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The Divergence Theorem · Tutorial 952 of 999

The Divergence Theorem for Rectangular Boxes

See exactly how the six outward face fluxes of a rectangular box combine into the volume integral of the divergence.

Advanced 10 min read

What You'll Learn

  • Express the outward flux through each face of an axis-aligned box as an iterated integral.
  • Track the signs of the outward normals on opposite faces.
  • Apply the one-variable Fundamental Theorem of Calculus to each vector-field component.
  • Derive the divergence theorem for a rectangular box by summing the six face contributions.
  • Check the theorem with complete flux and volume-integral calculations.
  • Identify and correct a common sign error when adding opposite-face fluxes.

Why Rectangular Boxes Make the Theorem Transparent

The Divergence Theorem relates flux across a closed boundary to the integral of divergence throughout the enclosed volume. For a rectangular box whose faces are parallel to the coordinate planes, the relationship can be proved directly: calculate the flux through each face, pair opposite faces, and use the one-variable Fundamental Theorem of Calculus.

This special case reveals the mechanism of the theorem. The flux in the positive \(x\)-direction through the right face is compared with the flux in the negative \(x\)-direction through the left face. Their difference records how the \(x\)-component of the field changes across the box. The same reasoning applies in the \(y\)- and \(z\)-directions.

Scope and assumptions: Let \(\mathbf{F}=(P,Q,R)\) be a vector field whose component functions have continuous first partial derivatives on an open set containing the box. The theorem below concerns the outward flux across the six faces of that axis-aligned box.

Set Up the Box and Its Face Fluxes

Let the box be

$$ B=[a,b]\times[c,d]\times[e,f], \qquad a<b,\quad c<d,\quad e<f. $$

Its boundary consists of the faces \(x=a\), \(x=b\), \(y=c\), \(y=d\), \(z=e\), and \(z=f\). The outward unit normal on each face is constant. For example, it is \(\mathbf{i}\) on \(x=b\) and \(-\mathbf{i}\) on \(x=a\), where \(\mathbf{i},\mathbf{j},\mathbf{k}\) are the coordinate unit vectors.

The flux through an oriented surface \(S\) is the surface integral \(\iint_S \mathbf{F}\cdot\mathbf{n}\,dS\), where \(\mathbf{n}\) is the chosen unit normal. On a coordinate-plane face, the dot product selects one component of \(\mathbf{F}\), and \(dS\) is the product of the two remaining coordinate differentials. The six outward fluxes are therefore

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

The minus signs on the faces \(x=a\), \(y=c\), and \(z=e\) come from their outward normals, which point in the negative coordinate directions. The total outward flux is the sum of these six face fluxes.

Derivation for the Pair of \(x\)-Faces

Add the flux through \(x=b\) to the flux through \(x=a\). Because the second face has outward normal \(-\mathbf{i}\), the result is the integral of the difference between the endpoint values of \(P\):

$$ \Phi_{x=b}+\Phi_{x=a} = \int_c^d\int_e^f \bigl(P(b,y,z)-P(a,y,z)\bigr)\,dz\,dy. $$

For each fixed pair \(y,z\), the function \(x\mapsto P(x,y,z)\) is continuously differentiable on \([a,b]\). The one-variable Fundamental Theorem of Calculus gives

$$ P(b,y,z)-P(a,y,z) = \int_a^b \frac{\partial P}{\partial x}(x,y,z)\,dx. $$

Substitute this identity into the face-flux expression. Since the partial derivative is continuous, the resulting iterated integral can be written over the whole box:

$$ \Phi_{x=b}+\Phi_{x=a} = \int_c^d\int_e^f\int_a^b \frac{\partial P}{\partial x}(x,y,z)\,dx\,dz\,dy = \iiint_B \frac{\partial P}{\partial x}\,dV. $$

The order of the three integrations does not affect the value here: the integrand is continuous on a rectangular box, so the iterated integrals agree. The same argument in the \(y\)- and \(z\)-directions gives the other two paired-face contributions.

1
Pair the \(y\)-faces.
The outward normals are \(\mathbf{j}\) on \(y=d\) and \(-\mathbf{j}\) on \(y=c\). Thus \(Q(x,d,z)-Q(x,c,z)=\int_c^d \frac{\partial Q}{\partial y}(x,y,z)\,dy\), and integrating over \(x,z\) gives the volume integral of \(\frac{\partial Q}{\partial y}\).
2
Pair the \(z\)-faces.
The outward normals are \(\mathbf{k}\) on \(z=f\) and \(-\mathbf{k}\) on \(z=e\). Thus \(R(x,y,f)-R(x,y,e)=\int_e^f \frac{\partial R}{\partial z}(x,y,z)\,dz\), and integrating over \(x,y\) gives the volume integral of \(\frac{\partial R}{\partial z}\).
3
Add the three pairs.
Each opposite-face pair contributes the volume integral of the corresponding component derivative.
Theorem: If \(\mathbf{F}=(P,Q,R)\) has continuous first partial derivatives on an open set containing \(B\), then $$ \iint_{\partial B}\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_B \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \right)dV = \iiint_B \nabla\cdot\mathbf{F}\,dV, $$ where \(\mathbf{n}\) is the outward unit normal and \(\nabla\cdot\mathbf{F}\) denotes the divergence of \(\mathbf{F}\).

This completes the proof for a rectangular box: the left side is the sum of the six face fluxes, and each opposite-face pair equals the integral of one component derivative over the box. Adding those three equalities gives the displayed identity.

Worked Examples

Worked Example: The Identity Field on a Box

Let \(\mathbf{F}=(x,y,z)\) and \(B=[0,2]\times[1,3]\times[-1,1]\). First compute the flux through every face. On \(x=2\), the outward normal is \(\mathbf{i}\), so

$$ \Phi_{x=2}=\int_1^3\int_{-1}^1 2\,dz\,dy =2\cdot 2\cdot 2=8. $$

On \(x=0\), the component \(P=x\) is zero, so \(\Phi_{x=0}=0\). On \(y=3\), the outward normal is \(\mathbf{j}\), and on \(y=1\) it is \(-\mathbf{j}\):

$$ \Phi_{y=3}=\int_0^2\int_{-1}^1 3\,dz\,dx=3\cdot2\cdot2=12, \qquad \Phi_{y=1}=-\int_0^2\int_{-1}^1 1\,dz\,dx=-1\cdot2\cdot2=-4. $$

On \(z=1\), the outward normal is \(\mathbf{k}\); on \(z=-1\), it is \(-\mathbf{k}\):

$$ \Phi_{z=1}=\int_0^2\int_1^3 1\,dy\,dx=1\cdot2\cdot2=4, \qquad \Phi_{z=-1}=-\int_0^2\int_1^3(-1)\,dy\,dx=4. $$

The total outward flux is \(8+0+12-4+4+4=24\). The divergence is \(1+1+1=3\), and the box has volume \(2\cdot2\cdot2=8\). Therefore \(\iiint_B \nabla\cdot\mathbf{F}\,dV=3\cdot8=24\), agreeing with the face-flux sum.

Worked Example: Components Depending on Other Coordinates

Let \(\mathbf{F}=(x^2+yz,\,3y+z,\,x+2z)\) on \(B=[0,1]\times[1,2]\times[0,2]\). On the two \(x\)-faces, direct integration gives

$$ \begin{aligned} \Phi_{x=1} &=\int_1^2\int_0^2(1+yz)\,dz\,dy =2+\left(\int_1^2y\,dy\right)\left(\int_0^2z\,dz\right) =2+\frac{3}{2}\cdot2=5,\\ \Phi_{x=0} &=-\int_1^2\int_0^2 yz\,dz\,dy =-\frac{3}{2}\cdot2=-3. \end{aligned} $$

Their combined contribution is \(5-3=2\). On the \(y\)-faces, integrate \(Q=3y+z\) over \(0\le x\le1\), \(0\le z\le2\):

$$ \Phi_{y=2}=\int_0^1\int_0^2(6+z)\,dz\,dx =12+2=14, \qquad \Phi_{y=1}=-\int_0^1\int_0^2(3+z)\,dz\,dx =-(6+2)=-8. $$

Thus the \(y\)-pair contributes \(14-8=6\). For the \(z\)-faces, \(R=x+2z\), giving

$$ \Phi_{z=2}=\int_0^1\int_1^2(x+4)\,dy\,dx =\frac{1}{2}+4=\frac{9}{2}, \qquad \Phi_{z=0}=-\int_0^1\int_1^2x\,dy\,dx=-\frac{1}{2}. $$

The \(z\)-pair contributes \(9/2-1/2=4\). The total flux is \(2+6+4=12\). Independently, \(\nabla\cdot\mathbf{F}=2x+3+2=2x+5\), so

$$ \iiint_B(2x+5)\,dV = \int_0^1\int_1^2\int_0^2(2x+5)\,dz\,dy\,dx = 2\int_0^1(2x+5)\,dx = 2(1+5)=12. $$

The two calculations agree. In particular, the terms involving \(yz\) on the \(x\)-faces cancel in the signed sum, as the component derivative \(\frac{\partial P}{\partial x}=2x\) predicts.

Worked Example: Checking Nonzero Flux on Every Face

Take \(\mathbf{F}=(2x+z,\,y^2+x,\,4z-xy)\) on \(B=[-1,1]\times[0,2]\times[1,3]\). The \(x\)-face fluxes are

$$ \begin{aligned} \Phi_{x=1} &=\int_0^2\int_1^3(2+z)\,dz\,dy =2\left[2z+\frac{z^2}{2}\right]_1^3 =2\left(\frac{21}{2}-\frac{5}{2}\right)=16,\\ \Phi_{x=-1} &=-\int_0^2\int_1^3(-2+z)\,dz\,dy =-2\left[-2z+\frac{z^2}{2}\right]_1^3=0. \end{aligned} $$

The \(y\)-face fluxes are

$$ \begin{aligned} \Phi_{y=2} &=\int_{-1}^1\int_1^3(4+x)\,dz\,dx =2\int_{-1}^1(4+x)\,dx=2(8)=16,\\ \Phi_{y=0} &=-\int_{-1}^1\int_1^3x\,dz\,dx =-2\int_{-1}^1x\,dx=0. \end{aligned} $$

For the \(z\)-faces, the \(xy\) term integrates to zero because \(\int_{-1}^1x\,dx=0\). Hence

$$ \Phi_{z=3}=\int_{-1}^1\int_0^2(12-xy)\,dy\,dx=12\cdot4=48, \qquad \Phi_{z=1}=-\int_{-1}^1\int_0^2(4-xy)\,dy\,dx=-4\cdot4=-16. $$

The total outward flux is \(16+0+16+0+48-16=64\). To check the volume side, the divergence is

$$ \nabla\cdot\mathbf{F}=2+2y+4=6+2y. $$

The \(x\)- and \(z\)-intervals each have length \(2\), so

$$ \iiint_B(6+2y)\,dV = 2\cdot2\int_0^2(6+2y)\,dy = 4[6y+y^2]_0^2 = 4(12+4)=64. $$

The volume integral and the sum of the six outward fluxes both equal \(64\).

A Common Sign Error

The fluxes through opposite faces must be added with their outward-normal signs, not as unsigned magnitudes. For a concrete example, let \(\mathbf{F}=(x+1,0,0)\) on \(B=[0,2]\times[0,1]\times[0,1]\). The right face \(x=2\) has flux \(3\), since \(P=3\) and its area is \(1\). The left face \(x=0\) has \(P=1\), but its outward normal is \(-\mathbf{i}\), so its flux is \(-1\). The pair contributes \(3-1=2\).

A mistaken calculation that adds the two face values as positive magnitudes gives \(3+1=4\). That is incorrect: it uses the wrong normal on \(x=0\). As a check, \(\nabla\cdot\mathbf{F}=1\) and the box has volume \(2\), so the theorem gives flux \(1\cdot2=2\), matching the signed calculation.

Why the Box Proof Matters

The box calculation is more than a special-purpose formula. It makes the relationship between local change and boundary flux explicit. For example, the \(x\)-component derivative measures the change of \(P\) as one moves from \(x=a\) to \(x=b\); the Fundamental Theorem of Calculus turns that change into the difference between the flux contributions on those two faces. Repeating this in all three coordinate directions accounts for the entire divergence integral.

When applying the result, check the conditions and the signs: the field must have continuous first partial derivatives on a neighborhood of the box, the boundary orientation must be outward, and each negative-coordinate face must carry its negative normal. The edges and corners, where face normals meet, do not alter the surface integral; they have zero area.

Check Your Understanding

Use the face-pairing argument to answer the following questions.

  1. What is the outward unit normal on each of the faces \(x=a\) and \(x=b\)?
  2. Why does the signed sum of the two \(x\)-face fluxes involve \(P(b,y,z)-P(a,y,z)\)?
  3. Which regularity condition lets the one-variable Fundamental Theorem of Calculus apply along each coordinate interval?
  4. For \(\mathbf{F}=(x+1,0,0)\) on \([0,2]\times[0,1]\times[0,1]\), what are the two \(x\)-face fluxes and their signed sum?
  5. In the box theorem, which three component derivatives make up \(\nabla\cdot\mathbf{F}\)?

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