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Flux and Circulation Applications · Tutorial 982 of 999

Flux Cancellation Across Internal Interfaces

Use outward orientations to identify and cancel internal fluxes, then account for the flux across the remaining exterior boundary.

Advanced 10 min read

What You'll Learn

  • Define flux across an oriented surface patch
  • Identify the outward normals on opposite sides of a shared interface
  • Derive the cancellation rule for a finite partition of solids
  • Apply cancellation to planar, curved, and slanted interfaces
  • Recognize when discontinuities or mismatched faces prevent cancellation

Why Internal Fluxes Disappear

When several solids fit together, each solid has its own closed boundary. A face where two solids meet belongs to both boundaries, but it is not part of the boundary of the combined solid. Its flux is counted once from each side, with opposite orientations. If the vector field agrees on the shared face, those two contributions cancel.

This is more than a bookkeeping convenience. It explains how a flux calculation over a complicated solid can be assembled from calculations over simpler pieces. The key is to keep the orientation of every face tied to the solid whose boundary is being considered.

Central idea: Across a shared interface, the outward normals of the two adjacent solids point in opposite directions. For a single vector field defined on both sides, the two interface fluxes are therefore equal in magnitude and opposite in sign.

Flux and Orientation

Let \(S\) be a smooth surface patch with a chosen unit normal vector field \(\mathbf{n}\), and let \(\mathbf{F}\) be a continuous vector field on \(S\). The flux of \(\mathbf{F}\) through \(S\), with that orientation, is the surface integral

$$ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS. $$

Reversing the orientation replaces \(\mathbf{n}\) by \(-\mathbf{n}\). Since \(\mathbf{F}\cdot(-\mathbf{n})=-(\mathbf{F}\cdot\mathbf{n})\), reversing the orientation changes the sign of the flux.

For a bounded solid, the outward orientation means that the normal points away from the solid. If two solids share a smooth face, the outward normal of the first points toward the second, while the outward normal of the second points back toward the first. This opposition holds at every regular point in the shared face, even if the face is curved.

Definition: An internal interface is a surface patch shared by two adjacent solids in a partition of a larger solid. It is internal because points on its regular part are not on the boundary of the combined solid.

The Cancellation Rule

Suppose a bounded solid \(D\) is divided into finitely many solids \(D_1,\ldots,D_m\) with disjoint interiors. Assume their boundaries are piecewise smooth and that any shared boundary pieces are matching surface patches. Let \(\mathbf{F}\) be continuous on the union. Write \(\partial D_i\) for the boundary of \(D_i\), oriented outward from \(D_i\).

When the flux integrals over the individual boundaries are added, each exterior patch appears once. Each internal interface appears twice, once for each adjacent solid. If \(S\) is such an interface, let \(\mathbf{n}_1\) and \(\mathbf{n}_2\) be the outward unit normals from its two sides. On its regular part, \(\mathbf{n}_2=-\mathbf{n}_1\), so

$$ \iint_S \mathbf{F}\cdot\mathbf{n}_1\,dS + \iint_S \mathbf{F}\cdot\mathbf{n}_2\,dS = \iint_S \mathbf{F}\cdot(\mathbf{n}_1+\mathbf{n}_2)\,dS =0. $$

Edges where smooth patches meet do not affect these surface integrals: they have surface area zero. Thus only the exterior patches remain after all internal pairs are canceled.

Theorem: Under the partition and regularity conditions above, the sum of the outward fluxes over the boundaries of the pieces equals the outward flux over the boundary of the combined solid:
$$ \sum_{i=1}^{m}\iint_{\partial D_i}\mathbf{F}\cdot\mathbf{n}_i\,dS = \iint_{\partial D}\mathbf{F}\cdot\mathbf{n}\,dS. $$

Here the normals on the left point outward from their respective pieces, while the normal on the right points outward from the combined solid. The equality follows by splitting each boundary integral into its exterior and shared patches, then canceling the two contributions on every shared patch.

Worked Example: Two Triangular Prisms

Worked Example: Two Triangular Prisms

Consider two triangular prisms, one below and one above the plane \(z=0\). Their common triangular face is

$$ S=\{(x,y,0):x\geq 0,\ y\geq 0,\ x+y\leq 1\}. $$

Take \(\mathbf{F}(x,y,z)=(0,0,1+x+y)\). The outward normal on this face for the lower prism is \(\mathbf{k}=(0,0,1)\), and for the upper prism it is \(-\mathbf{k}\). The lower prism's interface flux is

$$ \begin{aligned} \iint_S\mathbf{F}\cdot\mathbf{k}\,dS &=\int_0^1\int_0^{1-x}(1+x+y)\,dy\,dx\\ &=\int_0^1\left[(1+x)y+\frac{y^2}{2}\right]_{0}^{1-x}\,dx\\ &=\int_0^1\left((1+x)(1-x)+\frac{(1-x)^2}{2}\right)\,dx\\ &=\int_0^1\left(\frac{3}{2}-x-\frac{x^2}{2}\right)\,dx\\ &=\left[\frac{3x}{2}-\frac{x^2}{2}-\frac{x^3}{6}\right]_{0}^{1} =\frac{5}{6}. \end{aligned} $$

For the upper prism, the normal is the negative of the lower prism's normal, so its interface flux is \(-5/6\). The two internal contributions sum to \(5/6-5/6=0\).

As a check on how the interface fits into each closed boundary, note that \(\nabla\cdot\mathbf{F}=0\). The Divergence Theorem therefore gives total boundary flux zero for each prism. The flux through the non-interface faces of the lower prism must be \(-5/6\), balancing its \(5/6\) interface flux. For the upper prism, those faces must contribute \(5/6\), balancing its \(-5/6\) interface flux. When the prisms are joined, the exterior contributions also sum to zero, as expected.

Worked Example: Two Hemispheres

Worked Example: Two Hemispheres

Join the upper and lower unit hemispheres along their common unit disk \(z=0\). Let \(\mathbf{F}=(0,0,1)\). On the disk, the upper hemisphere's outward normal is \(-\mathbf{k}\), and the lower hemisphere's outward normal is \(\mathbf{k}\). Since the disk has area \(\pi\), their interface fluxes are

$$ \iint_S\mathbf{F}\cdot(-\mathbf{k})\,dS=-\pi, \qquad \iint_S\mathbf{F}\cdot\mathbf{k}\,dS=\pi. $$

They cancel. We can also check the remaining curved pieces directly. On the unit sphere, the outward normal at \((x,y,z)\) is \((x,y,z)\), so \(\mathbf{F}\cdot\mathbf{n}=z\). Parameterize the upper hemisphere by \(z=\cos\phi\), with \(0\leq\phi\leq\pi/2\) and \(0\leq\theta\leq2\pi\). Its surface element is \(\sin\phi\,d\phi\,d\theta\). The curved upper flux is

$$ \int_0^{2\pi}\int_0^{\pi/2} \cos\phi\sin\phi\,d\phi\,d\theta = 2\pi\left[\frac{\sin^2\phi}{2}\right]_{0}^{\pi/2} =\pi. $$

On the lower hemisphere, \(z\) is negative, and the corresponding integral over \(\pi/2\leq\phi\leq\pi\) is \(-\pi\). Each hemisphere's curved flux cancels its own disk flux: \(\pi+(-\pi)=0\) above, and \(-\pi+\pi=0\) below. After joining them, only the full sphere remains, and its total flux is \(\pi+(-\pi)=0\). This verifies the cancellation and the resulting exterior flux directly.

Worked Example: Tetrahedra Sharing a Slanted Face

Worked Example: Tetrahedra Sharing a Slanted Face

Let two tetrahedra share the triangular face with vertices \(A=(1,0,0)\), \(B=(0,1,0)\), and \(C=(0,0,1)\). One tetrahedron has its fourth vertex at the origin, and the other has its fourth vertex at \((1,1,1)\). Take the constant field \(\mathbf{F}=(1,2,3)\).

Two edge vectors of the shared triangle, starting at \(A\), are

$$ \overrightarrow{AB}=(-1,1,0), \qquad \overrightarrow{AC}=(-1,0,1). $$

Their cross product is \((1,1,1)\), which has length \(\sqrt{3}\). Hence the triangle's area is \(\sqrt{3}/2\), and a unit normal is \((1,1,1)/\sqrt{3}\). Orient it outward from the tetrahedron with vertex at the origin, toward the other tetrahedron. The flux through the shared face from that first tetrahedron is

$$ \mathbf{F}\cdot\frac{(1,1,1)}{\sqrt{3}} = \frac{1+2+3}{\sqrt{3}} = 2\sqrt{3}, \qquad \text{flux} = 2\sqrt{3}\cdot\frac{\sqrt{3}}{2} =3. $$

The second tetrahedron has the opposite outward normal on the same face, so its flux there is \(-3\). The field is constant and has divergence zero. By the Divergence Theorem, each tetrahedron's total boundary flux is zero; thus its other faces contribute \(-3\) for the first tetrahedron and \(3\) for the second. Adding the tetrahedra cancels both the internal fluxes and these opposite contributions, leaving total exterior flux zero. This is consistent with the zero divergence throughout the combined solid.

When Cancellation Applies—and When It Does Not

Cancellation requires more than two pieces that look adjacent in a diagram. Their interface patches must match, their outward normals must be opposite almost everywhere on those patches, and the same vector field must have the same value on both sides. If the field is continuous across the interface, the last condition is automatic.

A field defined separately on each piece may jump across the interface. In that case, opposing normals alone do not ensure cancellation, because the values of the field may differ. If the two limiting fields are \(\mathbf{F}_1\) and \(\mathbf{F}_2\), the paired contribution is generally

$$ \iint_S \mathbf{F}_1\cdot\mathbf{n}\,dS + \iint_S \mathbf{F}_2\cdot(-\mathbf{n})\,dS = \iint_S(\mathbf{F}_1-\mathbf{F}_2)\cdot\mathbf{n}\,dS, $$

which need not be zero. Cancellation is also not valid for mismatched patches, or if a supposed internal face is actually exposed on the outside of the combined solid.

Common mistake: Assigning the same normal direction to a shared face when computing the outward flux of both solids. Outward normals belong to individual solids, not to the face by itself; on a genuine shared interface they point in opposite directions.

Worked Example: Correcting a Sign Error

For the two triangular prisms above, suppose the lower prism's interface flux is correctly computed as \(5/6\), but the upper prism is mistakenly assigned the same normal \(\mathbf{k}\). The incorrect calculation would give

$$ \frac{5}{6}+\frac{5}{6}=\frac{5}{3}, $$

and falsely leave a nonzero internal contribution. The upper prism lies above the shared face, so its outward normal there is \(-\mathbf{k}\). Correcting that orientation gives

$$ \frac{5}{6}+\left(-\frac{5}{6}\right)=0. $$

The check is geometric: the upper prism's outward direction at its bottom face points downward, away from its interior. Using that direction restores cancellation.

Why the Rule Matters

Many solids are easier to describe as unions of simpler regions than as a single boundary. The cancellation rule lets you add flux calculations over those pieces without accidentally treating internal faces as part of the exterior boundary. It also clarifies why the Divergence Theorem can be applied piece by piece and then combined: the volume integrals add over regions with disjoint interiors, while the internal boundary fluxes cancel.

For a piecewise-smooth closed boundary, the outward normal may change from one smooth patch to another and may not be defined along edges. This causes no difficulty for flux integration: the normal is needed on the smooth patches, and their edges have surface area zero. The essential task is to orient each patch outward from the whole solid and to omit any shared internal patches after cancellation.

Check Your Understanding

Use orientation and flux additivity to answer these questions.

  1. Why do the outward normals on a shared smooth interface point in opposite directions?
  2. What two conditions on the field and interface are essential for the paired fluxes to cancel?
  3. If the flux through a shared face is \(4\) for one solid, what is its contribution for the adjacent solid when the same field is used?
  4. Why do edges where surface patches meet not affect the cancellation argument?
  5. If the field has different values on the two sides of an interface, what integral describes the remaining paired contribution?

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