Determining Whether \(\mathbf{F}\) Is a Conservative Vector Field

(a) \(\mathbf{F}=2xy\mathbf{i}+(x^{2}+1)\mathbf{j}\) is a conservative vector field on the entire \(xy\)-plane since \[\begin{equation*} \dfrac{\partial P}{\partial y}=\dfrac{\partial }{\partial y}(2xy)=2x\qquad \hbox{and}\qquad \dfrac{\partial Q}{\partial x}=\dfrac{\partial }{\partial x}(x^{2}+1)=2x \end{equation*}\]

are equal for any choice of \((x,y)\).

(b) The vector field \(\mathbf{F}=\dfrac{x}{y^{2}}\mathbf{i}-\dfrac{x^{2}}{y^{3}}\mathbf{j}\) is conservative on any connected region not containing points on the \(x\)-axis \((y=0)\) since \[\begin{equation*} \dfrac{\partial P}{\partial y}=\dfrac{\partial }{\partial y}\dfrac{x}{y^{2}} =-\dfrac{2x}{y^{3}} \qquad \hbox{and}\qquad \dfrac{\partial Q}{\partial x}=\dfrac{\partial }{\partial x}\left( -\dfrac{x^{2}}{y^{3}}\right) =-\dfrac{2x}{y^{3}} \end{equation*}\]

are equal, provided \(y\neq 0\). Because \(\mathbf{F}=\dfrac{x}{y^{2}}\mathbf{i}-\dfrac{x^{2}}{y^{3}}\mathbf{j}\) is conservative for \(y\neq 0,\) the line integral \(\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}\left[ \dfrac{x}{y^{2}}dx-\dfrac{x^{2}}{y^{3}}dy\right]\) is independent of the path in any simply connected region not containing points on the \(x\)-axis.