What is the corrected condition for the path independence of line integrals?

Conditions

• The domain $D$ is open and connected.
• The functions $P$ and $Q$ have continuous first partial derivatives on $D$.
• $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$ for all points in $D$.
• $D$ is simply connected.

Reasoning, step by step

1. State the standard condition: $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$.
2. Identify the missing topological constraint: simple connectivity.
3. Define simple connectivity: every closed loop in $D$ can be contracted to a point within $D$.
4. Combine these to form the complete theorem.
5. Explain that without simple connectivity, the equality of partial derivatives is insufficient.

Example

A solid disk is simply connected, so the theorem applies. An annulus (ring with a hole) is multiply connected, so the theorem may fail even if partial derivatives match.

Common misconceptions

• Thinking that simple connectivity is only a technicality with no practical impact.
• Believing that the condition $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$ is sufficient on its own.
• Confusing connectedness with simple connectedness.

Watch the explanation

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