A line integral of a vector field along a curve is a running total. It adds up the field's push in the direction of travel, all along the curve. Physically, that running total is the work done moving through the field. Some fields are conservative: the work done depends only on the endpoints, never on the path taken. That happens exactly when the field is a gradient, \mathbf{F}=\nabla f. Chapter 06 rewrites \mathbf{F}\cdot d\mathbf{r} as integrating a 1-form, and rewrites "conservative" as "exact."
A line integral is a running counter for "push along the way." Walk the curve, and at every step add in how much the field shoves you forward, or holds you back, right then. Work is what that running total means physically: how much energy the field actually spent moving you. Here is an example. Carry a heavy bag up a hill and back down the SAME hill. That trip takes zero net work, if nothing else changes, because you end up exactly where you started, at the same height. A conservative field is any field with that property for every possible path, not just the ones you happen to try.
The line integral \int_C \mathbf{F}\cdot d\mathbf{r} chops the curve C into tiny steps d\mathbf{r}. At each step, it dots that step against the field \mathbf{F} at that point — the piece of push actually along the direction of travel. Then it sums the results. This is the vector-field version of an ordinary integral: summing tiny contributions along a line instead of adding up numbers one by one. Work is exactly this integral, read physically: force dotted with displacement, added up along the whole path. The curl test \nabla\times\mathbf{F}=\mathbf{0} only guarantees a conservative field on a simply connected domain — a region with no holes to route a loop around. Chapter 07 makes this one exception the entire subject of its closing sections. There, a closed-but-not-exact 1-form on the punctured plane shows this test failing for a topological reason, not a computational one.
Theorem
Fundamental Theorem for Line Integrals
also known as: the gradient theorem
Recalled from Calculus III — Multivariable Calculus · The Fundamental Theorem for line integrals — first taught there.
Let C be a smooth curve from point A to point B, parametrized by \mathbf{r}(t) for t \in [a,b] with \mathbf{r}(a) = A, \mathbf{r}(b) = B. If f is differentiable and \nabla f is continuous on an open region containing C, then \int_C \nabla f \cdot d\mathbf{r} = f(B) - f(A).
By the chain rule, \dfrac{d}{dt}f(\mathbf{r}(t)) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t). So \int_C \nabla f \cdot d\mathbf{r} = \int_a^b \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt = \int_a^b \frac{d}{dt}f(\mathbf{r}(t))\, dt = f(\mathbf{r}(b)) - f(\mathbf{r}(a)) = f(B) - f(A), the last step being the ordinary single-variable Fundamental Theorem of Calculus applied to the scalar function t \mapsto f(\mathbf{r}(t)).
Theorem
The 3D Conservative-Field Test
also known as: the curl test
Recalled from Calculus III — Multivariable Calculus · The 3D conservative-field test — first taught there.
Let \mathbf{F}(x,y,z) = P\,\mathbf{i} + Q\,\mathbf{j} + R\,\mathbf{k} where P, Q, R have continuous first partial derivatives on an open, simply connected region D in space. Then \mathbf{F} is conservative on D if and only if \operatorname{curl}\mathbf{F} = \mathbf{0} \quad \text{throughout } D, i.e. all three component equations R_y = Q_z, P_z = R_x, Q_x = P_y hold at once.
(\Rightarrow) If \mathbf{F} = \nabla f, then P=f_x, Q=f_y, R=f_z. Each component of \operatorname{curl}\mathbf{F} is a difference of mixed second partials of f — e.g. the \mathbf{k}-component is f_{yx}-f_{xy} — which vanishes by Clairaut's Theorem since f's second partials are continuous. The same argument kills the other two components, so \operatorname{curl}\mathbf{F}=\mathbf{0}.
(\Leftarrow) The converse needs simple connectedness in space: every closed curve in D must be shrinkable to a point without leaving D (a stronger condition than merely "no holes" — a region built around a solid tube, for instance, is not simply connected even though it has no cavity). Given \operatorname{curl}\mathbf{F}=\mathbf{0} on such a D, Stokes' Theorem (later in this chapter) lets any closed curve C in D be spanned by a surface S entirely inside D, giving \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} = 0 — zero circulation around every closed loop, which is the closed-loop form of path independence.
Zero circulation around a conservative field
Drop the paddle-wheel instrument on this gradient field and drag it around a loop -- net circulation reads zero everywhere you try.
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Nonzero circulation: not conservative
This field has curl -- drop the paddle wheel anywhere and watch it spin, unlike the conservative field above.
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Worked example
Work along a conservative field
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1
Find the work done by \mathbf{F}=\langle 2x, 2y\rangle moving from (0,0) to (1,2), along any path.
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2
Check conservative: \mathbf{F}=\nabla f with f=x^2+y^2, since f_x=2x, f_y=2y. Confirmed.
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3
By the Fundamental Theorem for line integrals, work =f(1,2)-f(0,0).
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4
\boxed{\text{work} = (1+4) - 0 = 5} — no path needed at all.
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Where this goes next
Chapter 6 and 7 of this course
"Conservative" and "exact" are the same idea, in two different vocabularies. Chapter 06 proves this directly. Chapter 07 goes further with its broader Stokes theorem. It shows the curl test is really asking whether a 1-form is closed. Once you phrase it that way, it turns into a purely topological question.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
Find \int_C \mathbf{F}\cdot d\mathbf{r} for \mathbf{F}=(1,0) (constant, pointing right) along \mathbf{r}(t)=(t,\,t), 0\le t\le 3.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\mathbf{r}'(t)=(1,1), so \mathbf{F}\cdot\mathbf{r}'=1(1)+0(1)=1 at every t. Integrating: \int_0^3 1\,dt=3.
Correct answer
Also accepted:
Why
Dot the constant field against the path's derivative first — it should simplify to a constant.
From Calculus III — Multivariable Calculus
Find \int_C \mathbf{F}\cdot d\mathbf{r} for \mathbf{F}(x,y)=(0,\,x) along \mathbf{r}(t)=(\cos t,\,\sin t), 0\le t\le \pi (upper unit semicircle).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\mathbf{F}(\mathbf{r}(t))=(0,\cos t), \mathbf{r}'(t)=(-\sin t,\cos t), so \mathbf{F}\cdot\mathbf{r}'=\cos^2t. \int_0^{\pi}\cos^2t\,dt = \pi/2 \approx 1.5708.
Correct answer
Also accepted:
Why
Use the identity \cos^2t = \frac{1+\cos 2t}{2} to integrate.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
For f(x,y)=x^2+2xy+y^2, find \nabla f evaluated at (1,2), then report the sum of its two components.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
f_x=2x+2y, f_y=2x+2y. At (1,2): f_x=2+4=6, f_y=2+4=6. Sum =6+6=12.
Correct answer
Also accepted:
Why
Both partials of f=(x+y)^2 come out equal by symmetry — compute each and add.
From Calculus III — Multivariable Calculus
Which fact about a paddle wheel dropped anywhere in a gradient field is true?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
A gradient field's paddle wheel doesn't spin at all — this describes a rotational field instead.
Same issue — a gradient field's curl is zero, not a fixed rotation direction.
The curl of a gradient field is identically zero at every point, not a function of distance.
Why
A gradient field \mathbf{F}=\nabla f always has curl zero (because f_{xy}=f_{yx} by Clairaut's Theorem), so a paddle wheel dropped anywhere never spins.
Gradient fields have a special property tied to Clairaut's Theorem, f_{xy}=f_{yx}.
If f_{xy}=f_{yx} always holds, what does that force Q_x-P_y to equal?