Practice: Conservative Fields, Potentials and Path Independence

(a) One pair of derivatives decides both cases. Compute $\partial P/\partial y$ and $\partial Q/\partial x$. For ${\mathbf{F}}_1=(-y,x)$: $\partial P/\partial y=-1$ and $\partial Q/\partial x=1$. Unequal, so no potential exists anywhere and the integral is path dependent. Searching for a potential would be wasted effort, and the problem is not well posed without a stated path. For ${\mathbf{F}}_2=(2xy,x^2)$: $\partial P/\partial y=2x$ and $\partial Q/\partial x=2x$. Equal, and the domain is the whole plane, which is simply connected, so a potential does exist and is worth finding, since it replaces every path integral with one subtraction. The test costs two derivatives and determines the entire approach. Note that the second conclusion needed the domain as well as the equality; on a region with a hole, equality alone would not have licensed the search. (b) Two paths for ${\mathbf{F}}_1$. Diagonal. Parametrise $\mathbf{r}(t)=(t,t)$, $0\le t\le 1$, so $dx=dy=dt$:

Via $(0,1)$. On the vertical leg $x=0$, so $dx=0$ and $Q=x=0$: contribution $0$. On the horizontal leg $y=1$, so $dy=0$ and $P=-y=-1$:

Total $-1$. The comparison shows the integral is genuinely path dependent: $0$ against $-1$ for the same endpoints. This is not a discrepancy to be resolved. It is the correct behaviour of a field that fails the cross-partial test, and it confirms the prediction made in (a) without integrating anything. "The integral from $(0,0)$ to $(1,1)$" is not a well-defined quantity for this field. (c) ${\mathbf{F}}_2$ by potential. Part (a) recommends finding the potential. Integrate $P=2xy$ with respect to $x$:

where the constant of integration is an unknown function of $y$. Differentiating in $y$ gives $f_y=x^2+g^{\prime }(y)$, which must equal $Q=x^2$, so $g^{\prime }(y)=0$ and $g$ is a constant. Take $f=x^{2}y$. Verify the potential first. $f_x=2xy=P$ ✓ and $f_y=x^2=Q$ ✓. This step is cheap and catches sign errors before they propagate. Evaluate. $f(1,1)-f(0,0)=1-0=1$. How to check it. Integrate directly along a path and confirm agreement. Along the diagonal $(t,t)$:

And since the field is conservative, any other path must give $1$ as well. The route through $(0,1)$ gives $0+{\int }_0^{1}2xdx=1$ ✓. Agreement across paths is itself evidence that the potential was found correctly, because a wrong potential would generally disagree with a direct integration.

(a) The loop integral is $2\pi$. On the unit circle $x=\cos t$, $y=\sin t$, so $x^2+y^2=1$ and the denominators disappear. Also $dx=-\sin tdt$ and $dy=\cos tdt$. Then

so

A conservative field gives $0$ on every closed path, since the endpoints coincide and the integral is a difference of potential values. This gives $2\pi \ne 0$, so the field is not conservative on this domain, contradicting the student's conclusion while their derivatives stand. (b) The flaw is in the implication, not the computation. The cross-partial test runs in one direction unconditionally and in the other only with a hypothesis: - Conservative $\Rightarrow$ equal cross-partials. Always true, by Clairaut applied to the potential.
- Equal cross-partials $\Rightarrow$ conservative. True only on a simply connected domain. A region with no holes. The student used the second direction without checking its hypothesis. This field's domain is the plane minus the origin, since both components are undefined at $(0,0)$. That domain is not simply connected: a loop encircling the origin cannot be shrunk to a point without leaving the region. Why that matters is visible in the proof of the converse, which builds a potential by integrating along paths and needs any two paths between the same endpoints to be deformable into one another. With a hole in the way, that deformation is unavailable, and the construction fails, as the $2\pi$ demonstrates it must. So every derivative in the student's work is correct, and the inference from them is not licensed. Nothing about the arithmetic would have revealed this; only inspecting the domain does. (c) The question is missing a region. Conservativeness is a property of a vector field together with a region, not of a formula. Asking whether this field is conservative, without saying where, is like asking whether a function is increasing without saying on what interval. On the punctured plane: no, as part (a) shows. On the right half-plane : yes. That region excludes the origin and is simply connected, no loop within it encircles anything missing, so equal cross-partials now do imply a potential. One is $f=\arctan (y/x)$, the polar angle, and differentiating confirms $f_x={\displaystyle \frac{-y}{x^2+y^2}}$ and $f_y={\displaystyle \frac{x}{x^2+y^2}}$ ✓. The same formula, two regions, two answers. And the potential makes the earlier failure legible: the polar angle increases by $2\pi$ on a full circuit of the origin, so a single-valued potential cannot exist on any region containing such a loop. The $2\pi$ in part (a) is exactly that increment.

(a) ${\mathbf{F}}_1=(3x^{2}y,x^3+2y)$. Test first: $\partial P/\partial y=3x^2$ and $\partial Q/\partial x=3x^2$. Equal, and the domain is the whole plane, which is simply connected, so a potential exists.

Construct it. Integrating $P$ with respect to $x$, treating $y$ as constant:

where the constant of integration is an unknown function of $y$, not a constant: anything depending only on $y$ differentiates to zero in $x$. Differentiating in $y$ and matching against $Q$:

So $f=x^{3}y+y^2$.

Verify by differentiating back. $f_x=3x^{2}y=P$ ✓ and $f_y=x^3+2y=Q$ ✓. Both components, not one: an error in $g$ shows up only in the second.

(b) ${\mathbf{F}}_2=(y,-x)$. $\partial P/\partial y=1$ and $\partial Q/\partial x=-1$. They differ, so no potential exists and the integral is path dependent. One derivative each settles it, and no amount of searching would produce a potential, because a gradient field with continuous second partials must have equal mixed partials by Clairaut's theorem. A failed test is conclusive; that asymmetry is the practical content of the theorem.

(c) The punctured plane. Both cross-partials equal ${\displaystyle \frac{y^2-x^2}{(x^2+y^2{)}^2}}$ at every point where the field is defined. That establishes only that the field is locally a gradient: it rules nothing in globally, because the sufficiency direction requires the region to be simply connected, and the plane minus a point is not.

What decides it. Integrate around the unit circle, $\mathbf{r}(t)=(\cos t,\sin t)$. The integrand collapses to ${\sin }^{2}t+{\cos }^{2}t=1$, so the closed integral is ${\int }_0^{2\pi }1dt=2\pi$. A conservative field gives zero on every closed path, so this field is not conservative on the punctured plane, despite passing the test everywhere it is defined.

A region where it is. On the right half-plane , which is simply connected, the same formula is conservative with potential $f=\arctan (y/x)$. Nothing about the formula changed. Conservativeness is a property of a field together with a region, and reporting it without naming the region is reporting half a claim.