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When a line integral depends only on its endpoints: the fundamental theorem for line integrals, the cross-partials test and the hypothesis it needs, and how to recover a potential by partial integration.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state the fundamental theorem for line integrals and use it to replace an integral by a subtraction, test a planar or spatial field for being conservative, say why that test needs a simply connected region and give the field that shows it, and construct a potential function by integrating one component and correcting with the others.
Line integrals, and the observation at the end of the last lesson that the route usually matters: one field gave $\tfrac12$ along a diagonal and $0$ along two sides of the same square. This lesson is about the fields for which that never happens, and about the one-line reason why — they are gradients, and the fundamental theorem of calculus applies.
A field is conservative on a region when the line integral between two points is the same along every path in the region. A potential function for $\mathbf{F}$ is a scalar $f$ with $\nabla f = \mathbf{F}$; physics usually defines it with the opposite sign, so check the convention before quoting one. A region is open when every point has room around it, connected when any two points are joined by a path in it, and simply connected when it also has no holes — every loop can be shrunk to a point without leaving.
The fundamental theorem for line integrals. If $\mathbf{F} = \nabla f$ on a region and $C$ is a piecewise smooth path in it from $A$ to $B$, then
$$\int_C \nabla f \cdot d\mathbf{r} = f(B) - f(A).$$
The proof is one line: substituting the parametrisation makes the integrand $\frac{d}{dt}f(\mathbf{r}(t))$ by the chain rule, and the ordinary fundamental theorem of calculus does the rest. Everything else in this lesson is a consequence.
Three equivalent statements. On an open connected region, these say the same thing:
The third follows from the second because a loop starts and finishes at the same point, so $f(B) - f(A) = 0$.
The cross-partials test. For $\mathbf{F} = \langle P, Q \rangle$ in the plane, a potential requires $P = f_x$ and $Q = f_y$, and Clairaut's theorem then forces
$$P_y = Q_x.$$
That is necessary always. It is sufficient only on a simply connected region — and the hypothesis is not decoration: the standard vortex field has equal cross-partials everywhere it is defined and circulation $2\pi$ round the origin.
In space the same test reads $\nabla \times \mathbf{F} = \mathbf{0}$, which is the first identity of the last lesson used backwards.
Another way: picture
Think of $f$ as height on a landscape and $\mathbf{F} = \nabla f$ as the uphill push. Walking from $A$ to $B$, the total climb is the height at $B$ minus the height at $A$, whichever way you went; a round trip climbs nothing. A field that is not a gradient is a landscape that cannot exist — an Escher staircase where a loop leaves you higher than you started.
Another way: steps
To decide, and then to find the potential:
Take $\mathbf{F} = \langle 2xy + 3,\; x^2 - 4y \rangle$. The cross-partials are $P_y = 2x$ and $Q_x = 2x$: equal, and the plane is simply connected, so a potential exists.
Integrate $P$ with respect to $x$, holding $y$ fixed: $f = x^2y + 3x + g(y)$. The unknown is a function of $y$ alone, because anything depending on $x$ would have shown up in $P$.
Now differentiate with respect to $y$: $f_y = x^2 + g'(y)$. This must equal $Q = x^2 - 4y$, so $g'(y) = -4y$ and $g(y) = -2y^2 + C$.
So $f = x^2y + 3x - 2y^2 + C$. The step that makes the method work is the cancellation of $x^2$: if the $x$-terms had not cancelled, $g'(y)$ would have depended on $x$, which is impossible — and that failure is exactly what the cross-partials test predicts in advance. Running the construction on a non-conservative field does not produce a wrong potential; it produces a contradiction, which is a better outcome.
Quoting the test without the region. $P_y = Q_x$ is necessary everywhere and sufficient only where there are no holes. The vortex $\mathbf{F} = \langle -y, x \rangle / (x^2+y^2)$ passes the test at every point of its domain and is not conservative on the punctured plane.
Writing the constant of integration as a number. Integrating $P$ with respect to $x$ leaves an arbitrary function of $y$, not a constant. A number there loses most of the potentials.
Subtracting the wrong way round. The theorem says $f(B) - f(A)$, finish minus start.
Assuming the physics sign. Many texts define potential energy $U$ by $\mathbf{F} = -\nabla U$. Both conventions are in use, and mixing them in one piece of work reverses every answer. Green, Stokes and the divergence theorem each hold on a region of a particular shape, with a field defined everywhere inside it and a boundary oriented in a particular way. A theorem quoted where one of those fails has not been applied; it has been guessed with, and the answer it gives can be wrong by exactly the amount the missing hypothesis was carrying.
The sentence students carry away is equal cross-partials means conservative, and for most of the fields anybody meets it is true, which is what makes it dangerous. The correct sentence has a clause: equal cross-partials means conservative on a simply connected region.
The vortex field is not a curiosity invented to spoil the rule. It is the magnetic field round a wire, the velocity field of an ideal vortex, and the field whose circulation is what a residue computes in complex analysis. Every one of those applications turns on the circulation being non-zero round a loop enclosing the axis, and a learner who was taught the rule without its clause will compute zero and be confidently wrong.
So the habit worth building is to name the region out loud before applying the test, and to say what would have to be true for it to be simply connected.
Is $\mathbf{F} = \langle y^2, 2xy \rangle$ conservative on the plane? $P_y = 2y$ and $Q_x = 2y$, and the plane has no holes.
The test passes on a simply connected region.
Integrating $P$ with respect to $x$ gives $f = xy^2 + g(y)$; differentiating gives $f_y = 2xy + g'(y)$, which must equal $2xy$, so $g' = 0$ and $f = xy^2$.
The unknown function turned out to be constant.
The work from $(1,1)$ to $(3,2)$ is then $f(3,2) - f(1,1) = 12 - 1 = 11$ along any path whatever. One evaluation replaces an integral that would otherwise need a route to be chosen first.
A potential turns an integral into a subtraction.
Let $\mathbf{F} = \left\langle \dfrac{-y}{x^2+y^2}, \dfrac{x}{x^2+y^2} \right\rangle$, defined everywhere except the origin.
One point is missing from the domain.
A calculation gives $P_y = Q_x = \dfrac{y^2 - x^2}{(x^2+y^2)^2}$ at every point of the domain, so the necessary condition holds throughout.
The cross-partials agree everywhere it is defined.
But the circulation counterclockwise round the unit circle is $2\pi$, not zero, so the field is not conservative. Nothing is wrong with the test; the punctured plane is not simply connected, and the loop cannot be shrunk past the missing point. This single example is why the hypothesis is always stated.
A hole in the region is the whole difference.
Let $\mathbf{F} = \langle yz,\; xz,\; xy \rangle$. Its curl is $\langle x - x,\; y - y,\; z - z \rangle = \mathbf{0}$, and space has no holes.
The three-dimensional test is a zero curl.
Integrating the first component with respect to $x$ gives $f = xyz + g(y,z)$; the unknown now depends on the two variables that were held fixed.
Differentiating with respect to $y$ gives $xz + g_y$, which must equal $xz$, so $g_y = 0$; the same with $z$ gives $g_z = 0$. So $f = xyz$, and the work between any two points is a single subtraction. The pattern is identical to the plane, with one more round of the same step.
Match each fact about a planar field to what it settles. Take $3$ to be a non-zero measured circulation.
| Conservative: every path gives the same work | Not conservative: a loop does not come back to zero | Conservative there, by the cross-partials test | Not enough to conclude; the region has a hole | |
|---|---|---|---|---|
| $\mathbf{F} = \nabla f$ for some $f$ on the whole plane | ||||
| $\oint_C \mathbf{F} \cdot d\mathbf{r} = 3$ for one closed curve | ||||
| $P_y = Q_x$ everywhere on an open disc | ||||
| $P_y = Q_x$ everywhere except at one missing point |
A field has potential $f(x,y) = xy + x$. Fill in the table.
| Value | |
|---|---|
| The potential at the point one, zero | |
| The potential at the point one, one | |
| The potential at the point two, one | |
| The work from one, zero to two, one |
A field has potential $f(x,y) = 2x^{2}y + y^{3}$. How much work does it do along any path from $(1,1)$ to $(2,2)$?
Answer:
A field has potential $f(x,y) = 3x^{2} + 9y^{2}$. How much work does it do along a path from $(1,0)$ to $(0,1)$?
Answer:
Match each fact about a planar field to what it settles. Take $6$ to be a non-zero measured circulation.
| Conservative: every path gives the same work | Not conservative: a loop does not come back to zero | Conservative there, by the cross-partials test | Not enough to conclude; the region has a hole | |
|---|---|---|---|---|
| $\mathbf{F} = \nabla f$ for some $f$ on the whole plane | ||||
| $\oint_C \mathbf{F} \cdot d\mathbf{r} = 6$ for one closed curve | ||||
| $P_y = Q_x$ everywhere on an open disc | ||||
| $P_y = Q_x$ everywhere except at one missing point |
A planar field satisfies $P_y = Q_x$ at every point where it is defined, yet the circulation round one closed loop is measured as $9$. What follows?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A field has potential $f(x,y) = 8xy + x$. Fill in the table.
| Value | |
|---|---|
| The potential at the point one, zero | |
| The potential at the point one, one | |
| The potential at the point two, one | |
| The work from one, zero to two, one |
You can decide whether a field is conservative, name the region the decision depends on, and build a potential when one exists. Next: a theorem that computes the circulation a field does have, by turning a loop into the region it encloses.
10. Your turn: find a potential for a field in space, step 3