Why does the equality of cross-partial derivatives not guarantee path independence for the vortex field on a punctured plane?

Conditions

• The vector field is defined on a domain excluding the origin.
• The domain is multiply connected (has a hole).
• The cross-partial derivatives are equal wherever defined.

Reasoning, step by step

1. Verify that $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$ for the given field.
2. Identify the singularity at the origin $(0,0)$.
3. Recognize that the domain $\mathbb{R}^2 \setminus \{(0,0)\}$ is not simply connected.
4. Calculate the line integral along two different paths connecting the same endpoints.
5. Observe that the results differ ($\pi$ vs $-\pi$), proving path dependence.
6. Conclude that simple connectivity is a required hypothesis for the theorem.

Example

Traveling from (-1,0) to (1,0) along the upper semicircle yields $-\pi$, while the lower semicircle yields $+\pi$. A full counterclockwise loop yields $2\pi$.

Common misconceptions

• Believing that $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$ always implies path independence.
• Ignoring the domain's topology when applying calculus theorems.
• Thinking that a zero curl everywhere implies a conservative field without checking for holes.

Watch the explanation

Explore next

• What is the corrected statement of the path independence theorem?
• How does simple connectivity differ from multiply connectivity?
• Why does Green's Theorem fail for loops enclosing the origin?