Why Internal Faces Cancel When Solids Are Joined

Why Shared Faces Matter

Flux measures how much a vector field passes outward through a surface. If a solid is split into smaller pieces, some faces of those pieces lie inside the whole solid rather than on its boundary. Such a face is counted once from each adjoining piece, but with opposite outward directions. The two contributions cancel.

This cancellation is what allows a result proved for rectangular boxes to extend to a solid made from several boxes. When we add the fluxes of the individual boxes, the internal faces disappear from the sum. What remains is flux through the boundary of the joined solid.

Key idea: On a shared face, the vector field has the same values for both adjoining pieces, while the pieces’ outward unit normals are opposites. Therefore their flux contributions sum to zero.

Shared Faces and Opposite Normals

For a solid $E$ with outward unit normal $\mathbf{n}$ on its boundary $\partial E$, the outward flux of a vector field $\mathbf{F}$ is the surface integral $\iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS$. The dot product selects the component of the field in the outward-normal direction.

Suppose two boxes $E_1$ and $E_2$ meet along a complete face $S$, with their interiors on opposite sides of that face. At each point of $S$, let $\mathbf{n}_1$ and $\mathbf{n}_2$ be the outward unit normals for the two boxes. Then $\mathbf{n}_2=-\mathbf{n}_1$. If $\mathbf{F}$ is continuous on the face, its value there is the same whichever box is being considered. Thus the two flux contributions over $S$ are

\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.

The cancellation concerns the oriented flux, not the magnitude of the field on the face. The field may have a nonzero normal component there; the opposite orientations make the signed contributions cancel.

Worked Example: Two Adjacent Unit Boxes

Let $E_1=[0,1]\times[0,1]\times[0,1]$ and $E_2=[1,2]\times[0,1]\times[0,1]$. They share the face $x=1$. Take $\mathbf{F}(x,y,z)=(x^2,y,z)$.

On the shared face, the first box has outward normal $\mathbf{n}_1=(1,0,0)$, and the second has outward normal $\mathbf{n}_2=(-1,0,0)$. On that face, $\mathbf{F}(1,y,z)=(1,y,z)$. Since the face has unit area, the fluxes are

\iint_S \mathbf{F}\cdot\mathbf{n}_1\,dS =\int_0^1\int_0^1 1\,dz\,dy=1, \qquad \iint_S \mathbf{F}\cdot\mathbf{n}_2\,dS =\int_0^1\int_0^1 (-1)\,dz\,dy=-1.

Their sum is $1+(-1)=0$. To check the total flux, consider a box $[a,a+1]\times[0,1]\times[0,1]$. The net flux through its two $x$-faces is $(a+1)^2-a^2=2a+1$. Its $y$-faces contribute $1-0=1$, and its $z$-faces contribute $1-0=1$. The box flux is therefore $2a+3$. For $E_1$, $a=0$, giving $3$; for $E_2$, $a=1$, giving $5$. The two-box flux is $3+5=8$. The shared-face contributions cancel, and the remaining exterior flux is $8$: the outer $x$-faces contribute $4-0=4$, the outer $y$-faces contribute $2-0=2$, and the outer $z$-faces contribute $2-0=2$.

Finally, $\nabla\cdot\mathbf{F}=2x+2$, so the divergence integral over the joined box is

\iiint_{[0,2]\times[0,1]\times[0,1]}(2x+2)\,dV =\int_0^2(2x+2)\,dx =[x^2+2x]_0^2=8.

This agrees with the exterior flux. The face calculation also directly verifies the cancellation at $x=1$.

The Divergence Theorem for a Box-Tiled Solid

Consider a solid $E$ formed by joining finitely many rectangular boxes. Assume the boxes have disjoint interiors and meet face-to-face: any face lying inside $E$ is a whole face shared by two boxes. If needed, subdivide boxes into smaller boxes so that this condition holds. The exterior boundary $\partial E$ consists of the faces that are not shared. Assume $\mathbf{F}$ has continuous first partial derivatives on an open set containing $E$, so the Divergence Theorem for each rectangular box applies.

Theorem: If $E$ is a finite face-to-face union of rectangular boxes and $\mathbf{F}$ has continuous first partial derivatives on a neighborhood of $E$, then $$ \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_E \nabla\cdot\mathbf{F}\,dV. $$ Here $\mathbf{n}$ is outward from the joined solid, and $\nabla\cdot\mathbf{F}$ is the divergence of the field.

The proof uses the Divergence Theorem for rectangular boxes, together with additivity of integrals over pieces with disjoint interiors. Each box’s outward normal is used on its own boundary. On an internal face, those box normals point in opposite directions; on an exterior face, the box’s outward normal agrees with the outward normal of the joined solid.

Apply the box theorem.
For every box $E_i$, the outward flux through its boundary equals the integral of $\nabla\cdot\mathbf{F}$ over $E_i$.

Add the identities.
The boxes have disjoint interiors, so their volume integrals add to the volume integral over $E$.

Cancel shared faces.
Each internal face occurs twice, with opposite normals and the same field values, so its two flux integrals sum to zero.

Keep the exterior faces.
Every unpaired face lies on $\partial E$ and has the joined solid’s outward normal. The remaining flux is exactly the boundary flux of $E$.

In equation form, if the boxes are $E_1,\ldots,E_m$, the box theorem gives

\sum_{i=1}^m\iint_{\partial E_i}\mathbf{F}\cdot\mathbf{n}_i\,dS = \sum_{i=1}^m\iiint_{E_i}\nabla\cdot\mathbf{F}\,dV = \iiint_E\nabla\cdot\mathbf{F}\,dV.

On the left, paired internal-face integrals cancel, leaving $\iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS$. This proves the stated result. Edges and corners where faces meet do not affect these surface integrals: they have zero area.

Cancellation Across Several Internal Faces

Worked Example: Four Boxes in a Rectangular Block

Tile $E=[0,2]\times[0,2]\times[0,1]$ with four unit boxes, arranged in a $2$-by-$2$ grid in the $xy$-plane. Let $\mathbf{F}(x,y,z)=(x^2,y^2,z^2)$. There are two shared faces at $x=1$, one for each unit interval of $y$. On each face $F_x=1$ and the face area is $1$, so the two contributions across each shared face are $+1$ and $-1$. Their sum is zero. There are also two shared faces at $y=1$; on each, $F_y=1$, giving a pair $+1$ and $-1$. These internal contributions also sum to zero.

The exterior flux is computed face by face. At $x=2$, $F_x=4$ over area $2$, giving $8$; at $x=0$, it is zero. The $y=2$ face contributes $4\cdot2=8$, and $y=0$ contributes zero. The $z=1$ face contributes $1\cdot4=4$, and $z=0$ contributes zero. Thus the total exterior flux is $8+8+4=20$.

The divergence is $2x+2y+2z$. Its integral is

\begin{aligned} \iiint_E(2x+2y+2z)\,dV &=(4)(2)(1)+(2)(4)(1)+(1)(2)(2)\\ &=8+8+4\\ &=20. \end{aligned}

The three terms on the first line are the integrals of $2x$, $2y$, and $2z$, respectively. The divergence integral and the exterior flux both equal $20$, after all four shared faces have canceled in pairs.

A Tiled Solid Need Not Be a Single Box

The same argument works when the joined region has a stepped or notched outline. The exterior boundary need not be the six faces of one large box. The key requirement is that every internal face be paired with a matching face of another tile, while every unpaired face belongs to the solid’s exterior boundary.

Worked Example: Three Boxes Forming an L-Shape

Take three unit boxes, each extending from $z=0$ to $z=1$, with base squares at $[0,1]\times[0,1]$, $[1,2]\times[0,1]$, and $[0,1]\times[1,2]$. Let $\mathbf{F}(x,y,z)=(x,y,z)$. The first two boxes share a face at $x=1$. Its flux is $+1$ from the first box and $-1$ from the second, since $F_x=1$ there and the face area is $1$. The first and third boxes share a face at $y=1$, whose contributions are likewise $+1$ and $-1$, since $F_y=1$. Both internal faces cancel.

For the exterior faces perpendicular to the $x$-axis, the two faces at $x=0$ contribute zero. At $x=2$, the outward flux is $F_x=2$ over unit area, giving $2$. The exposed face at $x=1$ on the upper box contributes $F_x=1$, giving $1$. Thus the exterior $x$-face total is $3$. By the same calculation with $y$ in place of $x$, the exterior $y$-face total is $3$. The three top faces at $z=1$ each contribute $1$, for a total of $3$; the bottom faces at $z=0$ contribute zero. Total exterior flux is $3+3+3=9$.

Here $\nabla\cdot\mathbf{F}=1+1+1=3$, and the solid has volume $3$. Therefore

\iiint_E\nabla\cdot\mathbf{F}\,dV=3\cdot3=9,

which matches the exterior flux. The calculation confirms that the argument does not depend on the tiles forming one rectangular block.

A Common Sign Error

A frequent mistake is to assign the same outward normal to both boxes at a shared face. In the first example, this would incorrectly calculate the interface contribution as $1+1=2$. But the normal for the second box points in the negative $x$-direction, not the positive $x$-direction. Its contribution is $-1$, so the correct sum is $1+(-1)=0$.

Another way to keep the signs straight is to ask which side of the face contains the interior of each box. The outward normal points away from that box’s interior. For adjacent boxes, those directions necessarily oppose one another. Do not discard an internal face without checking that it is shared with matching area and opposite normals.

Why the Cancellation Is Useful

Adding local flux balances would be much less useful if every internal interface remained in the final result. Opposite normals remove those artificial interfaces, leaving only the boundary of the full region. This makes the theorem for boxes a building block for more complicated tiled solids.

The argument depends on its stated conditions: the tiles must fit together so internal faces pair, and the field must be regular enough for the box theorem and the integrals to apply. It establishes the theorem for finite box-tiled solids; extending the result to smooth boundaries requires an additional limiting argument.

Check Your Understanding

Use the normal directions and face areas to answer these questions.

Two boxes share a face where $\mathbf{F}\cdot\mathbf{n}_1=4$ for the first box’s outward normal. What is the second box’s flux contribution on that same face?
Why do the field values on a shared face agree when the field is continuous?
How many times does each internal face appear when the fluxes of all tiles are added?
In the L-shaped example, why does the exposed face at $x=1$ belong to the exterior boundary rather than canceling?
State the regularity condition on $\mathbf{F}$ used to apply the box theorem to every tile.