Line Integrals
Definition Let E be a subset in Rm. Avector field on E is a vector (or vector-valued) function F : E → Rn defined by
F (x) = (F1(x), F2(x), . . . , Fn(x)) ∈ Rn for each x = (x1, x2, . . . , xm) ∈ E.
Definition Let C be a plane curve defined by the parametric equations r(t) = x(t) , y(t), t ∈ [a, b]. Then
C is called a smooth curve if r′(t) is continuous and r′(t) ̸= 0 for all t ∈ [a, b].
C is called a piecewise smooth curve if there exists a partition P = {a = t0 < t1 < · · · < tn = b}
such that r′(t) ̸= 0 is continuous for all t ∈ (ti−1, ti) and lim
t→t±i
r′(t) exists for each 1 ≤ i ≤ n.
Definition Let C be a smooth plane curve given by the vector function r(t) = x(t) , y(t), t ∈ [a, b], let f be a function defined on C and let s(t) =
Z t a
|r′(u)| du. Thenthe line integral of f along C is defined by
Z
C
f (x, y) ds = Z b
a
f (r(t)) |r′(t)| dt = lim
n→∞
n
X
i=1
f (x∗i, yi∗) ∆si if this limit exists,
where ds = |r′(t)| dt =
Êdx dt
2
+
dy dt
2
dt, Pi∗(x∗i, yi∗) = r(t∗i) ∈ úPi−1Pi and ∆si is the length of the subarc úPi−1Pi from Pi−1= r(ti−1) to Pi = r(ti).
In general, if C is a (piecewise) smooth curve in Rm given by the vector function r(t) = (x1(t), . . . , xm(t)), t ∈ [a, b] and
if f is a function (scalar field) defined on C, then the line integral of f along C is defined by
Z
C
f (x1, x2, . . . , xm) ds = Z b
a
f (r(t)) |r′(t)| dt
= lim
n→∞
n
X
i=1
f (x∗1i, x∗2i, . . . , x∗mi) ∆si if this limit exists.
if F = (F1, . . . , Fm) is a continuous vector field defined on C, then the line integral of F along C is defined by
Z
C
F (r) · dr = Z b
a
F (r(t)) · r′(t) dt,
where F (r(t)) · r′(t) denotes the inner product of F (r(t)), r′(t) ∈ Rm. Remarks Let C (piecewise) smooth curve defined by r(t), t ∈ [a, b].
(a) If ϕ : [c, d] → [a, b] is a continuously differentiable, orientation preserving onto map such that ϕ′(u) > 0, ϕ(c) = a, ϕ(d) = b, then C is given by the vector function R(u) defined by
C : R(u) = r(ϕ(u)), u ∈ [c, d], and since
Z
C
F (R) · dR = Z d
c
F (R(u)) · R′(u) du
= Z d
c
[F (r(ϕ(u))) · r′(ϕ(u))] ϕ′(u) du
Set t = ϕ(u) =⇒ dt = ϕ′(u) du, ϕ(c) = a, ϕ(d) = b
= Z b
a
F (r(t)) · r′(t) dt = Z
C
F (r) · dr
the line integral is left invariant by every orientation-preserving change of parameter.
(b) For u ∈ [a, b], let ϕ(u) = a + b − u and let −C be a curve defined by
−C : R(u) = r(ϕ(u)) = r(a + b − u) for u ∈ [a, b].
Since ϕ : [a, b] → [a, b] is onto, ϕ′(u) = −1, ϕ(a) = b and ϕ(b) = a, the curve −C denotes the same curve traversed in the opposite direction and
Z
−C
F (R) · dR = Z b
a
F (R(u)) · R′(u) du
= Z b
a
F (r(ϕ(u))) · r′(ϕ(u)) ϕ′(u) du
Set t = ϕ(u) =⇒ dt = ϕ′(u) du, ϕ(a) = b, ϕ(b) = a
= Z a
b
F (r(t)) · r′(t) dt = − Z b
a
F (r(t)) · r′(t) dt =− Z
C
F (r) · dr
Examples
1. Evaluate Z
C
(2 + x2y) ds, where C is the upper half of the unit circle x2+ y2 = 1.
2. The center of mass of the wire C with density function ρ(x, y), (x, y) ∈ C, is located at the point
¯ x = 1
m Z
C
xρ(x, y) ds y =¯ 1 m
Z
C
yρ(x, y) ds, where m = Z
C
ρ(x, y) ds = total mass of C.
3. Evaluate Z
C1
y2dx + x dy, where C1 is the line segment from (−5, −3) to (0, 2), and evaluate Z
C2
y2dx + x dy, where C2 is the arc of the parabola x = 4 − y2 from (−5, −3) to (0, 2).
4. Evaluate Z
C
y sin z ds, where C is the circular helix given by the equations x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2π.
5. Find the work done Z
C
F (r) · dr by the force field F (x, y) = x2i − xy j in moving a particle along the quarter-circle C : r(t) = cos t i + sin t j, 0 ≤ t ≤ π/2.
Definition Let E be a subset of Rn and let F = (F1, F2, . . . , Fn) : E → Rn be a vector field defined on E. Then F is called a conservative vector field if there exists a function f : Rn → R such that
F (x) = ∇f (x) for all x ∈ E.
Note that if F is a continuously differentiable conservative vector field, then f has continuous 2nd order partial derivatives fxixj = fxjxi and
∂Fi
∂xj = fxixj = fxjxi = ∂Fj
∂xi =⇒ ∂Fi
∂xj = ∂Fj
∂xi for each 1 ≤ i, j ≤ n (integrability conditions) In this situation, the function f is called a potential function for F.
Definition Let F be a continuous vector field defined on D, and let C1 and C2 be paths in D having the same initial points and the same terminal points. Then the line integral
Z
C
F · dr is called independent of pathif
Z
C1
F · dr = Z
C2
F · dr
Fundamental Theorem for Line Integrals Let U be an open subset of Rm, let f : U → R be a continuously differentiable scalar field (function) and let C : r = r(u), u ∈ [a, b] be a (piecewise) smooth curve that begins at p = r(a) and ends at q = r(b). Then the line integral of (a conservative field) ∇f along C satisfies that
Z
C
∇f (r) · dr = f (r(b)) − f (r(a)) = f (q) − f (p) and is independent of the choice of paths in U joining from p to q.
Proof
Case 1: If C is smooth, then Z
C
∇f (r) · dr = Z b
a
∇f (r(u)) · r′(u) du = Z b
a
d
duf (r(u))
du = f (r(b)) − f (r(a)).
Case 2: If C a (piecewise) smooth and {a = a0 < a1 < · · · < an−1 < an = b} is a partition of [a, b] such that
C =
n
[
i=1
Ci where Ci = {r(t) | t ∈ [ai−1, ai]} is smooth for 1 ≤ i ≤ n,
then Z
C
∇f (r) · dr = Z
Sn i=1Ci
∇f (r) · dr =
n
X
i=1
Z
Ci
∇f (r) · dr =
n
X
i=1
Z ai
ai−1
∇f (r(u)) · r′(u) du
Case 1
=
n
X
i=1
f (r(ai)) − f (r(ai−1)) = f (r(b)) − f (r(a)).
Remarks
(a) This is a generalization of the Fundamental Theorem of Calculus since if g : [a, b] → R is a continuous function with an anti-derivative G(x) on (a, b) and if C is the line segment given by r(x) = x, x ∈ [a, b] ⊂ R, then
G(b) − G(a) = Z
C
∇G(r) · dr = Z b
a
d
dxG(x) dx = Z b
a
g(x) dx
(b) If C : r = r(u), u ∈ [a, b] is a (piecewise) smooth closed curve and if f is continuously differentiable on an open set U containing C, then
Z
C
∇f (r) · dr = f (r(b)) − f (r(a)) = 0 since r(b) = r(a).
This implies that if F = ∇f is conservative and continuous in U, the line integral Z
C
F (r) · dr = 0 along any (piecewise) smooth closed curve C in U.
Examples
1. Find the work done by the gravitational field F (X) = −mM G
|X|3 X, X = (x, y, z) ∈ R3, in moving a particle with mass m from the point (3, 4, 12) to the point (2, 2, 0) along a piecewise-smooth curve C.
2. Determine whether or not the given vector field is conservative.
• F (x, y) = (x − y) i + (x − 2) j.
• F (x, y) = (3 + 2xy) i + (x2− 3y2) j.
Definition A continuous curve C : r = r(u) u ∈ [a, b] is called simple if r(u) ̸= r(t) for all a ≤ t < u < b ⇐⇒ for all u ̸= t ∈ [a, b).
A Jordan curve is a plane curve that is both closed and simple.
Definition A region D is called (path) connected if any two points in D can be joined by a path that lies in D, i.e. D is (path) connected if for all p, q ∈ D, there exists a continuous map r : [a, b] → D from [a, b] into D such that r(a) = p and r(b) = q.
Definition Let D be a region in the plane. Then D is called simply-connected if every simple closed curve in D encloses only points that are in D.
RemarkIf F = ∇f is conservative and continuous in D, and if C1 and C2 are paths in D having the same initial points and the same terminal points. Then C1 ∪ (−C2) is a closed curve in D, and
0 =
Z
C1∪(−C2)
F · dr = Z
C1
F · dr + Z
−C2
F · dr = Z
C1
F · dr − Z
C2
F · dr
=⇒
Z
C1
F · dr = Z
C2
F · dr (independent of path)
Green’s TheoremLet F = (P, Q) be a vector field on an simply-connected region Ω. Suppose that P and Q have continuous first-order partial derivatives on an open set that contains Ω, then
Z Z
Ω
∂Q
∂x − ∂P
∂y
dx dy = I
C
P dx + Qdy,
where the boundary C = ∂Ω is oriented in the positive direction such that Ω is on the left- hand-side when traversing along C, and
I
C
denotes the line integral along C in the positively orientation.
Proof Suppose that Ω is an elementary region given by
{(x, y) | a ≤ x ≤ b, ϕ1(x) ≤ y ≤ ϕ2(x)} or {(x, y) | ψ1(y) ≤ x ≤ ψ2(y), c ≤ x ≤ d},
such that the boundary ∂Ω = C = C1 ∪ C2 = C3∪ C4, where
C1 = {(x, y) | y = ϕ1(x), a ≤ x ≤ b}, C2 = {(x, y) | y = ϕ2(x), a ≤ x ≤ b}, C3 = {(x, y) | x = ψ1(y), c ≤ y ≤ d}, C4 = {(x, y) | x = ψ2(y), c ≤ y ≤ d}.
Since Z Z
Ω
∂Q
∂x dx dy = Z d
c
Z ψ2(y) ψ1(y)
∂Q
∂x dx dy = Z d
c
Q(ψ2(y), y) dy − Z d
c
Q(ψ1(y), y) dy Z Z
Ω
−∂P
∂y dy dx = − Z b
a
Z ϕ2(x) ϕ1(x)
∂P
∂y dy dx = Z b
a
P (x, ϕ1(x)) dx − Z b
a
P (x, ϕ2(x)) dx and
I
C
Q dy = Z
C4∪ −C3
Q dy = Z
C4
Q dy − Z
C3
Q dy = Z d
c
Q(ψ2(y), y) dy − Z d
c
Q(ψ1(y), y) dy I
C
P dx = Z
C1∪ −C2 P dx = Z
C1
P dx − Z
C2
P dx = Z b
a
P (x, ϕ1(x)) dx − Z b
a
P (x, ϕ2(x)) dx
we have
Z Z
Ω
∂Q
∂x − ∂P
∂y
dx dy = I
C
P dx + Qdy.
RemarkSuppose that D1 and D2 are simply-connected regions with boundaries ∂D1 = C3∪ C1 and ∂D2 = C2∪ (−C3) respectively and suppose that D = D1∪ D2 has the boundary C = ∂D = C1∪ C2.
Since Z Z
D
∂Q
∂x − ∂P
∂y
dx =
Z Z
D1∪D2
∂Q
∂x − ∂P
∂y
dx =
Z Z
D1
∂Q
∂x − ∂P
∂y
dx +
Z Z
D2
∂Q
∂x −∂P
∂y
dx I
C
P dx + Q dy = I
C1∪C2∪C3∪(−C3)
P dx + Q dy = I
C1∪C3
P dx + Q dy + I
C2∪(−C3)
P dx + Q dy and since D1, D2 are simple regions as in the preceding proof, we have
Z Z
D1
∂Q
∂x − ∂P
∂y
dx = I
C1∪C3
P dx + Qdy Z Z
D2
∂Q
∂x − ∂P
∂y
dx = I
C2∪(−C3)
P dx + Qdy This proves the Green’s Theorem on a more general region D
Z Z
D
∂Q
∂x −∂P
∂y
dx dy = I
C1∪C2
P dx + Qdy = I
∂D
P dx + Qdy.
TheoremLet F = (P, Q) be a vector field on a simply-connected region D. Suppose that P and Q have continuous first-order partial derivatives and
∂P
∂y = ∂Q
∂x throughout D (integrability condition)
Then F is conservative, i.e. there is a continuously differentiable function f : D → R such that F = ∇f on D.
Proof Fix a point p = (a, b) ∈ D. For any (s, t) ∈ D, let f : D → R be defined by f (s, t) =
Z
C
P dx + Q dy, where C is a piecewise smooth path in D from p to (s, t).
This is well defined since if C1 is another piecewise smooth path in D from p to (s, t) such that C ∪ (−C1) is a simple closed curve enclosing a region E ⊂ D. By the Green’s Theorem, we have
Z
C
P dx + Q dy − Z
C1
P dx + Q dy = I
C∪(−C1)
P dx + Q dy = Z Z
E
∂Q
∂x − ∂P
∂y
dx dy = 0,
and Z
C
P dx + Q dy = Z
C1
P dx + Q dy (independent of path)
Hence we may simply rewrite the definition of f at each point (s, t) ∈ D as follows
f (s, t) = Z (s,t)
p
P dx + Q dy,
where the line integral is integrated along any piecewise smooth path in D from p to (s, t). For each (s, t) ∈ D and for any sufficiently small h, k such that (s + h, t + k) ∈ D, since
f (s + h, t) =
Z (s+h,t) p
P dx + Q dy
=
Z (s,t) p
P dx + Q dy +
Z (s+h,t) (s,t)
P dx + Qdy
since y = t (const.) on (s, t) → (s + h, t) =⇒ dy = 0
= f (s, t) +
Z (s+h,t) (s,t)
P dx
= f (s, t) + Z s+h
s
P (x, t) dx,
f (s, t + k) =
Z (s,t+k) p
P dx + Q dy
=
Z (s,t) p
P dx + Q dy +
Z (s,t+k) (s,t)
P dx+ Q dy
since x = s (const.) on (s, t) → (s, t + k) =⇒ dx = 0
= f (s, t) +
Z (s,t+k) (s,t)
Q dy
= f (s, t) + Z t+k
t
Q(s, y) dy,
by the Fundamental Theorem of Calculus, we have
fx(s, t) = lim
h→0
f (s + h, t) − f (s, t)
h = lim
h→0
Rs+h
s P (x, t) dx
h = P (s, t) fy(s, t) = lim
k→0
f (s, t + k) − f (s, t)
k = lim
k→0
Rt+k
t Q(s, y)dy
k = Q(s, t) Hence
F = (P, Q) = (fx, fy) = ∇f on D =⇒ F is conservative on D.
Examples 1. Evaluate
Z
C
x4dx + xy dy, where C is the triangular curve consisting of the line segments from (0, 0) to (1, 0), from (1, 0) to (0, 1) and from (0, 1) to (0, 0).
Solution: By the Green’s Theorem Z
C
x4dx + xy dy = Z 1
0
Z 1−x 0
(y − 0) dy dx.
2. Let C be a positively oriented, piecewise-smooth, simple closed curve in the plane and let D be the region bounded by C. By the Green’s Theorem,the area of D is given by
A(D) = I
C
x dy = − I
C
y dx = 1 2
I
C
x dy − y dx,
where C is positively oriented (i.e. move along C in the direction so that D is on the left).
Use the formula to find the area enclosed by the ellipse x2 a2 + y2
b2 = 1.
Solution: By the Green’s Theorem, A = 1 2
I
C
x dy − y dx = 1 2
Z 2π 0
a b dθ 3. Evaluate
Z
C
y2dx + 3xy dy, where C is the boundary of the semiannular region D in the upper half-plane between the circles x2+ y2 = 1 and x2+ y2 = 4.
Solution: By the Green’s Theorem Z
C
y2dx + 3xy dy = Z Z
D
y dA = Z π
0
Z 2 1
r2sin θ dr dθ
Curl and Divergence
DefinitionLet E be a subset of R3 and let F = (F1, F2, F3) be a vector field, f be a differentiable function defined on E. Let the vector differential operator ∇ (“del”) be defined by
∇ = i ∂
∂x1 + j ∂
∂x2 + k ∂
∂x3
Suppose that the partial derivatives of F1, F2, F3, and f all exist, then thecurl of F is the vector
field on R3 defined by
curl F = ∇ × F =
i j k
∂
∂x1
∂
∂x2
∂
∂x3
F1 F2 F3
=
∂F3
∂x2 −∂F2
∂x3
i +
∂F1
∂x3 − ∂F3
∂x1
j +
∂F2
∂x1 − ∂F1
∂x2
k
=
∂F3
∂x2 −∂F2
∂x3 , ∂F1
∂x3 − ∂F3
∂x1 , ∂F2
∂x1 − ∂F1
∂x2
the gradient of f is a vector field on R3 defined by grad f = ∇f = ∂f
∂x1i + ∂f
∂x2 j + ∂f
∂x3 k = fx1, fx2, fx3
and thedivergence of F is a function on R3 defined by
div F = ∇ · F = ∂F1
∂x1 +∂F2
∂x2 +∂F3
∂x3, where ∇ = i ∂
∂x1 + j ∂
∂x2 + k ∂
∂x3 ExampleLet F (x, y, z) = xz i + xyz j − y2k. (1) Find curl F. (2) Find div F.
Theorem Let F = (F1, F2, F3) be a vector field defined on R3. Suppose that the component functions have continuous partial derivatives. Then curl F = 0 if and only if F is a conservative vector field.
Proof If
curl F =
∂F3
∂x2 − ∂F2
∂x3 , ∂F1
∂x3 −∂F3
∂x1 , ∂F2
∂x1 − ∂F1
∂x2
= (0, 0, 0) on R3,
the integrability conditions hold and there is a continuously differentiable function f : R3 → R such that F = ∇f on R3, i.e. F is a conservative vector field.
Conversely, if F is a conservative vector field such that there is a continuously differentiable function f : R3 → R having has continuous 2nd order partial derivatives on R3, then
curl F = curl ∇f = 0 on R3.
Theorem Let F = (F1, F2, F3) be a vector field defined on R3. Suppose that the component functions have continuous 2nd order partial derivatives. Then
div curl F = 0 on R3.
Divergence Theorem Let W = {(x1, x2, x3) ∈ R3 | a ≤ x1 ≤ b, c ≤ x2 ≤ d, e ≤ x3 ≤ f } be a closed cell in R3 and let F = (F1, F2, F3) be a continuous vector field on W. Suppose that ∂F1
∂x1,
∂F2
∂x2 and ∂F3
∂x3 are continuous on an open set U containing W. Then Z Z Z
W
div F dV = Z Z
∂W
F · n dA,
where n = n(p) denotes the unit outward normal vector to ∂W at p ∈ ∂W.
Proof Since the boundary ∂W of W consists of 6 faces S1 ∪ S2∪ · · · ∪ S6, where S1 = {(x1, x2, x3) ∈ W | x1 = a} =⇒ if p ∈ S1 then n(p) = (−1, 0, 0) S2 = {(x1, x2, x3) ∈ W | x1 = b} =⇒ if p ∈ S2 then n(p) = (1, 0, 0) S3 = {(x1, x2, x3) ∈ W | x2 = c} =⇒ if p ∈ S3 then n(p) = (0, −1, 0) S4 = {(x1, x2, x3) ∈ W | x2 = d} =⇒ if p ∈ S4 then n(p) = (0, 1, 0) S5 = {(x1, x2, x3) ∈ W | x3 = e} =⇒ if p ∈ S5 then n(p) = (0, 0, −1) S6 = {(x1, x2, x3) ∈ W | x3 = f } =⇒ if p ∈ S6 then n(p) = (0, 0, 1) we have
Z Z Z
W
div F dV = Z Z Z
W
∂F1
∂x1 + ∂F2
∂x2 +∂F3
∂x3
dV
= Z f
e
Z d c
Z b a
∂F1
∂x1 dx1dx2dx3+ Z f
e
Z b a
Z d c
∂F2
∂x2 dx2dx1dx3+ Z b
a
Z d c
Z f e
∂F3
∂x3 dx3dx2dx1
= Z f
e
Z d c
(F1(b, x2, x3) − F1(a, x2, x3)) dx2dx3 + Z f
e
Z b a
(F2(x1, d, x3) − F2(x1, c, x3)) dx1dx3 +
Z b a
Z d c
(F3(x1, x2, f ) − F3(x1, x2, e)) dx2dx1
= Z Z
S2
F1(b, x2, x3) dx2dx3− Z Z
S1
F1(a, x2, x3) dx2dx3+ Z Z
S4
F2(x1, d, x3) dx1dx3
− Z Z
S3
F2(x1, c, x3) dx1dx3+ Z Z
S6
F3(x1, x2, f ) dx2dx1− Z Z
S5
F3(x1, x2, e)) dx2dx1
= Z Z
S2
F · (1, 0, 0) dx2dx3+ Z Z
S1
F · (−1, 0, 0) dx2dx3+ Z Z
S4
F · (0, 1, 0) dx1dx3 +
Z Z
S3
F · (0, −1, 0) dx1dx3+ Z Z
S6
F · (0, 0, 1) dx2dx1+ Z Z
S5
F · (0, 0, −1) dx2dx1
= Z Z
S2∪S1∪S4∪S3∪S6∪S5
F · n dA = Z Z
∂W
F · n dA
Remark In general, if R is a regular region in Rn with piecewise smooth boundary ∂R, and if F = (F1, . . . , Fn) is a continuously differentiable vector field on R ∪ ∂R, then
Z
R
div F dV = Z
∂R
F · ν dS
where ν = ν(x) denotes the unit outward normal vector to ∂R at x ∈ ∂R and dSx denotes the volume element of ∂R at x ∈ ∂R.
Corollary (Green’s Theorem) Let R be a regular region in R2 = x1x2-plane with piecewise smooth boundary ∂R, and let F = (F1, F2) : R ∪ ∂R → R2 be a continuously differentiable vector field on R ∪ ∂R. Then
Z
∂R
F · dx = Z Z
R
∂F2
∂x1 −∂F1
∂x2
dA, where dx = (dx1, dx2)
Proof Note that if r(t) = (x1(t), x2(t)) : [a, b] → ∂R is a parametrization (or coordinate func- tions) of ∂R, then
Z
∂R
F · dx = Z
r([a,b])
(F1, F2) · (dx1, dx2) = Z b
a
(F1, F2) ·
dx1 dt ,dx2
dt
dt
Since r′(t) =
dx1 dt ,dx2
dt
is tangent to ∂R at p = r(t), the vector ν(p) =
dx2
dt , −dx1 dt
is a normal vector there.
Let G = (G1, G2) = (F2, −F1) . Then G is a continuously differentiable vector field on R ∪ ∂R, and
Z
∂R
F · dx = Z
∂R
(F1, F2) · (dx1, dx2) = Z
∂R
(F2, −F1) · (dx2, −dx1)= Z b
a
(G1, G2) · ν dt,
and, by the divergence theorem, we have Z b
a
G1, G2 · ν dt = Z
∂R
G · ν = Z Z
R
div G dA = Z Z
R
∂F2
∂x1 − ∂F1
∂x2
dA
Definition If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the surface integral of F over S is
Z Z
S
F · dS = Z Z
S
F · n dS = the flux of F across S,
where dS is the vector area element of S, dS is the area element of S, n = n(p) is the unit outward normal vector to S at p. This integral is also called the flux of F across S.
Stokes’ Theorem Let S be an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve C with positive orientation. Let F be a vector field whose components have continuous partial derivatives on an open region in R3 that contains S.
Then Z
C
F · dr = Z Z
S
curl F · dS = Z Z
R
(∇ × F ) · ∂r
∂s1 × ∂r
∂s2 dA, where
r(s1, s2) = (x1(s1, s2), x2(s1, s2), x3(s1, s2)) : R → S is a smooth parametrization that maps a simple, closed, piecewise-smooth bounded region R, in s1s2-plane, to a surface S in x1x2x3- space and r : ∂R → C maps the boundary ∂R of R onto C,
dr = ∂r
∂s1 ds1+ ∂r
∂s2 ds2 is the tangent vector length element of C,
dS is the vector area element of S and dA is the area element of R.
Remark If C is a smooth simple closed curve given by the vector equation r(t) = x(t) i + y(t) j + z(t) k = x(t), y(t), 0
for a ≤ t ≤ b then
T (t) = x′(t)
|r′(t)|i + y′(t)
|r′(t)|j and n(t) = y′(t)
|r′(t)|i − x′(t)
|r′(t)|j.
are respectively the unit tangent vector and the outward unit normal vector to C at r(t), and
n(t) × T (t) = (x′(t))2+ (y′(t))2
|r′(t)|2 (i × j) = i × j = k for each a ≤ t ≤ b
=⇒ D ⊂ xy-plane =⇒ dz = 0 on D, Since
curl F =
∂F3
∂x2 −∂F2
∂x3 , ∂F1
∂x3 − ∂F3
∂x1 , ∂F2
∂x1 − ∂F1
∂x2
we have
Z Z
D
curl F = i Z Z
D
∂F3
∂y − ∂F2
∂z
dydz+ j Z Z
D
∂F1
∂z − ∂F3
∂x
dxdz+ k Z Z
D
∂F2
∂x −∂F1
∂y
dx dy
=
Z Z
D
∂F2
∂x −∂F1
∂y
dx dy
k since dz = 0 on D
=
I
C
F1dx + F2dy
k by the Green’s Theorem
=
I
C
F (r) · dr
n(t) × T (t)
perpendicular to the plane spanned by T, n
so, by setting N = T × n along C, we obtain a positively oriented basis {N, n, T } for R3 and note that the curl F at a point p = (x, y, z) can be defined by
curl F (p) = lim
A→0
1 A
Z Z
D
curl F = lim
A→0
1 A
I
C
F (r) · dr
N ⊥ the plane containing C
where A is the area of D and I
C
F (r) · dr, a line integral along the boundary of D,measures the velocity of particles move around the axis.
Proof of Stokes’ Theorem Since C = r(∂R) and dr = ∂r
∂s1ds1+ ∂r
∂s2 ds2, we have Z
C
F · dr = Z
∂R
F · ∂r
∂s1
ds1+ F · ∂r
∂s2
ds2
by the definition of line integral.
Setting
G1 = F · ∂r
∂s1 and G2 = F · ∂r
∂s2, and by the Green’s Theorem, we have
Z
C
F · dr = Z
∂R
(G1ds1+ G2ds2)= Z Z
R
∂G2
∂s1 − ∂G1
∂s2
ds1ds2,
On the other hand, since Z Z
S
curl F · dS = Z Z
R
curl F · ∂r
∂s1 × ∂r
∂s2 ds1ds2, and
curl F · ∂r
∂s1 × ∂r
∂s2 =
∂F3
∂x2 − ∂F2
∂x3
∂F1
∂x3 − ∂F3
∂x1
∂F2
∂x1 −∂F1
∂x2
∂x1
∂s1
∂x2
∂s1
∂x3
∂s1
∂x1
∂s2
∂x2
∂s2
∂x3
∂s2
=
3
X
i,j=1
∂Fj
∂xi − ∂Fi
∂xj
∂xi
∂s1
∂xj
∂s2 by the definition of determinant
=
3
X
i,j=1
∂Fj
∂xi
∂xi
∂s1
∂xj
∂s2 −
3
X
i,j=1
∂Fi
∂xj
∂xj
∂s2
∂xi
∂s1
= ∂F
∂s1 · ∂r
∂s2 − ∂F
∂s2 · ∂r
∂s1 by the Chain Rule
= ∂G2
∂s1 − ∂G1
∂s2 we have
Z Z
S
curl F · dS = Z Z
R
∂G2
∂s1 −∂G1
∂s2
ds1ds2 = Z
C
F · dr
Example 1. Evaluate Z
C
F · dr, where F (x, y, z) = −y2i + x j + z2k and C is the curve of intersection of the plane y + z = 2 and the cylinder x2+ y2 = 1. (Orient C to be counterclockwise when viewed from above.)
Solution: curl F = (1 + 2y) k, S = {z = g(x, y) = 2 − y}, curl F · dS = (0, 0, 1 + 2y) · (−gx, −gy, 1) dA = (1 + 2y) dA and
Z
C
F · dr = Z Z
S
curl F · dS = Z Z
D
(1 + 2y) dA = Z 2π
0
Z 1 0
(1 + 2r sin θ) r dr dθ.
Example 2. Find the flux of the vector field F (x, y, z) = z i + y j + x k over the unit sphere x2+ y2+ z2 = 1.
Solution: r(ϕ, θ) = (sin ϕ cos θ, sin ϕ sin θ, cos ϕ), (ϕ, θ) ∈ D = {0 ≤ ϕ ≤ π, 0 ≤ θ ≤ 2π}. Then Z Z
S
F · dS = Z Z
D
F · rϕ× rθdA = 4π/3.
