur so often that they are worth recording as named corollaries. Both are immediate from the theorem once one tracks which side vanishes and why. Recall that a form is closed if its exterior derivative is zero, and exact if it is itself the exterior derivative of another form.

Corollary: Integrals of Exact Forms over Closed Manifolds

Let \(M\) be a compact oriented smooth \(n\)-manifold without boundary, \(n \geq 1\). Then for any smooth \((n-1)\)-form \(\omega\) on \(M\), the integral of the exact \(n\)-form \(d\omega\) over \(M\) is zero: \[ \int_M d\omega = 0 \quad \text{whenever } \partial M = \varnothing. \]

Proof.

Apply the theorem to the \((n-1)\)-form \(\omega\). Compactness of \(M\) makes \(\omega\) automatically compactly supported, so the theorem applies and \(\int_M d\omega = \int_{\partial M}\omega\). The boundary is empty, so by the convention fixed in the statement the right-hand side is zero. Hence \(\int_M d\omega = 0\).

Corollary: Integrals of Closed Forms over Boundaries

Let \(M\) be a compact oriented smooth \(n\)-manifold with boundary, \(n \geq 1\), and let \(\omega\) be a closed \((n-1)\)-form on \(M\). Then the integral of \(\omega\) over the boundary is zero: \[ \int_{\partial M}\omega = 0 \quad \text{whenever } d\omega = 0 \text{ on } M. \]

Proof.

Again \(\omega\) is compactly supported because \(M\) is compact, so the theorem gives \(\int_{\partial M}\omega = \int_M d\omega\). But \(\omega\) is closed, meaning \(d\omega = 0\) identically, so the integrand on the right vanishes and the integral is zero.

The two corollaries are mirror images. The first sends the vanishing from the empty boundary back into the interior, and the second sends the vanishing of \(d\omega\) out to the boundary. Each isolates one side of the equality and exploits that the other side is forced to be zero.

Integration Detects When a Form Is Not Exact

The corollaries so far have used the theorem to prove that certain integrals vanish. Read in reverse, the same facts become a tool of an entirely different character. A single nonzero integral can certify that a form is not exact and that an oriented submanifold is not an oriented boundary, both of them global conclusions drawn from one number. This is the first appearance of the idea underlying de Rham cohomology, and it deserves to be stated carefully.

Corollary: A Nonzero Integral Obstructs Exactness and Bounding

Let \(M\) be a smooth manifold with or without boundary, let \(S \subseteq M\) be an oriented compact \(k\)-dimensional embedded submanifold without boundary, and let \(\omega\) be a closed \(k\)-form on \(M\). If \[ \int_S \omega \neq 0, \] then both of the following hold:

(a) \(\omega\) is not exact on \(M\).

(b) There is no oriented compact \((k+1)\)-dimensional embedded submanifold with boundary \(N\) in \(M\) with \(\partial N = S\) as oriented submanifolds, where \(\partial N\) carries the Stokes orientation and \(S\) its given orientation.

Proof.

Each part is proved by assuming the opposite and deriving that the integral would vanish, contradicting the hypothesis. The two arguments use the two vanishing corollaries above, one each.

Part (a). For \(k = 0\), part (a) holds under either reading of exactness in degree \(0\). If no forms of degree \(-1\) are admitted, no \(0\)-form is exact. If the forms of degree \(-1\) are taken to be \(\{0\}\), the only exact \(0\)-form is \(0\), whereas \(\omega \neq 0\) because its integral over \(S\) is nonzero. So let \(k \geq 1\), and suppose, for contradiction, that \(\omega\) were exact, say \(\omega = d\eta\) for some \((k-1)\)-form \(\eta\) on \(M\). Restrict attention to \(S\), which is compact, oriented, and has no boundary. The restriction \(\iota_S^*\omega\) is the integrand defining \(\int_S\omega\), and pullback commutes with the exterior derivative, so \(\iota_S^*\omega = \iota_S^*(d\eta) = d(\iota_S^*\eta)\) is exact on \(S\). By the corollary on integrals of exact forms over closed manifolds, applied to the closed manifold \(S\), \[ \int_S \omega = \int_S d(\iota_S^*\eta) = 0, \] contradicting \(\int_S\omega \neq 0\). Therefore \(\omega\) is not exact.

Part (b). Suppose, for contradiction, that \(S = \partial N\) for some oriented compact \((k+1)\)-dimensional embedded submanifold with boundary \(N \subseteq M\), with \(S\) carrying the Stokes orientation induced by \(N\). Here \(S = \partial N\) means equality as oriented submanifolds: the same subset, the same smooth structure, and the given orientation of \(S\) equal to the Stokes orientation. When \(M\) has no boundary, equality as subsets already forces equality of the smooth structures. Indeed \(\partial N\) is embedded in \(N\), which is embedded in \(M\), so \(\partial N\) is an embedded submanifold of \(M\), and by the uniqueness of embedded structures its smooth structure is that of \(S\). The form \(\omega\) is closed, so its restriction to \(N\) is a closed \(k\)-form on a compact manifold with boundary. By the corollary on integrals of closed forms over boundaries, applied to \(N\), \[ \int_S \omega = \int_{\partial N}\omega = 0, \] again contradicting the hypothesis. Therefore \(S\) bounds no such \(N\).

The orientation in part (b) matters. The conclusion is that \(S\) with its given orientation is not a boundary. Reversing the orientation of \(S\) flips the sign of \(\int_S\omega\) but not its vanishing, so the same conclusion holds for \(S\) with its orientation reversed.

Being non-exact and failing to bound are global properties, hard to verify directly. The force of this corollary is that a single nonzero integral, which is often a routine computation, certifies both at once. The next example is the canonical instance, on the punctured plane.

Example.

On the punctured plane \(\mathbb{R}^2 \setminus \{0\}\), consider the \(1\)-form \[ \omega = \frac{x\, dy - y\, dx}{x^2 + y^2}. \] A direct computation of the exterior derivative shows \(d\omega = 0\), so \(\omega\) is closed. Integrating it over the unit circle \(S^1\), parametrized counterclockwise, gives \(\int_{S^1}\omega = 2\pi \neq 0\). On any region where a polar angle \(\theta\) is defined smoothly, \(\omega = d\theta\), so the integral around the circle is the total angle swept.

By the corollary, the nonzero value forces two conclusions at once. First, \(\omega\) is not exact on the punctured plane. There is no globally defined smooth function whose differential is \(\omega\), even though \(\omega\) looks locally like \(d\theta\). The angle \(\theta\) cannot be defined consistently all the way around the puncture. Second, \(S^1\) is not the boundary of any compact surface lying within \(\mathbb{R}^2 \setminus \{0\}\), and since the punctured plane has no boundary, equality as sets is all that this requires. Such a surface inherits an orientation from the plane. The circle is connected, so the Stokes orientation it would receive is the counterclockwise one or its reverse, and the corollary excludes both, since reversing the orientation of the circle only flips the sign of the integral. Any disk the circle would bound must enclose the missing origin and so cannot fit in the punctured plane. A single integral has detected the hole.

Green's Theorem

The classical theorems of vector calculus are specializations of the theorem to low dimensions, obtained by choosing \(\omega\) suitably. The simplest is Green's theorem, which is nothing more than the two-dimensional case written out in coordinates. We state it for a compact regular domain, that is, for a compact embedded \(2\)-dimensional submanifold with boundary sitting inside \(\mathbb{R}^2\). A closed disk and a closed annulus are typical, and compactness makes the properness demanded by that definition automatic, since such a subset is properly embedded exactly when it is closed.

Theorem: Green's Theorem

Let \(D \subseteq \mathbb{R}^2\) be a compact regular domain, oriented by the standard orientation of \(\mathbb{R}^2\) (at each point \(T_pD = T_p\mathbb{R}^2\)), and let \(P\) and \(Q\) be smooth real-valued functions on \(D\). Then \[ \int_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx\, dy = \int_{\partial D} P\, dx + Q\, dy, \] where \(\partial D\) carries the Stokes orientation. That orientation traverses each boundary curve so as to keep \(D\) on the left, counterclockwise around an outer boundary and clockwise around the boundary of a hole, because a tangent vector \(T\) is positively oriented exactly when \((N, T)\) is positively oriented for an outward vector \(N\). Taking \(N\) to be the outward unit normal, this says that \(T\) is a positive multiple of \(N\) rotated counterclockwise by a right angle, so that rotating \(T\) counterclockwise points into \(D\).

Proof.

Apply the theorem to the \(1\)-form \(\omega = P\, dx + Q\, dy\) on \(D\). Its exterior derivative is computed term by term. Differentiating \(P\) gives \(dP = \frac{\partial P}{\partial x}\,dx + \frac{\partial P}{\partial y}\,dy\), and wedging with \(dx\) annihilates the \(dx\) term while \(dy \wedge dx = -\,dx \wedge dy\), so \[ d(P\, dx) = \frac{\partial P}{\partial y}\, dy \wedge dx = -\frac{\partial P}{\partial y}\, dx \wedge dy. \] Similarly \(d(Q\, dy) = \frac{\partial Q}{\partial x}\, dx \wedge dy\), the \(dy \wedge dy\) term being zero. Adding, \[ d\omega = \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \wedge dy. \]

The theorem equates \(\int_D d\omega\) with \(\int_{\partial D}\omega\), and the right-hand side is the line integral of \(P\, dx + Q\, dy\) around the boundary. It remains to see that the left-hand side, the integral over the oriented manifold \(D\) of \(g\, dx \wedge dy\) with \(g = \partial_x Q - \partial_y P\), is the ordinary double integral of \(g\) over \(D\). Compute it with charts \((U_i, \varphi_i)\) of signs \(\varepsilon_i\), for \(i = 1, \dots, m\), and a partition of unity \(\{\psi_0, \psi_1, \dots, \psi_m\}\) as in the definition of the integral, so that \(\psi_0\) vanishes on the support of \(g\) and \(\sum_{i \geq 1} \psi_i\, g = g\). The manifold interior of \(D\) is its topological interior, an open subset of \(\mathbb{R}^2\) on which the identity is a positively oriented chart. The map \(\varphi_i^{-1}\), from \(\varphi_i(U_i \cap \operatorname{Int} D)\) onto \(U_i \cap \operatorname{Int} D\), is a diffeomorphism between open subsets of \(\mathbb{R}^2\) of sign \(\varepsilon_i\). By the diffeomorphism invariance on open sets, together with the convention that discards \(\partial\mathbb{H}^2\) in a boundary chart, each term \(\int_D \psi_i\, g\, dx \wedge dy\) is \(\varepsilon_i \cdot \varepsilon_i = 1\) times the ordinary integral of \(\psi_i\, g\) over \(U_i \cap \operatorname{Int} D\), which is its integral over \(\operatorname{Int} D\). Summing over \(i \geq 1\), \(\int_D d\omega\) is the ordinary integral of \(g\) over \(\operatorname{Int} D\). This is the double integral over \(D\), because \(\partial D\) is embedded in \(D\), hence in \(\mathbb{R}^2\), as a one-dimensional submanifold, and therefore has measure zero. This is precisely the stated identity.

The mechanism is entirely contained in the sign \(dy \wedge dx = -\,dx \wedge dy\). The antisymmetry of the wedge product is what produces the difference \(\partial_x Q - \partial_y P\) rather than a sum. The divergence theorem and the classical Stokes theorem for the curl arise the same way, in three dimensions, once the correspondence between forms and the vector-calculus operators is set up. That correspondence requires a metric and is developed in the treatment of integration on Riemannian manifolds.