Proving the FTLI, line by line
Let C be parametrised by \mathbf{r}(t),
a \le t \le b, and let
\mathbf{F} = \nabla f. We follow the integral straight to its
endpoints.
Step 1 — reduce to an ordinary integral. Parametrising the vector line
integral (as on the previous page) gives
\int_C \nabla f \cdot d\mathbf{r} = \int_a^b \nabla f\big(\mathbf{r}(t)\big) \cdot \mathbf{r}'(t)\, dt.
Step 2 — recognise the chain rule. By the
multivariable chain rule,
the derivative of the composite f(\mathbf{r}(t)) is exactly that
dot product:
\frac{d}{dt}\, f\big(\mathbf{r}(t)\big) = f_x\, x'(t) + f_y\, y'(t) = \nabla f\big(\mathbf{r}(t)\big) \cdot \mathbf{r}'(t).
Step 3 — substitute. The integrand is a perfect derivative in
t:
\int_a^b \nabla f\big(\mathbf{r}(t)\big)\cdot\mathbf{r}'(t)\, dt = \int_a^b \frac{d}{dt}\, f\big(\mathbf{r}(t)\big)\, dt.
Step 4 — apply the ordinary fundamental theorem. The integral of a
derivative is the net change of the antiderivative:
\int_a^b \frac{d}{dt}\, f\big(\mathbf{r}(t)\big)\, dt = f\big(\mathbf{r}(b)\big) - f\big(\mathbf{r}(a)\big).
Step 5 — read off the result. Since
\mathbf{r}(b) is the end and
\mathbf{r}(a) the start,
\int_C \nabla f \cdot d\mathbf{r} = f(\text{end}) - f(\text{start}).
The parametrisation has vanished entirely. Any two paths sharing the same endpoints give the
same answer — the integral is path-independent — and around a
closed loop, where end equals start, the work is 0.
The test, and reconstructing the potential
Given \mathbf{F} = (P, Q), how do we know a potential exists — and
how do we find it? In a simply connected region (one with no holes) there is
a clean test.
Step 1 — apply the cross-partial test. A conservative field must satisfy
\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x},
and on a simply connected region this condition is also sufficient. Take
\mathbf{F} = (2xy,\; x^2 + 1). Then
P_y = 2x and Q_x = 2x — they agree, so a
potential exists.
Step 2 — integrate P in x.
We need f_x = P = 2xy, so integrate treating
y as a constant:
f(x, y) = \int 2xy \, dx = x^2 y + g(y),
where the "constant" of integration g(y) may depend on
y (it vanishes under \partial_x).
Step 3 — match the other partial. Differentiate this
f by y and set it equal to
Q:
f_y = x^2 + g'(y) \;=\; Q = x^2 + 1 \quad\Rightarrow\quad g'(y) = 1.
Step 4 — solve for the leftover. Integrating,
g(y) = y + C, so
f(x, y) = x^2 y + y + C.
Step 5 — check. Indeed
\nabla f = (2xy,\; x^2 + 1) = \mathbf{F}. The potential is
reconstructed, and now any line integral of \mathbf{F} is just a
difference of f-values.
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Equivalences. On a connected open region, these are equivalent:
\mathbf{F} = \nabla f for some potential
f; the line integral
\int_C \mathbf{F}\cdot d\mathbf{r} is path-independent; and
\oint_C \mathbf{F}\cdot d\mathbf{r} = 0 around every closed loop.
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Fundamental theorem for line integrals. If
\mathbf{F} = \nabla f, then
\int_C \mathbf{F}\cdot d\mathbf{r} = f(\text{end}) - f(\text{start}).
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The cross-partial test. If
\partial P/\partial y = \partial Q/\partial x throughout a
simply connected region, then \mathbf{F} = (P, Q) is
conservative — and the potential is recovered by integrating
P in x and matching against
Q.
"Conservative" is borrowed from physics: a conservative force conserves
energy. Define potential energy
U = -f; then the work the field does moving a particle from
A to B is
f(B) - f(A) = U(A) - U(B) — the drop in potential energy,
turned into kinetic energy. Total energy stays fixed. Gravity and electrostatics are
conservative, which is why "potential energy" and "voltage" are well-defined numbers at
each point. Friction is not conservative: drag a box around a loop and you never
get the work back.
The "simply connected" caveat is not pedantry. Consider the swirl
\mathbf{F} = \left( \frac{-y}{x^2 + y^2},\; \frac{x}{x^2 + y^2} \right).
A direct computation gives
P_y = Q_x = \dfrac{y^2 - x^2}{(x^2+y^2)^2} everywhere it is
defined — the test passes. Yet integrating once counterclockwise around the unit
circle yields
\oint_C \mathbf{F}\cdot d\mathbf{r} = 2\pi \ne 0,
so the field is not conservative. What went wrong? The field blows up at
the origin, and the disk has a hole there: the region is not simply connected, and
the test's sufficiency fails. This single example is the seed of de Rham cohomology and the
winding number — the hole is "felt" by the integral even though the test is locally blind to
it.