Example 2.4.4: Stokes' theorem for a square

Example 2.4.4

Stokes' theorem for a square

Verify Stokes' theorem for the vector field

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

and the unit square in the $y z$ -plane ( $x = 0$ , $0 \leq y \leq 1$ , $0 \leq z \leq 1$ ), traversed along $C_{1}$ , $C_{2}$ , $C_{3}$ , $C_{4}$ as shown.

Try it yourself first.

Strategy

On each side of the square only one coordinate changes, so $d ℓ$ is $a_{y} d y$ or $a_{z} d z$ . For the surface side we only need the component of the curl normal to the square, $(\nabla \times A)_{x}$ , evaluated on $x = 0$ .

Orientation

The path runs O $\to (0, 1, 0) \to (0, 1, 1) \to (0, 0, 1) \to$ O, i.e. counter-clockwise when the $y z$ -plane is viewed from the positive $x$ axis (along $+ y$ , then $+ z$ , and $a_{y} \times a_{z} = a_{x}$ ). By the right-hand rule $d S = + a_{x} d y d z$ .

Line integral

On the plane $x = 0$ the field is $A = (2 y + 2 y z^{2}) a_{y} + (2 z + 4 y z) a_{z}$ .

$C_{1}$ : $z = 0$ , $y : 0 \to 1$ , $d ℓ = a_{y} d y$ : $\int_{0}^{1} 2 y d y = 1$ ,
$C_{2}$ : $y = 1$ , $z : 0 \to 1$ , $d ℓ = a_{z} d z$ : $\int_{0}^{1} (2 z + 4 z) d z = 3$ ,
$C_{3}$ : $z = 1$ , $y : 1 \to 0$ , $d ℓ = a_{y} d y$ : $\int_{1}^{0} (2 y + 2 y) d y = - [2 y^{2}]_{0}^{1} = - 2$ , and
$C_{4}$ : $y = 0$ , $z : 1 \to 0$ , $d ℓ = a_{z} d z$ : $\int_{1}^{0} 2 z d z = - 1$ .

(On $C_{3}$ and $C_{4}$ the vector $d ℓ = a_{y} d y$ or $a_{z} d z$ automatically points in the direction of travel because $d y < 0$ or $d z < 0$ there, so the limits take care of the sign.) Hence $\oint_{C} A \cdot d ℓ = 1 + 3 - 2 - 1 = 1$ .

Surface integral

Only the $x$ component of the curl is needed:

(\nabla \times A)_{x} = \frac{\partial A _{z}}{\partial y} - \frac{\partial A _{y}}{\partial z} = (4 z - 3 x^{2}) - 4 y z,

which on the surface $x = 0$ equals $4 z (1 - y)$ . (For completeness, $(\nabla \times A)_{y} = - \partial A_{z} / \partial x = 6 x y$ and $(\nabla \times A)_{z} = 0$ , but they do not contribute because $d S ∥ a_{x}$ .)

\int_{S} (\nabla \times A) \cdot d S = \int_{0}^{1} \int_{0}^{1} 4 z (1 - y) d y d z = 4 \cdot \frac{1}{2} \cdot \frac{1}{2} = 1.

Answer

\oint_{C} A \cdot d ℓ = \int_{S} (\nabla \times A) \cdot d S = 1

The positive sign means that the net circulation is counter-clockwise as seen from the $+ x$ axis. Note also that the term $- 3 x^{2} y$ of $A_{z}$ played no role: it vanishes on the plane $x = 0$ , and so does its contribution $- 3 x^{2}$ to $(\nabla \times A)_{x}$ . Figure 2.4.9 shows where the positive sign comes from.

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