Why Interior Boundary Terms Cancel

Why a Surface Can Be Treated Patch by Patch

A surface may be easier to describe as a collection of simpler pieces than as one whole. For example, it may consist of flat regions joined along edges, or smooth pieces joined at a crease. Stokes’ theorem can be applied to each piece and the resulting equations added together.

That addition works because every shared interior edge occurs twice in the patch boundaries: once for each adjacent patch. The key is that the two induced directions along the edge are opposite. Consequently, the two line integrals cancel. The surfaces themselves need not meet in a single plane, and at a crease their normal vectors need not agree.

Central idea: For patches that are compatibly oriented as parts of one oriented surface, each shared interior edge is traversed in opposite directions in the two patch boundaries. Its line-integral contributions therefore sum to zero.

Patchwise Stokes’ Theorem

Let \(S\) be an oriented, piecewise smooth surface divided into finitely many oriented patches \(S_1,\ldots,S_N\). Assume the patch orientations fit together to give the chosen orientation of \(S\), and that the vector field \(\mathbf{F}\) has continuous first partial derivatives on an open set containing the patches. Stokes’ theorem applies to each patch, with its induced boundary orientation:

\iint_{S_i}(\nabla\times\mathbf{F})\cdot\mathbf{n}_i\,dS = \int_{\partial S_i}\mathbf{F}\cdot d\mathbf{r}.

Here \(\mathbf{n}_i\) is the chosen unit normal on patch \(S_i\), and \(\partial S_i\) is its oriented boundary curve or collection of boundary curves. When the equations are added, the surface integrals add over the patches. The boundary integrals require more care: they include both the exterior edges of \(S\) and the shared interior edges.

Definition: The orientations of adjacent patches are compatible when they are restrictions of one orientation on the assembled surface. In particular, their induced boundary directions on a shared interior edge are opposite. Compatibility concerns the orientation of the assembled surface, not equality of the patches’ normal vectors.

Why a Shared Edge Gets Opposite Directions

First consider two patches that meet smoothly along an edge. Let \(\mathbf{n}\) be their common unit normal there, and let \(\mathbf{m}\) denote the outward unit conormal: the vector tangent to the patch, perpendicular to the edge, and pointing out of the patch. With the usual boundary convention, the induced unit tangent is \(\mathbf{T}=\mathbf{n}\times\mathbf{m}\). Along a smooth join, the two patches lie on opposite sides of the edge within the common tangent plane. Their outward conormals are opposite, so the induced tangents are opposite as well.

That vector argument is useful for a smooth join, but it does not establish the general case: adjacent patches can meet at a crease, where their tangent planes and normal vectors differ. The general reason for cancellation is instead the compatibility of the orientations. Locally, regard the two patches as the two sides of a single oriented surface divided by the shared edge. Choose a coordinate \(u\) along the edge and a transverse coordinate \(v\) across it. One patch lies on the \(v\geq 0\) side and the other on the \(v\leq 0\) side. The same surface orientation induces opposite directions along \(v=0\) from those two sides. This description uses the orientation of the assembled surface; it does not require the geometric tangent planes to match at the edge.

Once the directions are opposite, cancellation follows from a basic property of line integrals. If \(\gamma\) describes an edge from one endpoint to the other, the reverse traversal is \(\widetilde{\gamma}(t)=\gamma(1-t)\). Reversing the parameter changes the sign of the derivative, so

\int_{\widetilde{\gamma}}\mathbf{F}\cdot d\mathbf{r} = \int_0^1\mathbf{F}(\gamma(1-t))\cdot\bigl(-\gamma'(1-t)\bigr)\,dt = -\int_{\gamma}\mathbf{F}\cdot d\mathbf{r}.

Thus the two contributions from a shared edge sum to zero, regardless of the values of the line integrals individually.

Worked Example: Two Rectangles in One Plane

Consider the rectangles \(S_1=[-1,0]\times[0,2]\) and \(S_2=[0,1]\times[0,2]\), both in the plane \(z=0\), with upward orientation. They share the edge \(x=0\), \(0\leq y\leq 2\). Let \(\mathbf{F}(x,y,z)=(0,x+1,0)\). Its curl is \(\nabla\times\mathbf{F}=(0,0,1)\).

Each rectangle has area \(1\cdot 2=2\), so its curl flux is \(2\). Check the boundary circulation on each patch. With upward orientation, each rectangle’s boundary runs counterclockwise when viewed from above. On horizontal edges \(d\mathbf{r}\) has only an \(x\)-component, while \(\mathbf{F}\) has only a \(y\)-component; their dot product is zero.

For \(S_1\), the edge at \(x=-1\) contributes zero because \(x+1=0\). Its shared edge at \(x=0\) is traversed upward, with \(\mathbf{F}=(0,1,0)\) and \(d\mathbf{r}=(0,dy,0)\). Its contribution is \(\int_0^2 1\,dy=2\). The total boundary circulation is \(2\), matching its curl flux.

For \(S_2\), the outer edge at \(x=1\) is traversed upward. There \(\mathbf{F}=(0,2,0)\), so the contribution is \(\int_0^2 2\,dy=4\). The shared edge is traversed downward, contributing \(\int_2^0 1\,dy=-2\). The horizontal edges contribute zero, so the total is \(4-2=2\), again matching the curl flux.

The shared-edge contributions are \(2\) and \(-2\), which cancel. The exterior boundary contributes \(4\), equal to the total curl flux \(2+2=4\).

Worked Example: Cancellation at a Crease

Consider a rectangular patch \(S_1\) in the plane \(z=0\), with \(0\leq x\leq 1\) and \(0\leq y\leq 1\), oriented by \(\mathbf{n}_1=(0,0,1)\). Let \(S_2\) be a rectangle in the plane \(y=0\), with \(0\leq x\leq 1\) and \(0\leq z\leq 1\), oriented by \(\mathbf{n}_2=(0,1,0)\). They share the \(x\)-axis segment from \((0,0,0)\) to \((1,0,0)\). The patches meet at a right-angle crease.

On the shared edge, the outward conormal for \(S_1\) is \(\mathbf{m}_1=(0,-1,0)\); its induced tangent is \(\mathbf{n}_1\times\mathbf{m}_1=(1,0,0)\). For \(S_2\), the outward conormal is \(\mathbf{m}_2=(0,0,-1)\); its induced tangent is \(\mathbf{n}_2\times\mathbf{m}_2=(-1,0,0)\). The edge directions are opposite, even though \(\mathbf{n}_1\ne\mathbf{n}_2\) and \(\mathbf{m}_1\ne-\mathbf{m}_2\).

Take \(\mathbf{F}=(1,0,0)\). In the direction from \(x=0\) to \(x=1\), the edge integral is \(\int_0^1 1\,dx=1\). In the reverse direction it is \(\int_1^0 1\,dx=-1\). Their sum is zero. This example shows why the general argument must use compatible induced edge orientations rather than assume that adjacent patches share a normal or have opposite conormals.

Worked Example: Two Triangles Sharing a Diagonal

In the plane \(z=0\), split the unit square into two triangles along the diagonal from \((0,0,0)\) to \((1,1,0)\). Give both triangles the upward orientation and take \(\mathbf{F}(x,y,z)=(0,x,0)\). Then \(\nabla\times\mathbf{F}=(0,0,1)\). The diagonal, parameterized in one direction by \(\gamma(t)=(t,t,0)\) for \(0\leq t\leq 1\), has integral

\int_{\gamma}\mathbf{F}\cdot d\mathbf{r} = \int_0^1 (0,t,0)\cdot(1,1,0)\,dt = \int_0^1 t\,dt = \frac12.

The adjacent triangle traverses the diagonal in reverse, so its contribution is \(-\frac12\). To check the patchwise Stokes equations, each triangle has area \(\frac12\), and hence curl flux \(\frac12\). For the triangle with vertices \((0,0),(1,0),(1,1)\) in counterclockwise order, the bottom edge contributes zero, the right edge contributes \(\int_0^1 1\,dy=1\), and the diagonal contributes \(-\frac12\). Its circulation is \(1-\frac12=\frac12\). For the other triangle, the diagonal contributes \(\frac12\), and the top and left edges contribute zero. Its circulation is also \(\frac12\). Adding the two patch equations cancels the diagonal terms; the remaining exterior circulation is \(1\), equal to the combined curl flux \(\frac12+\frac12=1\).

The Patchwise Cancellation Argument

The examples illustrate the general calculation. Suppose each interior edge is shared by exactly two patches, with compatible orientations, and divide each patch boundary into its exterior edges and its shared edges. Adding the patchwise Stokes equations gives the following sequence.

Apply Stokes’ theorem on each patch.
For every \(S_i\), equate its curl flux with the line integral around its induced boundary.

Add the surface integrals.
The patches partition \(S\), so the sum of their curl fluxes is the curl flux over \(S\).

Pair the shared-edge terms.
Each interior edge occurs twice, with opposite directions. Reversal changes the sign of its line integral, so the pair sums to zero.

Keep the exterior-edge terms.
Each edge on the exterior boundary occurs once, with the direction induced by the orientation of \(S\).

Therefore the patchwise equations combine to give Stokes’ theorem for the assembled surface:

\sum_{i=1}^{N}\iint_{S_i}(\nabla\times\mathbf{F})\cdot\mathbf{n}_i\,dS = \sum_{i=1}^{N}\int_{\partial S_i}\mathbf{F}\cdot d\mathbf{r} = \int_{\partial S}\mathbf{F}\cdot d\mathbf{r}.

The last equality is precisely where the interior-edge cancellation is used. Endpoints where edges meet do not contribute separate terms to a line integral.

A Common Mistake: Requiring Opposite Conormals

A tempting but incorrect general argument is: “The two patches lie on opposite sides of their shared edge, so their outward conormals are opposite; since they have the same normal, their boundary tangents are opposite.” This is valid for a smooth join with a common tangent plane, but not for a crease. In the crease example, the conormals are \((0,-1,0)\) and \((0,0,-1)\): they are not negatives of each other, and the normals are different. The conclusion that the induced edge directions are opposite is still true, but it follows from compatibility of the patch orientations, not from those vector comparisons.

Practical check: Before adding patchwise boundary integrals, identify each shared edge and compare its two induced traversal directions. If they are opposite, its contributions cancel. Do not infer the directions by comparing normal or conormal vectors unless the patches meet smoothly and the common-tangent-plane argument applies.

Why the Cancellation Matters

Patchwise reasoning lets Stokes’ theorem be used on surfaces assembled from pieces, without counting internal seams as part of the final boundary. The cancellation is a bookkeeping principle grounded in orientation: internal edges separate patches but do not belong to the exterior boundary of the assembled surface. It also explains why the orientation condition matters. If the patches are assigned incompatible orientations, a shared edge may be traversed in the same direction twice, and the terms will add rather than cancel. In that situation, the patch orientations do not represent one consistent orientation of the assembled surface.

Check Your Understanding

Use the definitions and examples above to answer these questions.

Why does reversing the direction of a curve change the sign of its line integral?
What does orientation compatibility imply about the induced directions along a shared interior edge?
At a crease, why is it not generally valid to argue that adjacent patches have the same normal and opposite outward conormals?
In the two-rectangle example, what are the two shared-edge contributions, and why do they cancel?
When patchwise Stokes equations are added, which boundary terms remain after all shared-edge terms cancel?