Example 2.4.5: Stokes' theorem for a circle

Example 2.4.5

Stokes' theorem for a circle

Verify Stokes' theorem for the vector field

A = - y a_{x} + 3 x a_{y}

and the circle of radius $3$ in the $x y$ -plane centred at the origin, traversed counter-clockwise when viewed from the positive $z$ axis, as shown.

Try it yourself first.

Strategy

Parametrise the circle by the azimuth $ϕ$ for the line integral. The curl turns out to be uniform, so its flux through the flat disc is just "curl $\times$ area". As a bonus we check that a different surface with the same rim gives the same flux.

Line integral

Parametrise the circle by the azimuth $ϕ$ : $x = 3 cos ϕ$ , $y = 3 sin ϕ$ , $0 \leq ϕ \leq 2 π$ (increasing $ϕ$ is counter-clockwise seen from $+ z$ ). Then $d x = - 3 sin ϕ d ϕ$ , $d y = 3 cos ϕ d ϕ$ , and

A \cdot d ℓ = - y d x + 3 x d y = (- 3 sin ϕ) (- 3 sin ϕ d ϕ) + 3 (3 cos ϕ) (3 cos ϕ d ϕ) = (9 sin^{2} ϕ + 27 cos^{2} ϕ) d ϕ .

Since $\int_{0}^{2 π} sin^{2} ϕ d ϕ = \int_{0}^{2 π} cos^{2} ϕ d ϕ = π$ ,

\oint_{C} A \cdot d ℓ = 9 π + 27 π = 36 π .

(Alternatively, in cylindrical coordinates $d ℓ = ρ d ϕ a_{ϕ}$ with $ρ = 3$ , and $A_{ϕ} = - A_{x} sin ϕ + A_{y} cos ϕ = ρ (sin^{2} ϕ + 3 cos^{2} ϕ)$ by the transformation of Section 2.3, and the same integral results.)

Surface integral over the disc

The counter-clockwise sense corresponds, by the right-hand rule, to $d S = + a_{z} d S$ on the flat disc $S$ of radius $3$ . The curl is

\nabla \times A = a_{x} \partial / \partial x - y a_{y} \partial / \partial y 3 x a_{z} \partial / \partial z 0

\nabla \times A = [\frac{\partial ( 3 x )}{\partial x} - \frac{\partial ( - y )}{\partial y}] a_{z} = 4 a_{z},

a uniform field. Its flux through the disc is simply $4$ times the area $π \cdot 3^{2}$ :

\int_{S} (\nabla \times A) \cdot d S = 4 \int_{0}^{2 π} \int_{0}^{3} ρ d ρ d ϕ = 4 \cdot 9 π = 36 π .

Answer

\oint_{C} A \cdot d ℓ = \int_{S} (\nabla \times A) \cdot d S = 36 π \approx 113.1

A different surface with the same rim

To illustrate the first consequence of Stokes' theorem, take instead the upper hemisphere $r = 3$ , $0 \leq θ \leq π /2$ , whose rim is the same circle (Figure 2.4.10(b)). With the outward normal ( $d S = a_{r} 9 sin θ d θ d ϕ$ , which agrees with the right-hand rule for the counter-clockwise rim) and $a_{z} \cdot a_{r} = cos θ$ ,

\int_{hemisphere} (4 a_{z}) \cdot d S = 4 \int_{0}^{2 π} \int_{0}^{π /2} cos θ \cdot 9 sin θ d θ d ϕ = 4 \cdot 9 \cdot \frac{1}{2} \cdot 2 π = 36 π,

the same value, as it must be.

Check. Figure 2.4.10(a) shows that the field circulates counter-clockwise, in the chosen direction of travel, so a positive circulation is expected.

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