Conservative Vector Fields — Question 4

Question 4 – Solution

Strategy. The cross-partial test needs a suitable domain; use a closed-loop integral to detect the hole.

Step 1: Cross partials For $(x,y)\ne(0,0)$, $$P_y=\frac{y^2-x^2}{(x^2+y^2)^2}, \qquad Q_x=\frac{y^2-x^2}{(x^2+y^2)^2}.$$ Thus the local equality holds throughout $D$.

Step 2: Test a closed circle Set $\mathbf r(t)=\langle\cos t,\sin t\rangle$, $0\le t\le 2\pi$. Then $$\mathbf F(\mathbf r(t))=\langle-\sin t,\cos t\rangle =\mathbf r'(t),$$ so $$\oint_C\mathbf F\cdot d\mathbf r=\int_0^{2\pi}1\,dt=\boxed{2\pi}.$$ A conservative field must have zero integral around every closed curve. Therefore $$\boxed{\mathbf F\text{ is not conservative on }D}.$$

Step 3: Reconcile The punctured plane is not simply connected: the unit circle encloses the missing origin and cannot contract to a point within $D$. Consequently $P_y=Q_x$ is not sufficient here.