Topic 16 / 17advanced

Line Integrals, Conservative Fields & Green's Theorem

Compute scalar and vector line integrals; recognize work, circulation, and flux setups; use the Fundamental Theorem of Line Integrals; apply Green's Theorem in circulation and flux forms.

A line integral accumulates a quantity along a curve $C$ in $R^{2}$ or $R^{3}$ . There are two flavors: **scalar** (e.g., mass of a wire) and **vector** (e.g., work done by a force). Conservative fields and Green's Theorem provide powerful shortcuts.

Scalar line integrals

\int_{C} f d s = \int_{a}^{b} f (r (t)) ∣ r^{'} (t) ∣ d t

where

C

is parameterized by

r (t), a \leq t \leq b

. **Independent of orientation** —

d s > 0

always. Application: mass of a wire with linear density

f

Vector line integrals (work, circulation)

\int_{C} F \cdot d r = \int_{a}^{b} F (r (t)) \cdot r^{'} (t) d t = \int_{C} P d x + Q d y + R d z

. **Orientation-dependent**: reverse direction → flip sign. Interpretation: work done by force field along the path.

Key takeaways

Vector form depends on orientation; scalar form does not.
Component form: $\int_{C} P d x + Q d y = \int_{a}^{b} (P x^{'} (t) + Q y^{'} (t)) d t$ .

Fundamental Theorem of Line Integrals

F = \nabla f

and

C

goes from

A

B

, then

\int_{C} F \cdot d r = f (B) - f (A)

. **Path-independent**: only endpoints matter. Closed curve ⇒ integral is 0.

Strategy tips

Test $F$ for conservativity (curl = 0) before parametrizing — saves huge effort.
If conservative, find $f$ and just evaluate $f$ at endpoints.

Green's Theorem — circulation form

For a positively oriented (counterclockwise), simple closed curve

C

bounding a region

D

\oint_{C} P d x + Q d y = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A .

Green's Theorem — flux form

Outward flux across

C

\oint_{C} F \cdot n d s = \oint_{C} P d y - Q d x = \iint_{D} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) d A .

Useful for outward flux through a 2D closed curve.

Area via Green's Theorem

Picking

P, Q

so that

Q_{x} - P_{y} = 1

A = \oint_{C} x d y = - \oint_{C} y d x = \frac{1}{2} \oint_{C} (x d y - y d x)

Choosing direct vs Green's

If the boundary is messy but the interior is simple → Green's. If the integrand is simple along

C

but the partial derivatives explode → direct line integral. If the field is conservative and curve is closed → integral is 0; no work needed.

Worked examples

Example 1

Compute

\int_{C} y z cos x d s

where

C : x = t, y = 3 cos t, z = 3 sin t, 0 \leq t \leq π

1
$r^{'} (t) = ⟨ 1, - 3 sin t, 3 cos t ⟩$ .
$∣ r^{'} (t) ∣ = 1 + 9 sin^{2} t + 9 cos^{2} t = 10$
2
Substitute.
$\int_{0}^{π} (3 cos t) (3 sin t) cos t \cdot 10 d t = 910 \int_{0}^{π} cos^{2} t sin t d t$
3
Substitute $u = cos t$ .
$= 910 [- \frac{c o s ^{3} t}{3}]_{0}^{π} = 910 \cdot \frac{2}{3} = 610$

Answer

610

Example 2

Use Green's Theorem to compute

\oint_{C} x y^{2} d x - x^{2} d y

where

C

consists of the line from

(0, - 3)

(0, 3)

and the right semicircular arc back.

1
Region $D$ : right half disk of radius 3.
2
Curl integrand.
$Q_{x} - P_{y} = - 2 x - 2 x y = - 2 x (1 + y)$
3
Polar (right half): $- π /2 \leq θ \leq π /2$ , $0 \leq r \leq 3$ .
$\iint_{D} - 2 x (1 + y) d A = \int_{- π /2}^{π /2} \int_{0}^{3} - 2 r cos θ (1 + r sin θ) \cdot r d r d θ$
4
Split, use symmetry. The $r sin θ$ term integrates to zero over $θ$ symmetric.
$= - 2 \int_{- π /2}^{π /2} cos θ d θ \int_{0}^{3} r^{2} d r = - 2 \cdot 2 \cdot 9 = - 36$
5
Note: orientation must be CCW; check geometry of the curve to confirm or insert sign.

Answer

- 36

(subject to orientation check).

Example 3

\int_{C} F \cdot d r

for

F = ⟨ e^{y}, x e^{y} + e^{z}, y e^{z} ⟩

C

any curve from

(0, 2, 0)

(4, 0, 3)

1
Test conservativity: curl all components agree (verify $P_{y} = e^{y} = Q_{x}$ , etc.). Conservative.
2
Integrate to find potential. $f_{x} = e^{y} \Rightarrow f = x e^{y} + g (y, z)$ .
3
$f_{y} = x e^{y} + g_{y} = x e^{y} + e^{z} \Rightarrow g_{y} = e^{z} \Rightarrow g = y e^{z} + h (z)$ .
4
$f_{z} = y e^{z} + h^{'} (z) = y e^{z} \Rightarrow h (z) = C$ .
5
$f (x, y, z) = x e^{y} + y e^{z} + C$ .
6
Apply FTLI.
$\int_{C} F \cdot d r = f (4, 0, 3) - f (0, 2, 0) = (4 + 0) - (0 + 2) = 2$

Answer

2

Interactive visualizations

Loading visualization…

t3.14

F = ⟨−y, x⟩ along C: r(t) = ⟨2cos t, 2 sin t⟩; current point (-2.00, 0.00)

\int_{C_{[t_{0}, t]}} F \cdot d r \approx 12.535

\int_{C} F \cdot d r \approx 25.133

Move

t

along the circle to watch the cumulative line integral

\int_{C} F \cdot d r

build up. Field

F = ⟨ - y, x ⟩

Loading visualization…

a0.00

b-1.00

c1.00

d0.00

radius R2.00

\oint_{C} F \cdot d r = \int_{0}^{2 π} (P x^{'} + Q y^{'}) d t \approx 25.133

\iint_{D} (\partial_{x} Q - \partial_{y} P) d A = (2.00) π R^{2} \approx 25.133

Both sides should match — that's Green's theorem. Vary the field and radius to confirm.

Green's Theorem: line integral around the red boundary equals the double integral over the green region. Tune the field and radius to see both sides match.

Formulas in this topic

Scalar line integral

\int_{C} f d s = \int_{a}^{b} f (r (t)) ∣ r^{'} (t) ∣ d t

Mass of a wire if f is density.

Vector line integral

\int_{C} F \cdot d r = \int_{a}^{b} F (r (t)) \cdot r^{'} (t) d t

Work / circulation along a curve.

Vector form (component)

\int_{C} P d x + Q d y = \int_{a}^{b} (P x^{'} + Q y^{'}) d t

Component-form alternative.

Fundamental Theorem of Line Integrals

\int_{C} \nabla f \cdot d r = f (B) - f (A)

Path-independent for conservative F.

Green's Theorem (circulation)

\oint_{C} P d x + Q d y = \iint_{D} (Q_{x} - P_{y}) d A

Closed-curve integral via curl.

Green's Theorem (flux)

\oint_{C} F \cdot n d s = \iint_{D} (P_{x} + Q_{y}) d A

Outward flux via divergence.

Area via Green's

A = \oint_{C} x d y

Pick P = 0, Q = x; gives Q_x - P_y = 1.

Scalar line integrals

Vector line integrals (work, circulation)

Fundamental Theorem of Line Integrals

Green's Theorem — circulation form

Green's Theorem — flux form

Area via Green's Theorem

Choosing direct vs Green's

Worked examples

Interactive visualizations

Formulas in this topic

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