Section 2.4 · Examples · 2 of 5

Example 2.4.2: Divergence theorem for a unit cube

Example 2.4.2

Divergence theorem for a unit cube

Verify the divergence theorem for the vector field

A = (x^{2} + 4 y z) a_{x} + (3 y^{2} + 4 x z) a_{y} + (z^{2} + 4 x y) a_{z}

and the unit cube $0 \leq x \leq 1$ , $0 \leq y \leq 1$ , $0 \leq z \leq 1$ shown in the figure. The faces are numbered (i) $x = 1$ , (ii) $x = 0$ , (iii) $y = 1$ , (iv) $y = 0$ , (v) $z = 1$ , (vi) $z = 0$ .

Try it yourself first.

Strategy

Both sides are easy in Cartesian coordinates: the divergence is a polynomial, and on each face of the cube $d S$ is $\pm$ a Cartesian unit vector, so only one component of $A$ crosses each face.

Volume integral

The divergence is

\nabla \cdot A = \frac{\partial ( x ^{2} + 4 y z )}{\partial x} + \frac{\partial ( 3 y ^{2} + 4 x z )}{\partial y} + \frac{\partial ( z ^{2} + 4 x y )}{\partial z} = 2 x + 6 y + 2 z .

Over the cube, using $\int_{0}^{1} x d x = \frac{1}{2}$ and $\int_{0}^{1} d x = 1$ (and the same for $y$ and $z$ ),

\int_{V} (\nabla \cdot A) d V = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (2 x + 6 y + 2 z) \times d x d y d z = 2 \cdot \frac{1}{2} + 6 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} = 5.

Surface integral

On each face $d S$ points outwards along $\pm$ one Cartesian unit vector, so only one component of $A$ contributes. The mixed terms $4 y z$ , $4 x z$ , $4 x y$ average to $4 \cdot \frac{1}{2} \cdot \frac{1}{2} = 1$ over a face:

(i) $x = 1$ , $d S = + a_{x} d y d z$ : $\int_{0}^{1} \int_{0}^{1} A_{x}_{x = 1} d y d z = \int_{0}^{1} \int_{0}^{1} (1 + 4 y z) d y d z = 1 + 1 = 2$ ,
(ii) $x = 0$ , $d S = - a_{x} d y d z$ : $- \int_{0}^{1} \int_{0}^{1} A_{x}_{x = 0} d y d z = - \int_{0}^{1} \int_{0}^{1} 4 y z d y d z = - 1$ ,
(iii) $y = 1$ , $d S = + a_{y} d x d z$ : $\int_{0}^{1} \int_{0}^{1} A_{y}_{y = 1} d x d z = \int_{0}^{1} \int_{0}^{1} (3 + 4 x z) d x d z = 3 + 1 = 4$ ,
(iv) $y = 0$ , $d S = - a_{y} d x d z$ : $- \int_{0}^{1} \int_{0}^{1} A_{y}_{y = 0} d x d z = - \int_{0}^{1} \int_{0}^{1} 4 x z d x d z = - 1$ ,
(v) $z = 1$ , $d S = + a_{z} d x d y$ : $\int_{0}^{1} \int_{0}^{1} A_{z}_{z = 1} d x d y = \int_{0}^{1} \int_{0}^{1} (1 + 4 x y) d x d y = 1 + 1 = 2$ , and
(vi) $z = 0$ , $d S = - a_{z} d x d y$ : $- \int_{0}^{1} \int_{0}^{1} A_{z}_{z = 0} d x d y = - \int_{0}^{1} \int_{0}^{1} 4 x y d x d y = - 1$ .

Adding the six faces, $\oint_{S} A \cdot d S = 2 - 1 + 4 - 1 + 2 - 1 = 5$ (Figure 2.4.5).

Answer

\oint_{S} A \cdot d S = \int_{V} (\nabla \cdot A) d V = 5

Notice how the two sides correspond term by term. The $2 x$ , $6 y$ and $2 z$ parts of the divergence produce the net fluxes $2 - 1 = 1$ , $4 - 1 = 3$ and $2 - 1 = 1$ of the three pairs of opposite faces. The mixed terms $4 y z$ , $4 x z$ , $4 x y$ do not change along their own direction: each of them carries the same flux $1$ into the cube through one face and out through the opposite face, so it contributes to every single face but not to the total, just as it contributes nothing to the divergence. Six surface integrals were replaced by one volume integral.

Check. Both numbers are equal. The three faces through the origin have negative flux (the field enters there), the three faces through the corner $(1, 1, 1)$ have positive flux, and the outflow wins because the divergence $2 x + 6 y + 2 z$ is positive everywhere inside the cube.

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