Determine Whether the Following Vector Field Is Conservative on ℝ²

Answer

A field $\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}$ is conservative exactly when it is the gradient of some potential function, and on an open, simply connected region that happens precisely when the cross partials agree:

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}\quad\text{throughout the region.}$

Every field below is built from polynomials, exponentials, sines and cosines, so each component is defined and differentiable on all of $\mathbb{R}^{2}$. The domain is open and simply connected, no point has to be excluded, and the test is decisive for all six.

The two fields that fail the test

107. Differentiating $P=-y+e^{x}\sin y$ in $y$ gives $-1+e^{x}\cos y$, while differentiating $Q=(x+2)e^{x}\cos y$ in $x$ gives $(x+3)e^{x}\cos y$. The extra $x$-dependence in $Q$ cannot be produced by $P$, and the constant term $-1$ has nothing to match, so the cross partials differ and $\mathbf{F}$ is not conservative. Choosing the test point $y=0$ makes this visible immediately: $P_{y}(x,0)=-1+e^{x}$ but $Q_{x}(x,0)=(x+3)e^{x}$.

108. Here $P=e^{2x}\sin y$ and $Q=e^{2x}\cos y$, so $P_{y}=e^{2x}\cos y$ while $Q_{x}=2e^{2x}\cos y$. The two components differ by a factor of $2$ in how fast they change in $x$, and the cross partials are equal only where $\cos y=0$ — at isolated curves, not throughout the domain. The field is not conservative.

Not-belonging-to-the-test cases like these are why the comparison has to be made as functions on the whole region. Agreeing at a few sampled points proves nothing.

Recovering a potential when the test passes

Take 109, where $P=6x+5y$ and $Q=5x+4y$. Integrating $P$ in $x$ while treating $y$ as a constant gives

$f(x,y)=\int(6x+5y)\,dx=3x^{2}+5xy+g(y),$

where $g$ is the "constant" of integration in $x$. Differentiating this in $y$ and matching the second component,

$\frac{\partial f}{\partial y}=5x+g'(y)=5x+4y\quad\Longrightarrow\quad g'(y)=4y\quad\Longrightarrow\quad g(y)=2y^{2}+C,$

so $f(x,y)=3x^{2}+5xy+2y^{2}+C$. Any constant works, because a constant has zero gradient; the potential is determined only up to that constant.

The same procedure on 111 gives $f=ye^{x}+x\sin y+C$: integrating $P=ye^{x}+\sin y$ in $x$ produces $ye^{x}+x\sin y+g(y)$, and matching $\partial f/\partial y$ to $Q=e^{x}+x\cos y$ forces $g'(y)=0$.

A useful cross-check

A potential can be verified without recomputing anything: differentiate the candidate. For 106, $f=x^{2}y^{3}$ gives $f_{x}=2xy^{3}=P$ and $f_{y}=3x^{2}y^{2}=Q$, which confirms both the potential and the conservative verdict in one step.