Example 2.3.15: A closed line integral in cylindrical coordinates

Path and line element

The circle (in the plane $z = 0$ ) is $ρ = 3$ . Along it $d ρ = d z = 0$ , so from Eq. (2.3.24)

d ℓ = ρ d ϕ a_{ϕ} = 3 d ϕ a_{ϕ},

and travelling “along $a_{ϕ}$ ” means that $ϕ$ increases, from $0$ to $2 π$ (Figure 2.3.15).

Integrand

The field has only a $ϕ$ -component, which is tangent to the path. With $a_{ϕ} \cdot a_{ϕ} = 1$ and $ρ = 3$ ,

(4 ρ cos \frac{ϕ}{4} a_{ϕ}) \cdot (3 d ϕ a_{ϕ}) = 4 \cdot 3 \cdot 3 cos \frac{ϕ}{4} d ϕ = 36 cos \frac{ϕ}{4} d ϕ .

Integration

\oint_{C} A \cdot d ℓ = 36 \int_{0}^{2 π} cos \frac{ϕ}{4} d ϕ = 36 [4 sin \frac{ϕ}{4}]_{0}^{2 π} = 144 (sin \frac{π}{2} - sin 0),

Remark

Here the dot product is simple because $a_{ϕ}$ appears in both factors at the same point of the path, so we never needed to know how $a_{ϕ}$ is oriented in space. Note also that $cos (ϕ /4)$ does not repeat after $2 π$ : the given field takes the value $4 ρ$ at $ϕ = 0$ but approaches $0$ as $ϕ \to 2 π$ , so it is discontinuous across the half-plane $ϕ = 0$ . The integral is therefore evaluated with the standard range $0 \leq ϕ < 2 π$ , starting and ending the loop on that half-plane, as stated.