Why does reversing the path direction change the sign of the line integral of the second kind?

Conditions

• The path $L$ is traversed in the reverse direction.
• $\vec{T}$ is the unit tangent vector for the forward direction.
• $ds$ is the scalar arc length element.

Reasoning, step by step

1. Consider the unit tangent vector $\vec{T}$ for the forward direction.
2. Determine the unit tangent vector for the reverse direction, which is $-\vec{T}$.
3. Calculate the new direction cosines: $\cos(\pi - \alpha) = -\cos \alpha$ and $\cos(\pi - \beta) = -\cos \beta$.
4. Substitute these into the integrand: $P(-\cos \alpha) + Q(-\cos \beta) = -(P \cos \alpha + Q \cos \beta)$.
5. Note that $ds$ is invariant under reversal (it is a scalar measure of length).
6. Conclude that the integral value is negated: $\int_{-L} \vec{F} \cdot d\vec{r} = -\int_L \vec{F} \cdot d\vec{r}$.
7. Interpret this as the change in the projection of the field onto the direction of travel.

Example

The script states: 'Reversing traversal replaces the unit tangent by $-T$, negating every direction cosine... The vector line integral therefore changes sign. Arc length $ds$ stays nonnegative; the right-hand integrand changes because it contains the oriented tangent.'

Common misconceptions

• Thinking that $ds$ changes sign when the path is reversed.
• Believing that the magnitude of the integral changes, rather than just the sign.
• Confusing the line integral of the second kind with the scalar line integral (first kind), which is orientation-independent.

Watch the explanation

Explore next

• Is the scalar line integral (first kind) affected by path orientation?
• How does this relate to the concept of conservative vector fields?
• What is the physical interpretation of the sign change in terms of work?