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Flux Integrals and Orientation · Tutorial 890 of 999

Direct Flux Through the Faces of a Box

Learn to calculate box flux face by face, track each outward normal, and obtain inward flux by reversing all six orientations.

Advanced 10 min read

What You'll Learn

  • Write the outward unit normal for each of the six faces of a rectangular box.
  • Set up and evaluate each face’s flux integral using its two varying coordinates.
  • Add the six signed contributions to find total outward flux.
  • Reverse every face normal to calculate inward flux.
  • Check orientation signs and verify that inward flux is the negative of outward flux.

Why Calculate Box Flux Face by Face?

A rectangular box has six flat faces, so its flux can be found directly without parametrizing a curved surface. The essential work is to evaluate the vector field on each face, use that face’s outward normal, and add the six signed contributions.

The signs matter. A field pointing out of a face contributes positive outward flux; a field pointing into that face contributes negative outward flux. Once all six outward contributions are known, the inward flux follows by reversing every normal, which changes the sign of each contribution.

Definition and Face Formula

Let the box be \(B=[a,b]\times[c,d]\times[e,f]\), where \(a<b\), \(c<d\), and \(e<f\). Let \(\mathbf F(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z))\) be continuous on the box. Each face is a planar patch. Its outward flux is the integral of the field dotted with the outward unit normal, multiplied by the face’s area element.

Definition: The outward flux through the box is the sum of the outward fluxes through its six faces. The outward unit normals are \(+\mathbf i\) on \(x=b\), \(-\mathbf i\) on \(x=a\), \(+\mathbf j\) on \(y=d\), \(-\mathbf j\) on \(y=c\), \(+\mathbf k\) on \(z=f\), and \(-\mathbf k\) on \(z=e\).

On a face of constant \(x\), the area element is \(dy\,dz\), and only the \(P\)-component contributes to the dot product. The corresponding statements hold for constant \(y\) and constant \(z\). Thus the complete outward-flux formula is

$$ \begin{aligned} \Phi_{\mathrm{out}} ={}&\int_c^d\int_e^f P(b,y,z)\,dz\,dy -\int_c^d\int_e^f P(a,y,z)\,dz\,dy\\ &+\int_a^b\int_e^f Q(x,d,z)\,dz\,dx -\int_a^b\int_e^f Q(x,c,z)\,dz\,dx\\ &+\int_a^b\int_c^d R(x,y,f)\,dy\,dx -\int_a^b\int_c^d R(x,y,e)\,dy\,dx . \end{aligned} $$

For each term, the component is evaluated at the fixed coordinate of that face. The plus or minus sign comes from the outward normal, not from whether the component itself happens to be positive or negative.

Orientation rule: The inward unit normal on each face is the negative of its outward unit normal. Therefore, for the same box and field, \(\Phi_{\mathrm{in}}=-\Phi_{\mathrm{out}}\).

Why the Face Formula Works

Consider the face \(x=b\), parametrized by \(\mathbf r(y,z)=(b,y,z)\), with \(c\le y\le d\) and \(e\le z\le f\). The oriented area element in this parameter order is \(\mathbf r_y\times\mathbf r_z\,dy\,dz\). Since \(\mathbf r_y=(0,1,0)\) and \(\mathbf r_z=(0,0,1)\), their cross product is \((1,0,0)=\mathbf i\), the outward unit normal on \(x=b\). Consequently, the flux through this face is

$$ \int_c^d\int_e^f \mathbf F(b,y,z)\cdot\mathbf i\,dz\,dy = \int_c^d\int_e^f P(b,y,z)\,dz\,dy. $$

On the opposite face \(x=a\), the outward normal is \(-\mathbf i\), so the dot product is \(-P(a,y,z)\). The other four terms follow by fixing \(y\) or \(z\) and taking the corresponding outward normal. Adding the six integrals gives the total because the faces cover the box boundary; their shared edges have zero area and do not add any flux.

Worked Example: A Box with Different Contributions on Its Faces

Worked Example: A Box with Different Contributions on Its Faces

Calculate the outward and inward flux of \(\mathbf F(x,y,z)=(xy,0,1)\) through \(B=[0,2]\times[1,3]\times[-1,1]\). We evaluate the six outward contributions separately.

Face \(x=2\). Its outward normal is \(+\mathbf i\), and \(P(2,y,z)=2y\). Therefore,

$$ \int_1^3\int_{-1}^1 2y\,dz\,dy =\int_1^3 2y(1-(-1))\,dy =4\int_1^3 y\,dy =4\left[\frac{y^2}{2}\right]_1^3 =4\left(\frac{9-1}{2}\right) =16. $$

Face \(x=0\). Here \(P(0,y,z)=0\), so the outward contribution, including its normal \(-\mathbf i\), is \(0\).

Faces \(y=3\) and \(y=1\). The \(Q\)-component is zero everywhere, so both contributions are \(0\), regardless of whether the outward normal is \(+\mathbf j\) or \(-\mathbf j\).

Face \(z=1\). Its outward normal is \(+\mathbf k\), and \(R=1\). Its area is \((2-0)(3-1)=4\), so its contribution is \(\int_0^2\int_1^3 1\,dy\,dx=4\).

Face \(z=-1\). Its outward normal is \(-\mathbf k\), while \(R=1\). Its contribution is \(-\int_0^2\int_1^3 1\,dy\,dx=-4\).

Adding the six contributions gives \(16+0+0+0+4-4=16\) for outward flux. Reversing all the normals changes each contribution’s sign, so the inward flux is \(-16\). As a check, the total inward contributions are \(-16,0,0,0,-4,+4\), which sum to \(-16\).

Worked Example: Nonzero Flux from Several Face Pairs

Worked Example: Nonzero Flux from Several Face Pairs

Find the outward and inward flux of \(\mathbf F(x,y,z)=(x+z,y,2z)\) through \(B=[-1,1]\times[0,2]\times[1,3]\).

The two \(x\)-faces. At \(x=1\), the outward normal is \(+\mathbf i\), so the contribution is

$$ \int_0^2\int_1^3(1+z)\,dz\,dy =2\left[z+\frac{z^2}{2}\right]_1^3 =2\bigl((3+9/2)-(1+1/2)\bigr) =12. $$

At \(x=-1\), the outward normal is \(-\mathbf i\), and \(P(-1,y,z)=z-1\). The contribution is

$$ -\int_0^2\int_1^3(z-1)\,dz\,dy =-2\left[\frac{(z-1)^2}{2}\right]_1^3 =-2\left(\frac{4}{2}-0\right) =-4. $$

The two \(y\)-faces. At \(y=2\), the outward normal is \(+\mathbf j\). Since \(Q=2\) there and the face has area \((1-(-1))(3-1)=4\), the contribution is \(2\cdot4=8\). At \(y=0\), \(Q=0\), so the contribution is \(0\).

The two \(z\)-faces. At \(z=3\), \(R=6\), and the area is \((1-(-1))(2-0)=4\), giving \(6\cdot4=24\). At \(z=1\), the outward normal is \(-\mathbf k\), and \(R=2\), giving \(-2\cdot4=-8\).

The outward total is \(12-4+8+0+24-8=32\). Hence the inward total is \(-32\). Checking by reversing the six contributions gives \(-12+4-8+0-24+8=-32\), as required.

Worked Example: Keeping the Signs with Mixed Contributions

Worked Example: Keeping the Signs with Mixed Contributions

Calculate both orientations for \(\mathbf F(x,y,z)=(x-y,2y+z,x-z)\) on \(B=[0,2]\times[1,2]\times[-1,1]\).

Face \(x=2\). Here \(P=2-y\), the outward normal is \(+\mathbf i\), and the \(z\)-interval has length \(2\). Thus

$$ \int_1^2\int_{-1}^1(2-y)\,dz\,dy =2\int_1^2(2-y)\,dy =2\left[2y-\frac{y^2}{2}\right]_1^2 =2\left(2-\frac32\right)=1. $$

Face \(x=0\). Here \(P=-y\), and the outward normal is \(-\mathbf i\), so the signed integrand is \(y\). Its contribution is \(2\int_1^2y\,dy=2(3/2)=3\).

Face \(y=2\). The outward normal is \(+\mathbf j\), and \(Q=4+z\). Therefore,

$$ \int_0^2\int_{-1}^1(4+z)\,dz\,dx =2\left[4z+\frac{z^2}{2}\right]_{-1}^1 =2\bigl((4+1/2)-(-4+1/2)\bigr) =16. $$

Face \(y=1\). The outward normal is \(-\mathbf j\), and \(Q=2+z\). Its contribution is

$$ -\int_0^2\int_{-1}^1(2+z)\,dz\,dx =-2\left[2z+\frac{z^2}{2}\right]_{-1}^1 =-2\bigl((2+1/2)-(-2+1/2)\bigr) =-8. $$

Face \(z=1\). The outward normal is \(+\mathbf k\), and \(R=x-1\). The contribution is \(\int_1^2\int_0^2(x-1)\,dx\,dy=1\cdot[x^2/2-x]_0^2=0\).

Face \(z=-1\). The outward normal is \(-\mathbf k\), and \(R=x+1\). The contribution is \(-\int_1^2\int_0^2(x+1)\,dx\,dy=-[x^2/2+x]_0^2=-4\).

The outward flux is \(1+3+16-8+0-4=8\). Reversing the six contributions gives inward flux \(-1-3-16+8+0+4=-8\). The two totals are negatives, confirming the orientation rule.

Common Sign and Area Pitfalls

A frequent error is to choose the same normal direction on opposite faces. For example, on \(x=a\) the outward normal is \(-\mathbf i\), not \(+\mathbf i\). In the last example, the correct contribution from \(x=0\) is \(+3\), because the component there is \(-y\) and the normal contributes another minus sign:

$$ \mathbf F\cdot(-\mathbf i)=-(x-y)\big|_{x=0}=y, \qquad 2\int_1^2y\,dy=3. $$

If one mistakenly uses \(+\mathbf i\), that face’s contribution becomes \(-3\) instead of \(+3\). The resulting total would be \(8-6=2\), not \(8\), because changing \(+3\) to \(-3\) changes the sum by \(-6\).

Another error is to forget the width of the integration interval in the other coordinate. On the face \(x=2\) in the first example, integrating \(2y\) over \(z\in[-1,1]\) multiplies it by the interval length \(2\). Omitting that factor would produce \(8\) instead of \(16\). For every face, identify both varying coordinates and their bounds before evaluating the integral.

Using the Result

Direct face integration is useful when a box’s six face integrals are simple to evaluate, or when the contribution of each individual face is important. Keeping a six-entry list of signed contributions makes the calculation auditable: record the face, its outward normal, the relevant field component, and the resulting integral. Only after all six entries are computed should they be added.

For inward flux, do not recompute the field or the face areas. Reverse every normal, equivalently negate every outward contribution. This provides both a quick calculation and a reliable sign check.

Check Your Understanding

For each question, keep the face’s outward normal separate from the value of the field component on that face.

  1. What are the outward unit normals on \(x=a\) and \(x=b\)?
  2. Which component of \(\mathbf F=(P,Q,R)\) contributes to the flux on a face \(y=d\)?
  3. Why does the contribution from \(x=a\) have a minus sign in the outward-flux formula?
  4. If the outward flux through a box is \(-7\), what is its inward flux?
  5. For \(\mathbf F=(x,y,z)\) on \([0,1]\times[0,1]\times[0,1]\), set up the six outward face contributions before evaluating them.

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