Why a Missing Piece Flux Can Be Recovered
A closed boundary can be made from several pieces: the faces of a box, the caps and curved side of a cylinder, or the outer and inner surfaces of a region with a cavity. Sometimes the flux through one piece is difficult to calculate directly, while the flux through the other pieces and the divergence over the enclosed volume are accessible.
The Divergence Theorem provides a balance equation for the entire closed boundary. Once the total outward flux is known, the flux through a missing piece is the difference between that total and the fluxes through all the remaining pieces. The subtraction is simple; the important care is to use the correct orientation for every piece.
The Flux Balance for a Closed Boundary
Let \(E\) be a bounded solid in \(\mathbb{R}^3\), and let \(\partial E\) denote its closed boundary. Suppose \(\mathbf{F}\) is a vector field whose components have continuous first partial derivatives on an open set containing \(E\). Suppose also that \(\partial E\) is piecewise smooth. These are standard conditions for applying the Divergence Theorem.
Write \(\mathbf{n}\) for the unit normal pointing outward from the solid \(E\). The outward flux of \(\mathbf{F}\) through a boundary piece \(S\) is the surface integral of the normal component of the field.
The divergence theorem states that the total outward flux through the closed boundary equals the integral of the divergence over the solid. If the boundary is divided into pieces \(S_1,\dots,S_m\), with each piece given the normal pointing outward from \(E\), then additivity of the surface integral gives
Intersections between adjacent pieces, such as the edges of a box, do not contribute to these surface integrals: they have surface area zero. If \(S_k\) is the piece whose flux is unknown, the balance equation becomes
This is a recovery formula, not a separate theorem. It follows by rearranging the Divergence Theorem after the total flux and the known piece fluxes have been identified.
A Reliable Sign and Subtraction Procedure
Before subtracting, confirm that every known flux is measured with the normal pointing outward from the solid. A piece can have a natural normal in the opposite direction, or a supplied flux may use an orientation different from the required one. Reversing a unit normal reverses the sign of the flux because \(\mathbf{F}\cdot(-\mathbf{n})=-\mathbf{F}\cdot\mathbf{n}\).
Check that the boundary pieces together make a closed surface enclosing the region of interest.
For each piece, use the normal pointing out of the solid. On an inner boundary around a cavity, this points into the cavity.
Evaluate \(\nabla\cdot\mathbf{F}\) and integrate it over the volume, checking that the field and region meet the theorem’s conditions.
Subtract the known outward fluxes from the total. Verify the result by adding all piece fluxes back together.
Worked Examples
Worked Example: Recovering a Face Flux on a Rectangular Box
Let \(E=[0,2]\times[0,1]\times[0,3]\) and let \(\mathbf{F}(x,y,z)=(x,y,z)\). Find the outward flux through the face \(x=2\), using the fluxes through the other five faces.
First, calculate the divergence:
The box has side lengths \(2\), \(1\), and \(3\), so its volume is \(2\cdot1\cdot3=6\). The total outward flux is therefore
Now compute the five known face fluxes. On \(x=0\), the outward normal is \((-1,0,0)\), and \(x=0\), so \(\mathbf{F}\cdot\mathbf{n}=0\); this face contributes \(0\). On \(y=0\), the outward normal is \((0,-1,0)\), and \(y=0\), so the contribution is also \(0\). On \(z=0\), the outward normal is \((0,0,-1)\), and \(z=0\), giving \(0\).
On \(y=1\), the outward normal is \((0,1,0)\), so \(\mathbf{F}\cdot\mathbf{n}=1\). The face area is \(2\cdot3=6\), giving flux \(1\cdot6=6\). On \(z=3\), the outward normal is \((0,0,1)\), so \(\mathbf{F}\cdot\mathbf{n}=3\). The face area is \(2\cdot1=2\), giving flux \(3\cdot2=6\). The sum of the known outward fluxes is \(0+0+0+6+6=12\).
Thus the missing flux through \(x=2\) is
Check the result directly: on \(x=2\), the outward normal is \((1,0,0)\), so \(\mathbf{F}\cdot\mathbf{n}=2\). The face area is \(1\cdot3=3\), and its flux is \(2\cdot3=6\). Adding all six face fluxes gives \(6+6+6=18\), which agrees with the divergence integral.
Worked Example: Recovering the Curved-Side Flux of a Cylinder
Let \(E\) be the solid cylinder \(x^2+y^2\leq4\), \(0\leq z\leq3\), and let \(\mathbf{F}(x,y,z)=(x,y,2z)\). Use the fluxes through the two caps to find the outward flux through the curved side.
The divergence is
The cylinder has radius \(2\) and height \(3\), so its volume is \(\pi(2)^2(3)=12\pi\). Its total outward flux is \(4(12\pi)=48\pi\).
At the top cap \(z=3\), the outward normal is \((0,0,1)\). The normal component is \(\mathbf{F}\cdot\mathbf{n}=2z=6\), constant across the cap, whose area is \(4\pi\). The top flux is \(6(4\pi)=24\pi\). At the bottom cap \(z=0\), the outward normal is \((0,0,-1)\). There \(\mathbf{F}\cdot\mathbf{n}=-2z=0\), so the bottom flux is \(0\).
Subtract the two cap fluxes from the total:
For a check, the curved side has outward unit normal \((x/2,y/2,0)\) because \(x^2+y^2=4\). There, \(\mathbf{F}\cdot\mathbf{n}=(x^2+y^2)/2=2\). The curved side has area \(2\pi(2)(3)=12\pi\), so its flux is \(2(12\pi)=24\pi\). The three piece fluxes sum to \(24\pi+0+24\pi=48\pi\), as required.
Worked Example: Recovering Flux on an Inner Boundary
Consider the spherical shell \(E=\{(x,y,z):1\leq x^2+y^2+z^2\leq4\}\) and the field \(\mathbf{F}(x,y,z)=(x,y,z)\). Find the outward flux through the inner spherical boundary, given the outward flux through the outer sphere.
The field is smooth throughout the shell, which excludes the origin. Its divergence is \(1+1+1=3\). The shell volume is the volume of the radius-\(2\) ball minus the volume of the radius-\(1\) ball:
The total outward flux from the shell is therefore \(3(28\pi/3)=28\pi\). The known outer-boundary flux is \(32\pi\): on the radius-\(2\) sphere, \(\mathbf{F}\cdot\mathbf{n}=2\) and the area is \(16\pi\), giving \(2(16\pi)=32\pi\).
The outward normal for the shell on its inner boundary points toward the center, not away from it. Let \(\Phi_{\text{inner}}\) denote the flux with this outward-from-the-shell orientation. Then
Check the sign directly. At radius \(1\), the normal pointing out of the shell is \(-\mathbf{e}_r\), while \(\mathbf{F}=\mathbf{e}_r\). Their dot product is \(-1\), and the inner sphere has area \(4\pi\). Its flux is \((-1)(4\pi)=-4\pi\). The sum \(-4\pi+32\pi=28\pi\) matches the divergence integral.
Why the Subtraction Works
The justification combines two facts. First, under the stated smoothness and boundary conditions, the Divergence Theorem equates the total outward flux to the integral of divergence over the solid. Second, a surface integral over a boundary divided into pieces is the sum of its integrals over those pieces. Therefore, the volume integral equals the sum of all piece fluxes. Moving every known piece flux to the other side leaves exactly the missing flux.
The result depends on using a genuinely closed boundary and a field with no singularity in the region where the theorem is applied. If the field is undefined or fails the required regularity inside the solid, the usual theorem cannot be applied to that solid without further analysis. Likewise, omitting a boundary piece from the accounting does not make its flux zero; it makes the recovery equation incomplete.
Common Sign Error: Subtracting an Inward Flux as if It Were Outward
Suppose the field is \(\mathbf{F}(x,y,z)=(x,y,z)\) on the unit cube \([0,1]^3\), and the missing piece is \(x=1\). The divergence integral is \(3\), and the outward fluxes on \(x=0\), \(y=0\), \(z=0\), and \(z=1\) are respectively \(0,0,0,1\). The outward flux on \(y=1\) is \(1\), so the missing flux is \(3-(0+0+0+1+1)=1\).
Now suppose the flux reported for \(y=1\) was instead computed using the inward normal. That reported value is \(-1\). A common wrong calculation is to insert \(-1\) directly as though it were the outward flux, giving \(3-(0+0+0+1-1)=3\). This is incorrect because the given orientation does not match the flux balance. Reverse the sign first: the outward flux on \(y=1\) is \(1\), and the missing flux is \(3-(0+0+0+1+1)=1\). The check is that all six outward face fluxes then sum to \(3\).
Interpreting the Recovered Value
A positive answer means the field has net flow outward across the missing piece with the chosen outward normal. A negative answer means the net flow is inward across that piece. A negative value is not automatically an error, as the inner-boundary example shows. The sign records the direction of the normal component relative to the required orientation.
Recovering a piece flux is useful when the total divergence integral is manageable and the remaining boundary pieces have known fluxes. It can avoid a separate surface-integral calculation for the missing piece. The final check is always the same: add every piece flux using consistent outward orientations and confirm that the sum equals the volume integral of the divergence.
Check Your Understanding
Use outward normals relative to the solid in each question.
- If the total outward flux is \(20\) and the known piece fluxes are \(4\), \(-3\), and \(8\), what is the missing flux?
- Why must a flux supplied with the inward normal have its sign reversed before it is used in the outward-flux sum?
- For a region with a cavity, which way does the outward normal of the region point on the inner boundary?
- What two conditions about the field and boundary are needed for the standard Divergence Theorem setup used here?
- If the recovered flux is negative, what does that say about the net flow across the piece?