Why Polyhedral Flux Requires Care
A polyhedral solid has a boundary made from flat faces. Its outward flux is the sum of the fluxes across those faces, but the calculation is only reliable if each face is given the correct outward normal. A reversed normal changes the sign of that face’s contribution.
There are two useful ways to find the total. One is to integrate the vector field over every face, keeping track of its orientation. The other is to integrate the divergence throughout the solid. The divergence theorem says these totals agree when its hypotheses hold. Comparing the two methods is also a practical way to catch errors in face normals, bounds, or arithmetic.
Normals and Flux on a Planar Face
Suppose a face is parametrized by \(\mathbf{r}(u,v)\). The two tangent vectors \(\mathbf{r}_u\) and \(\mathbf{r}_v\) span the face, and their cross product is perpendicular to it. Its magnitude gives the area scaling factor: a small rectangle of parameter area \(du\,dv\) corresponds to a surface patch of area \(\lVert\mathbf{r}_u\times\mathbf{r}_v\rVert\,du\,dv\).
Choose the order of the parameters so that \(\mathbf{r}_u\times\mathbf{r}_v\) points outward. Then the oriented area vector is \((\mathbf{r}_u\times\mathbf{r}_v)\,du\,dv\), and the flux is computed directly with this vector. Equivalently, use the outward unit normal and the surface-area element.
This formula follows from the geometry of a small patch: its outward area vector is approximately \((\mathbf{r}_u\times\mathbf{r}_v)\,du\,dv\). The field’s component along that vector is \(\mathbf{F}\cdot(\mathbf{r}_u\times\mathbf{r}_v)\,du\,dv\). Adding these contributions over the parameter domain gives the surface integral. If the cross product points inward, reverse its sign.
For a plane written as \(ax+by+cz=d\), a normal vector is \(\langle a,b,c\rangle\). The solid’s defining inequalities determine which of the two opposite directions is outward. On a face \(g(x,y,z)=d\) bounding a region where \(g\le d\), the outward direction is the direction in which \(g\) increases, provided the face is a regular level plane.
The Divergence Theorem for a Polyhedral Solid
For \(\mathbf{F}=\langle P,Q,R\rangle\), the divergence is \(\nabla\cdot\mathbf{F}=P_x+Q_y+R_z\). The divergence theorem relates the flux through a closed boundary to the integral of this scalar quantity over the enclosed volume.
Edges and vertices do not need individual normals: they have zero surface area and do not contribute separate face integrals. Each face is integrated with its outward orientation, and the face contributions are added. The theorem does require a closed boundary and a sufficiently regular field throughout the solid; a field singular inside \(V\) does not meet the stated hypotheses.
Worked Example: Flux Through a Tetrahedron
Worked Example: Flux Through a Tetrahedron
Let \(V\) be the tetrahedron \(x\ge0,\ y\ge0,\ z\ge0,\ x+y+z\le1\), and let \(\mathbf{F}=\langle x,y,z\rangle\). Find the outward flux by faces and check it using divergence.
The coordinate faces have outward normals \(\langle-1,0,0\rangle\), \(\langle0,-1,0\rangle\), and \(\langle0,0,-1\rangle\). On \(x=0\), the field’s \(x\)-component is zero, so the flux is zero. On \(y=0\), its \(y\)-component is zero, so the flux is zero. On \(z=0\), its \(z\)-component is zero, so the flux is zero.
The slanted face is \(x+y+z=1\). The solid lies on the side \(x+y+z\le1\), so an outward unit normal is \(\langle1,1,1\rangle/\sqrt{3}\). On the face, \(\mathbf{F}\cdot\mathbf{n}=(x+y+z)/\sqrt{3}=1/\sqrt{3}\). Its projection onto the \(xy\)-plane is the triangle \(0\le x\le1,\ 0\le y\le1-x\). Since \(z=1-x-y\), the area factor is \(\sqrt{1+z_x^2+z_y^2}=\sqrt{3}\). Thus the slanted-face flux is $$ \int_0^1\int_0^{1-x} \frac{1}{\sqrt{3}}\sqrt{3}\,dy\,dx = \int_0^1(1-x)\,dx = \left[x-\frac{x^2}{2}\right]_0^1 = \frac12. $$ The total face flux is \(0+0+0+\frac12=\frac12\).
For the check, \(\nabla\cdot\mathbf{F}=1+1+1=3\). The tetrahedron’s volume is \(\int_0^1\int_0^{1-x}\int_0^{1-x-y} dz\,dy\,dx=\int_0^1(1-x)^2/2\,dx=1/6\). Therefore the volume integral is \(3(1/6)=1/2\), agreeing with the face calculation.
As a final check, the three coordinate faces contribute zero and the only nonzero face contributes \(1/2\); the divergence calculation also gives \(1/2\).
Worked Example: A Rectangular Box
Worked Example: A Rectangular Box
Let \(V=[0,1]\times[0,2]\times[0,1]\) and \(\mathbf{F}=\langle x^2,y,z\rangle\). Compute the flux across all six faces, then compare it with the divergence integral.
The outward normals on the faces \(x=1\) and \(x=0\) are \(\langle1,0,0\rangle\) and \(\langle-1,0,0\rangle\). At \(x=1\), \(\mathbf{F}\cdot\mathbf{n}=1\), and the face has area \(2\cdot1=2\), so its flux is \(2\). At \(x=0\), the normal component is \(-x^2=0\), so the flux is \(0\).
The outward normals on \(y=2\) and \(y=0\) are \(\langle0,1,0\rangle\) and \(\langle0,-1,0\rangle\). Their normal components are \(y=2\) and \(-y=0\), respectively. Each face has area \(1\cdot1=1\), so their fluxes are \(2\) and \(0\).
The outward normals on \(z=1\) and \(z=0\) are \(\langle0,0,1\rangle\) and \(\langle0,0,-1\rangle\). The normal components are \(z=1\) and \(-z=0\). Each face has area \(1\cdot2=2\), so their fluxes are \(2\) and \(0\). Adding all six contributions gives \(2+0+2+0+2+0=6\).
For the volume check, \(\nabla\cdot\mathbf{F}=2x+1+1=2x+2\). Hence $$ \iiint_V(2x+2)\,dV = \int_0^1\int_0^2\int_0^1(2x+2)\,dz\,dy\,dx = 2\int_0^1(2x+2)\,dx = 2[x^2+2x]_0^1 = 6. $$ This matches the sum of the face fluxes.
Worked Example: A Triangular Prism
Worked Example: A Triangular Prism
Take \(V=\{(x,y,z):x\ge0,\ y\ge0,\ x+y\le1,\ 0\le z\le2\}\) and \(\mathbf{F}=\langle x,y,z\rangle\). Compute and check the outward flux.
The bottom face \(z=0\) has outward normal \(\langle0,0,-1\rangle\) and contributes zero. The top face \(z=2\) has outward normal \(\langle0,0,1\rangle\). Its triangular area is \(1/2\), and the normal component is \(2\), so its flux is \(2(1/2)=1\).
The side \(x=0\) has outward normal \(\langle-1,0,0\rangle\) and zero normal component, as \(x=0\) there. The side \(y=0\) likewise has outward normal \(\langle0,-1,0\rangle\) and zero normal component.
The remaining side is \(x+y=1\). Since the solid satisfies \(x+y\le1\), its outward unit normal is \(\langle1,1,0\rangle/\sqrt{2}\). Parametrize it by \(\mathbf{r}(x,z)=\langle x,1-x,z\rangle\), with \(0\le x\le1\) and \(0\le z\le2\). Then \(\mathbf{r}_x=\langle1,-1,0\rangle\), \(\mathbf{r}_z=\langle0,0,1\rangle\), and \(\mathbf{r}_x\times\mathbf{r}_z=\langle-1,-1,0\rangle\), which points inward. Reverse the order to obtain the outward area vector \(\langle1,1,0\rangle\,dx\,dz\). On this face, \(\mathbf{F}\cdot\langle1,1,0\rangle=x+(1-x)=1\). Its flux is $$ \int_0^2\int_0^1 1\,dx\,dz = \int_0^2 1\,dz = 2. $$ The total is \(0+1+0+0+2=3\).
Here \(\nabla\cdot\mathbf{F}=3\). The volume is the triangular base area \(1/2\) times prism length \(2\), giving \(1\). The divergence integral is \(3(1)=3\), confirming the face sum.
Orientation Errors and Other Pitfalls
A frequent mistake is to use an inward normal on one face while treating the other faces as outward. For the tetrahedron above, the slanted-face unit normal must point in the direction \(\langle1,1,1\rangle\), not \(\langle-1,-1,-1\rangle\). Reversing it would turn its contribution into \(-1/2\), and the resulting total would be \(-1/2\), not \(1/2\).
Another common error is to use a unit normal but forget the area factor, or to use a non-unit normal as though it were a unit normal while still multiplying by the full surface-area element. Either approach can introduce an extra factor. Use one consistent form: \(\mathbf{F}\cdot\mathbf{n}\,dS\) with a unit normal, or \(\mathbf{F}\cdot(\mathbf{r}_u\times\mathbf{r}_v)\,du\,dv\) with the oriented area vector.
For a solid with more complicated polygonal faces, split each face into simpler pieces, such as triangles, and add their fluxes using the same outward orientation. The divergence theorem offers an independent check, not permission to ignore the boundary: the boundary must be closed, the field must be regular throughout the solid, and the volume integral must use the correct region.
Check Your Understanding
Use the definitions and procedures in this tutorial to answer the following.
- For the face \(x+y+z=2\) bounding the region \(x+y+z\le2\), which direction is outward?
- Why does the face \(x=0\) contribute zero flux for \(\mathbf{F}=\langle x,y,z\rangle\)?
- What must change in a parametrized face-flux calculation if \(\mathbf{r}_u\times\mathbf{r}_v\) points inward?
- For what class of fields and solids does the divergence theorem equate boundary flux with the volume integral of divergence?
- What is the main risk of using a non-unit normal together with the full surface-area element?