Why the Domain Matters
A planar vector field is described by two component functions, \( \mathbf{F}(x,y)=(P(x,y),Q(x,y)) \). For a continuously differentiable field, the scalar curl is \(Q_x-P_y\). A zero curl is an important test for a potential, but the test does not have the same conclusion on every domain.
On a disk, every radial segment from its center remains inside the disk, and a zero-curl field has a potential. On a punctured plane or an annulus, a loop can wind around the missing region. A field can have zero curl at every point of such a domain and still have a nonzero integral around one of those loops. The distinction is global: it depends on the domain as well as the field.
For a piecewise smooth curve \(C\), the line integral of the field is \( \int_C P\,dx+Q\,dy \). If \(f\) is a potential and \(C\) runs from \(A\) to \(B\), the chain rule gives \( \int_C P\,dx+Q\,dy=f(B)-f(A) \). Thus a potential makes the integral depend only on the endpoints. In particular, its integral around every closed curve is zero.
The Curl Test and Its Domain Condition
If \(P\) and \(Q\) have continuous first partial derivatives and \( \mathbf{F}=\nabla f \), then \(P=f_x\) and \(Q=f_y\). Equality of mixed partial derivatives gives \(P_y=f_{xy}=f_{yx}=Q_x\), so the scalar curl vanishes. This is a necessary condition for a potential. On a general domain, it is not by itself sufficient.
A disk is an example of a simply connected domain: it has no holes, and every loop can be continuously contracted to a point while staying in the domain. The following construction proves directly that zero curl is sufficient on an open disk centered at the origin. It also shows why the disk’s geometry matters.
For \((x,y)\in D\), define \( f(x,y)=\int_0^1 [xP(tx,ty)+yQ(tx,ty)]\,dt \). The segment from \((0,0)\) to \((x,y)\) lies in \(D\), so every field value in this integral is defined.
Differentiate under the integral sign, which is valid because the first partial derivatives are continuous. Then use \(Q_x=P_y\) to obtain \( f_x=\int_0^1[P(tx,ty)+t xP_x(tx,ty)+t yP_y(tx,ty)]\,dt \).
The integrand in the preceding expression is \( \frac{d}{dt}[tP(tx,ty)] \), by the chain rule. Hence \( f_x=[tP(tx,ty)]_0^1=P(x,y) \).
In the same way, \( f_y=\int_0^1[t xQ_x(tx,ty)+Q(tx,ty)+t yQ_y(tx,ty)]\,dt \). Since \(Q_x=P_y\), the integrand is \( \frac{d}{dt}[tQ(tx,ty)] \), and therefore \( f_y=Q(x,y) \).
Both partial derivatives of the constructed function match the field components, so \(f\) is a potential. The radial segments used in the definition are available throughout a disk centered at the origin. This proof establishes the result on that disk, without claiming that the same construction works on a domain with a hole.
Worked Example: A Curl-Free Field on a Disk
Worked Example: A Curl-Free Field on a Disk
Let \(D=\{(x,y):x^2+y^2<4\}\), and let \( \mathbf{F}(x,y)=(2x+y,x+2y) \). Test the field and construct a potential.
Here \(P=2x+y\) and \(Q=x+2y\). Their relevant derivatives are \(P_y=1\) and \(Q_x=1\), so \(Q_x-P_y=1-1=0\) throughout \(D\). The disk theorem applies. A potential is \( f(x,y)=x^2+xy+y^2 \), since \( f_x=2x+y=P \) and \( f_y=x+2y=Q \).
As a path check, take \(C(t)=(t,t)\), \(0\leq t\leq1\). Every point on this path lies in \(D\), because \(x^2+y^2=2t^2\leq2<4\). Along the path, \(P=3t\), \(Q=3t\), and \(dx=dy=dt\). Thus \( \int_C P\,dx+Q\,dy=\int_0^1(3t+3t)\,dt=\int_0^1 6t\,dt=3 \). The endpoints are \((0,0)\) and \((1,1)\), and the potential difference is \( f(1,1)-f(0,0)=(1+1+1)-0=3 \). The integral agrees with the endpoint difference.
A Hole Can Defeat the Curl Test
The punctured plane \( \mathbb{R}^2\setminus\{(0,0)\} \) is not simply connected: a loop can go around the missing origin and cannot contract to a point without crossing it. Consider the field \( \mathbf{F}(x,y)=(-y/(x^2+y^2),x/(x^2+y^2)) \), defined everywhere on this punctured plane.
Worked Example: Zero Curl but No Potential on the Punctured Plane
Worked Example: Zero Curl but No Potential on the Punctured Plane
Write \(s=x^2+y^2\), which is positive on the domain. Then \(P=-y/s\) and \(Q=x/s\). Differentiation gives \( Q_x=(s-2x^2)/s^2=(y^2-x^2)/s^2 \) and \( P_y=(-s+2y^2)/s^2=(y^2-x^2)/s^2 \). Therefore \(Q_x-P_y=0\) at every point of the punctured plane.
Now integrate around the circle of radius \(a>0\), parametrized once counterclockwise by \(x=a\cos t,\ y=a\sin t\), for \(0\leq t\leq2\pi\). Along it, \(P=-\sin t/a\), \(Q=\cos t/a\), \(dx=-a\sin t\,dt\), and \(dy=a\cos t\,dt\). Consequently, \(P\,dx+Q\,dy=(\sin^2t+\cos^2t)\,dt=dt\), and the closed-loop integral is \( \int_0^{2\pi}dt=2\pi \).
If this field had a potential on the punctured plane, its integral around every closed curve would be zero by the chain rule. The computed value \(2\pi\) contradicts that consequence. Thus the field is curl-free but has no potential on this domain.
The nonzero loop integral also demonstrates path dependence: traversing the circle returns to its starting point but accumulates a nonzero integral. The obstruction is not a failure of the local derivative calculation. It is the presence of loops around the missing point.
Annuli: Both Outcomes Are Possible
An annulus centered at the origin is a domain of the form \( \{(x,y):a^2<x^2+y^2<b^2\} \), where \(0<a<b\). Like the punctured plane, it has a hole. A zero-curl test on an annulus is therefore inconclusive by itself: the field may have a potential, or it may fail to have one.
Worked Example: A Curl-Free Field Without a Potential on an Annulus
Worked Example: A Curl-Free Field Without a Potential on an Annulus
Restrict the field \( \mathbf{F}(x,y)=(-y/(x^2+y^2),x/(x^2+y^2)) \) to the annulus \(1<x^2+y^2<9\). The derivative calculation above applies at every point of this annulus, so its curl is zero there.
The unit circle is not in this annulus, but the circle of radius \(2\) is: on that circle \(x^2+y^2=4\), and \(1<4<9\). Using \(x=2\cos t,\ y=2\sin t\), \(0\leq t\leq2\pi\), gives \(P=-\sin t/2\), \(Q=\cos t/2\), \(dx=-2\sin t\,dt\), and \(dy=2\cos t\,dt\). Thus \(P\,dx+Q\,dy=(\sin^2t+\cos^2t)\,dt=dt\), and the integral is \(2\pi\). Since a potential would force this closed-loop integral to be zero, the restricted field has no potential on the annulus.
Worked Example: A Potential on the Same Annulus
Worked Example: A Potential on the Same Annulus
On \(1<x^2+y^2<9\), consider \( \mathbf{G}(x,y)=(2x/(x^2+y^2),2y/(x^2+y^2)) \). Again put \(s=x^2+y^2>0\). Here \(P_y=-4xy/s^2\) and \(Q_x=-4xy/s^2\), so \(Q_x-P_y=0\).
Unlike the angular field, this field has the explicit potential \(g(x,y)=\ln(x^2+y^2)\). It is defined throughout the annulus, and \( g_x=2x/(x^2+y^2)=P \), \( g_y=2y/(x^2+y^2)=Q \). Thus the field is conservative there.
For a direct loop check, use the circle of radius \(2\). Along it, \(P=\cos t\), \(Q=\sin t\), \(dx=-2\sin t\,dt\), and \(dy=2\cos t\,dt\). Hence \(P\,dx+Q\,dy=-2\cos t\sin t\,dt+2\sin t\cos t\,dt=0\), so the loop integral is zero. The potential also has the constant value \(g=\ln4\) on the circle, giving endpoint difference \(\ln4-\ln4=0\), as required.
How to Combine the Tests
When classifying a planar field on one of these domains, keep the logical roles of the tests separate. The curl calculation is local: it checks a derivative relation at each point. A loop integral tests a global consequence: if a potential exists, every closed-loop integral must vanish. A potential formula, when found, verifies conservativeness directly by differentiating it.
| Domain and evidence | Conclusion |
|---|---|
| Open disk centered at the origin; continuous first partial derivatives and zero curl | A potential exists; the radial-segment construction gives one. |
| Punctured plane or annulus; a closed loop has nonzero integral | No potential exists on that domain. |
| Punctured plane or annulus; zero curl alone | No conclusion about existence of a potential follows from that test alone. |
| Any domain; an explicit differentiable function has the field as its gradient | The field has a potential on that domain. |
There is also a useful general loop criterion. If a domain \(D\) is open and path connected, the components \(P\) and \(Q\) are continuous on \(D\), and every piecewise smooth closed curve in \(D\) has integral zero, then integrals from a fixed base point to a variable point are independent of the chosen path. Define \(f(X)\) to be that integral. To see why this gives a potential, change the endpoint \(X\) a small distance \(h\) in the \(x\)-direction. The difference in \(f\) is the integral along that short horizontal segment, so \( [f(x+h,y)-f(x,y)]/h \) is the average of \(P\) along the segment. As \(h\) tends to zero, continuity gives \(f_x=P\); the corresponding vertical-segment argument gives \(f_y=Q\). Thus vanishing integrals on all closed curves are sufficient for a potential, while any one nonzero closed-loop integral rules it out.
A Common Mistake
A frequent error is to treat zero curl as sufficient on every domain. The faulty argument is: “\(Q_x-P_y=0\), therefore the field is conservative.” The punctured-plane example disproves this inference: its curl is zero everywhere in its domain, but its integral around a circle is \(2\pi\), so it cannot have a potential. The correction is to check the domain condition. On a disk, the zero-curl theorem applies; on a punctured plane or annulus, zero curl alone is not decisive, and a potential or loop-integral argument is needed.
Another essential check is that every path and loop used in an argument must lie wholly within the stated domain. A field or potential need not be defined outside it, so an integral along a curve leaving the domain cannot establish path independence there.
Check Your Understanding
Use the domain as well as the derivative calculations when answering these questions.
- Why does zero curl imply the existence of a potential for a continuously differentiable field on an open disk centered at the origin?
- What does a nonzero integral around one closed curve prove about the field on its domain?
- For the angular field \( (-y/(x^2+y^2),x/(x^2+y^2)) \), why does the zero-curl calculation not establish a potential on the punctured plane?
- Give the potential for \( (2x/(x^2+y^2),2y/(x^2+y^2)) \) on the annulus \(1<x^2+y^2<9\), and verify one component by differentiation.
- Why must a path used in a path-integral check remain inside the stated domain?