The Stokes Theorem. (Sect. 16.7)
I
The curl of a vector field in space.
I
The curl of conservative fields.
I
Stokes’ Theorem in space.
I
Idea of the proof of Stokes’ Theorem.
The curl of a vector field in space.
Definition
The curl of a vector field F = hF1 , F2 , F3 i in R3 is the vector field
curl F = (∂2 F3 − ∂3 F2 ), (∂3 F1 − ∂1 F3 ), (∂1 F2 − ∂2 F1 ) .
Remark: Since the following formula holds,
i
j
k
curl F = ∂1 ∂2 ∂3 F 1 F 2 F 3 curl F = (∂2 F3 − ∂3 F2 ) i − (∂1 F3 − ∂3 F1 ) j + (∂1 F2 − ∂2 F1 ) k,
then one also uses the notation
curl F = ∇ × F.
The curl of a vector field in space.
Example
Find the curl of the vector field F = hxz, xyz, −y 2 i.
Solution: Since curl F = ∇ × F, we get,
i
j
k
∂z =
∇ × F = ∂x ∂y
xz xyz −y 2 ∂y (−y 2 )−∂z (xyz) i− ∂x (−y 2 )−∂z (xz) j + ∂x (xyz)−∂y (xz) k,
= −2y − xy i − 0 − x j + yz − 0 k,
We conclude that
∇ × F = h−y (2 + x), x, yzi.
The Stokes Theorem. (Sect. 16.7)
I
The curl of a vector field in space.
I
The curl of conservative fields.
I
Stokes’ Theorem in space.
I
Idea of the proof of Stokes’ Theorem.
C
The curl of conservative fields.
Recall: A vector field F : R3 → R3 is conservative iff there exists a
scalar field f : R3 → R such that F = ∇f .
Theorem
If a vector field F is conservative, then ∇ × F = 0.
Remark:
I
This Theorem is usually written as ∇ × (∇f ) = 0.
I
The converse is true only on simple connected sets.
That is, if a vector field F satisfies ∇ × F = 0 on a simple
connected domain D, then there exists a scalar field
f : D ⊂ R3 → R such that F = ∇f .
Proof of the Theorem:
∇×F =
∂y ∂z f − ∂z ∂y f , − ∂x ∂z f − ∂z ∂x f , ∂x ∂y f − ∂y ∂x f
The curl of conservative fields.
Example
Is the vector field F = hxz, xyz, −y 2 i conservative?
Solution: We have shown that ∇ × F = h−y (2 + x), x, yzi.
Since ∇ × F 6= 0, then F is not conservative.
C
Example
Is the vector field F = hy 2 z 3 , 2xyz 3 , 3xy 2 z 2 i conservative in R3 ?
Solution: Notice that
i
j
k
∂y
∂z ∇ × F = ∂x
y 2 z 3 2xyz 3 3xy 2 z 2 = (6xyz 2 − 6xyz 2 ), −(3y 2 z 2 − 3y 2 z 2 ), (2yz 3 − 2yz 3 ) = 0.
Since ∇ × F = 0 and R3 is simple connected, then F is
conservative, that is, there exists f in R3 such that F = ∇f .
C
The Stokes Theorem. (Sect. 16.7)
I
The curl of a vector field in space.
I
The curl of conservative fields.
I
Stokes’ Theorem in space.
I
Idea of the proof of Stokes’ Theorem.
Stokes’ Theorem in space.
Theorem
The circulation of a differentiable vector field F : D ⊂ R3 → R3
around the boundary C of the oriented surface S ⊂ D satisfies the
equation
I
ZZ
F · dr =
C
(∇ × F) · n dσ,
S
where dr points counterclockwise when the unit vector n normal to
S points in the direction to the viewer (right-hand rule).
n
S
C
r’ (t)
r (t)
Stokes’ Theorem in space.
Example
Verify Stokes’ Theorem for the field F = hx 2 , 2x, z 2 i on the ellipse
S = {(x, y , z) : 4x 2 + y 2 6 4, z = 0}.
I
ZZ
Solution: We compute both sides in
F · dr =
(∇ × F) · n dσ.
C
We start computing the circulation
2
integral on the ellipse x 2 + y22 = 1.
We need to choose a counterclockwise
parametrization, hence the normal to S
points upwards.
We choose, for t ∈ [0, 2π],
z
−1
S
−2
2
C
S
y
1
x
z
n
−1
−2
r(t) = hcos(t), 2 sin(t), 0i.
S
2
C
y
Therefore, the right-hand rule normal n
to S is n = h0, 0, 1i.
1
x
Stokes’ Theorem in space.
Example
Verify Stokes’ Theorem for the field F = hx 2 , 2x, z 2 i on the ellipse
S = {(x, y , z) : 4x 2 + y 2 6 4, z = 0}.
I
ZZ
Solution: Recall:
F · dr =
(∇ × F) · n dσ, with
C
S
r(t) = hcos(t), 2 sin(t), 0i, t ∈ [0, 2π] and n = h0, 0, 1i.
The circulation integral is:
I
Z 2π
F(t) · r0 (t) dt
F · dr =
0
C
Z
=
2π
hcos2 (t), 2 cos(t), 0i · h− sin(t), 2 cos(t), 0i dt.
0
I
Z
F · dr =
C
0
2π − cos2 (t) sin(t) + 4 cos2 (t) dt.
Stokes’ Theorem in space.
Example
Verify Stokes’ Theorem for the field F = hx 2 , 2x, z 2 i on the ellipse
S = {(x, y , z) : 4x 2 + y 2 6 4, z = 0}.
I
Z 2π
Solution:
F · dr =
− cos2 (t) sin(t) + 4 cos2 (t) dt.
0
C
The substitution on the first term u = cos(t) and du = − sin(t) dt,
Z 2π
Z 1
2
− cos (t) sin(t) dt =
u 2 du = 0.
implies
0
1
I
Z
2π
Z
2
F · dr =
2π
2 1 + cos(2t) dt.
4 cos (t) dt =
0
C
0
2π
Z
Since
I
F · dr = 4π.
cos(2t) dt = 0, we conclude that
0
C
Stokes’ Theorem in space.
Example
Verify Stokes’ Theorem for the field F = hx 2 , 2x, z 2 i on the ellipse
S = {(x, y , z) : 4x 2 + y 2 6 4, z = 0}.
I
Solution:
F · dr = 4π and n = h0, 0, 1i.
C
We now compute the right-hand side in Stokes’ Theorem.
z
ZZ
n
−1
(∇ × F) · n dσ.
I =
S
S
−2
2
C
y
1
x
i
j
∇ × F = ∂x ∂y
x 2 2x
S is the flat surface {x 2 +
k ∂z z 2
y2
22
⇒
∇ × F = h0, 0, 2i.
6 1, z = 0}, so dσ = dx dy .
Stokes’ Theorem in space.
Example
Verify Stokes’ Theorem for the field F = hx 2 , 2x, z 2 i on the ellipse
S = {(x, y , z) : 4x 2 + y 2 6 4, z = 0}.
I
Solution:
F · dr = 4π, n = h0, 0, 1i, ∇ × F = h0, 0, 2i, and
C
dσ = dx dy .
ZZ
Z
Then,
(∇ × F) · n dσ =
1
−1
S
Z
√
2 1−x 2
h0, 0, 2i
√
−2 1−x 2
· h0, 0, 1i dy dx.
The right-hand side above is twice the area of the ellipse. Since we
know that an ellipse x 2 /a2 + y 2 /b 2 = 1 has area πab, we obtain
ZZ
(∇ × F) · n dσ = 4π.
S
This verifies Stokes’ Theorem.
C
Stokes’ Theorem in space.
Remark: Stokes’ Theorem implies that for any smooth field F and
any two surfaces S1 , S2 having the same boundary curve C holds,
ZZ
ZZ
(∇ × F) · n1 dσ1 =
(∇ × F) · n2 dσ2 .
S1
S2
Example
Verify Stokes’ Theorem for the field F = hx 2 , 2x, z 2 i on any
y2 z2
2
half-ellipsoid S2 = {(x, y , z) : x + 2 + 2 = 1, z > 0}.
2
a
Solution: (The previous example was the case a → 0.)
We must verify Stokes’ Theorem on S2 ,
I
ZZ
F · dr =
(∇ × F) · n2 dσ2 .
z
n2
n2
a
S2
2
C
1
x
S1
y
C
S2
Stokes’ Theorem in space.
Example
Verify Stokes’ Theorem for the field F = hx 2 , 2x, z 2 i on any
y2 z2
half-ellipsoid S2 = {(x, y , z) : x 2 + 2 + 2 = 1, z > 0}.
2
a
I
ZZ
Solution:
F · dr = 4π, ∇ × F = h0, 0, 2i, I =
C
(∇ × F) · n2 dσ2 .
S2
z
S2 is the level surface F = 0 of
n2
n2
a
S2
2
C
1
y2 z2
F(x, y , z) = x + 2 + 2 − 1.
2
a
2
y
S1
x
∇F
n2 =
,
|∇F|
|∇F|
|∇F|
=
2z/a2
|∇F · k|
dσ2 =
2z/a2
(∇ × F) · n2 = 2
.
|∇F|
y 2z E
∇F = 2x, , 2 ,
2 a
D
⇒
(∇ × F) · n2 dσ2 = 2.
Stokes’ Theorem in space.
Example
Verify Stokes’ Theorem for the field F = hx 2 , 2x, z 2 i on any
y2 z2
2
half-ellipsoid S2 = {(x, y , z) : x + 2 + 2 = 1, z > 0}.
2
a
I
F · dr = 4π and (∇ × F) · n2 dσ2 = 2.
Solution:
C
Therefore,
ZZ
ZZ
(∇ × F) · n2 dσ2 =
2 dx dy = 2(2π).
S1
S2
ZZ
(∇ × F) · n2 dσ2 = 4π, no matter what is
We conclude that
the value of a > 0.
S2
C
The Stokes Theorem. (Sect. 16.7)
I
The curl of a vector field in space.
I
The curl of conservative fields.
I
Stokes’ Theorem in space.
I
Idea of the proof of Stokes’ Theorem.
Idea of the proof of Stokes’ Theorem.
S
Split the surface S into n surfaces Si ,
for i = 1, · · · , n, as it is done in the
figure for n = 9.
C
I
F · dr =
C
n I
X
F · dri
Ci
'
=
i=1
n I
X
F · dr̃i
i=1 C̃ i
n ZZ
X
Zi=1
Z
(C̃i the border of small rectangles);
(∇ × F) · ni dA (Green’s Theorem on a plane);
R̃ i
(∇ × F) · n dσ.
'
S
The Divergence Theorem. (Sect. 16.8)
I
The divergence of a vector field in space.
I
The Divergence Theorem in space.
I
The meaning of Curls and Divergences.
Applications in electromagnetism:
I
I
I
Gauss’ law. (Divergence Theorem.)
Faraday’s law. (Stokes Theorem.)
The divergence of a vector field in space.
Definition
The divergence of a vector field F = hFx , Fy , Fz i is the scalar field
div F = ∂x Fx + ∂y Fy + ∂z Fz .
Remarks:
I
It is also used the notation div F = ∇ · F.
I
The divergence of a vector field measures the expansion
(positive divergence) or contraction (negative divergence) of
the vector field.
I
A heated gas expands, so the divergence of its velocity field is
positive.
I
A cooled gas contracts, so the divergence of its velocity field
is negative.
The divergence of a vector field in space.
Example
Find the divergence and the curl of F = h2xyz, −xy , −z 2 i.
Solution: Recall: div F = ∂x Fx + ∂y Fy + ∂z Fz .
∂x Fx = 2yz,
∂y Fy = −x,
∂z Fz = −2z.
Therefore ∇ · F = 2yz − x − 2z, that is ∇ · F = 2z(y − 1) − x.
Recall: curl F = ∇ × F.
i
j
k
∂y
∂z = h(0 − 0), −(0 − 2xy ), (−y − 2xz)i
∇ × F = ∂x
2xyz −xy −z 2 We conclude: ∇ × F = h0, 2xy , −(2xz + y )i.
C
The divergence of a vector field in space.
Example
r
Find the divergence of F = 3 , where r = hx, y , zi, and
ρ
p
ρ = |r| = x 2 + y 2 + z 2 . (Notice: |F| = 1/ρ2 .)
y
z
x
F
=
,
F
=
.
,
y
z
ρ3
ρ3
ρ3
−3/2 ∂x Fx = ∂x x x 2 + y 2 + z 2
Solution: The field components are Fx =
∂x Fx = x 2 + y 2 + z 2
−3/2
−5/2
3
− x x2 + y2 + z2
(2x)
2
1
y2
1
x2
∂x Fx = 3 − 3 5 ⇒ ∂y Fy = 3 − 3 5 ,
ρ
ρ
ρ
ρ
1
z2
∂z Fz = 3 − 3 5 .
ρ
ρ
3
(x 2 + y 2 + z 2 )
3
ρ2
3
3
∇·F= 3 −3
=
−
3
=
−
.
ρ
ρ5
ρ3
ρ5
ρ 3 ρ3
We conclude: ∇ · F = 0.
C
The Divergence Theorem. (Sect. 16.8)
I
The divergence of a vector field in space.
I
The Divergence Theorem in space.
I
The meaning of Curls and Divergences.
Applications in electromagnetism:
I
I
I
Gauss’ law. (Divergence Theorem.)
Faraday’s law. (Stokes Theorem.)
The Divergence Theorem in space.
Theorem
The flux of a differentiable vector field F : R3 → R3 across a
closed oriented surface S ⊂ R3 in the direction of the surface
outward unit normal vector n satisfies the equation
ZZ
ZZZ
F · n dσ =
(∇ · F) dV ,
S
V
where V ⊂ R3 is the region enclosed by the surface S.
Remarks:
I
The volume integral of the divergence of a field F in a volume
V in space equals the outward flux (normal flow) of F across
the boundary S of V .
I
The expansion part of the field F in V minus the contraction
part of the field F in V equals the net normal flow of F across
S out of the region V .
The Divergence Theorem in space.
Example
Verify the Divergence Theorem for the field F = hx, y , zi over the
sphere x 2 + y 2 + z 2 = R 2 .
ZZ
ZZZ
F · n dσ =
(∇ · F) dV .
Solution: Recall:
S
V
We start with the flux integral across S. The surface S is the level
surface f = 0 of the function f (x, y , z) = x 2 + y 2 + z 2 − R 2 . Its
outward unit normal vector n is
n=
∇f
,
|∇f |
∇f = h2x, 2y , 2zi,
p
|∇f | = 2 x 2 + y 2 + z 2 = 2R,
1
hx, y , zi, where z = z(x, y ).
R
|∇f |
R
Since dσ =
dx dy , then dσ = dx dy , with z = z(x, y ).
|∇f · k|
z
We conclude that n =
The Divergence Theorem in space.
Example
Verify the Divergence Theorem for the field F = hx, y , zi over the
sphere x 2 + y 2 + z 2 = R 2 .
1
R
Solution: Recall: n = hx, y , zi, dσ = dx dy , with z = z(x, y ).
R ZZ z
ZZ
1
F · n dσ =
hx, y , zi · hx, y , zi dσ.
R
S
S
ZZ
ZZ
ZZ
1
2
2
2
F · n dσ =
x + y + z dσ = R
dσ.
R
S
S
S
The integral on the sphere S can be written as the sum of the
integral on the upper half plus the integral on the lower half, both
integrated on the disk R = {x 2 + y 2 6 R 2 , z = 0}, that is,
ZZ
ZZ
R
F · n dσ = 2R
dx dy .
S
R z
The Divergence Theorem in space.
Example
Verify the Divergence Theorem for the field F = hx, y , zi over the
sphere x 2 + y 2 + z 2 = R 2 .
ZZ
ZZ
R
F · n dσ = 2R
dx dy .
Solution:
z
S
R
Using polar coordinates on {z = 0}, we get
ZZ
Z 2π Z R
R2
√
r dr dθ.
F · n dσ = 2
R2 − r 2
S
0
0
The substitution u = R 2 − r 2 implies du = −2r dr , so,
ZZ
F · n dσ = 4πR 2
Z
u −1/2
R2
S
ZZ
0
F · n dσ = 2πR 2
S
R 2 1/2 2u 0
(−du)
= 2πR 2
2
ZZ
⇒
Z
R2
u −1/2 du
0
F · n dσ = 4πR 3 .
S
The Divergence Theorem in space.
Example
Verify the Divergence Theorem for the field F = hx, y , zi over the
sphere x 2 + y 2 + z 2 = R 2 .
ZZ
F · n dσ = 4πR 3 .
Solution:
S
ZZZ
We now compute the volume integral
∇ · F dV . The
V
divergence of F is ∇ · F = 1 + 1 + 1, that is, ∇ · F = 3. Therefore
ZZZ
ZZZ
4
3
∇ · F dV = 3
dV = 3 πR
3
V
V
ZZZ
We obtain
∇ · F dV = 4πR 3 .
V
We have verified the Divergence Theorem in this case.
C
The Divergence Theorem in space.
Example
r
across the boundary of the region
ρ3
between the spheres
of radius R1 > R0 > 0, where r = hx, y , zi,
p
2
and ρ = |r| = x + y 2 + z 2 .
Find the flux of the field F =
Solution: We use the Divergence Theorem
ZZ
ZZZ
F · n dσ =
(∇ · F) dV .
S
V
ZZZ
Since ∇ · F = 0, then
(∇ · F) dV = 0. Therefore
V
ZZ
F · n dσ = 0.
S
The flux along any surface S vanishes as long as 0 is not included
in the region surrounded by S. (F is not differentiable at 0.)
C
The Divergence Theorem. (Sect. 16.8)
I
The divergence of a vector field in space.
I
The Divergence Theorem in space.
I
The meaning of Curls and Divergences.
Applications in electromagnetism:
I
I
I
Gauss’ law. (Divergence Theorem.)
Faraday’s law. (Stokes Theorem.)
The meaning of Curls and Divergences.
Remarks: The meaning of the Curl and the Divergence of a vector
field F is best given through the Stokes and Divergence Theorems.
I
1
I ∇ × F = lim
F · dr,
S →{P } A(S)
C
I
where S is a surface containing the point P with boundary
given by the loop C and A(S) is the area of that surface.
ZZ
1
∇ · F = lim
F · ndσ,
R →{P } V (R)
S
where R is a region in space containing the point P with
boundary given by the closed orientable surface S and V (R) is
the volume of that region.
The Divergence Theorem. (Sect. 16.8)
I
The divergence of a vector field in space.
I
The Divergence Theorem in space.
I
The meaning of Curls and Divergences.
Applications in electromagnetism:
I
I
I
Gauss’ law. (Divergence Theorem.)
Faraday’s law. (Stokes Theorem.)
Applications in electromagnetism: Gauss’ Law.
Gauss’ law: Let q : R3 → R be the charge density in space, and
E : R3 → R3 be the electric field generated by that charge. Then
ZZZ
ZZ
q dV = k
E · n dσ,
R
S
that is, the total charge in a region R in space with closed
orientable surface S is proportional to the integral of the electric
field E on this surface S.
The Divergence Theorem
relates Zsurface
ZZ
Z Z integrals with volume
integrals, that is,
E · n dσ =
(∇ · E) dV .
S
R
Using the Divergence Theorem we obtain the differential form of
Gauss’ law,
1
∇ · E = q.
k
Applications in electromagnetism: Faraday’s Law.
Faraday’s law: Let B : R3 → R3 be the magnetic field across an
orientable surface S with boundary given by the loop C , and let
E : R3 → R3 measured on that loop. Then
ZZ
I
d
B · n dσ = − E · dr,
dt
S
C
that is, the time variation of the magnetic flux across S is the
negative of the electromotive force on the loop.
The Stokes
I Theorem
Z Zrelates line integrals with surface integrals,
that is,
E · dr =
(∇ × E) · n dσ.
C
S
Using the Stokes Theorem we obtain the differential form of
Faraday’s law,
∂t B = −∇ × E.
Review for Exam 4.
I
Sections 16.1-16.5, 16.7, 16.8.
I
50 minutes.
I
5 problems, similar to homework problems.
I
No calculators, no notes, no books, no phones.
I
No green book needed.
Review for Exam 4.
I
(16.1) Line integrals.
I
(16.2) Vector fields, work, circulation, flux (plane).
I
(16.3) Conservative fields, potential functions.
I
(16.4) The Green Theorem in a plane.
I
(16.5) Surface area, surface integrals.
I
(16.7) The Stokes Theorem.
I
(16.8) The Divergence Theorem.
Line integrals (16.1).
Example
Integrate the function f (x, y ) = x 3 /y along the plane curve C
given by y = x 2 /2 for x ∈ [0, 2], from the point (0, 0) to (2, 2).
Z
Solution: We have to compute I = f ds, by that we mean
C
Z
t1
Z
f ds =
f x(t), y (t) |r0 (t)| dt,
t0
C
where r(t) = hx(t), y (t)i for t ∈ [t0 , t1 ] is a parametrization of the
path C . In this case the path is given by the parabola y = x 2 /2, so
a simple parametrization is to use x = t, that is,
D t2 E
r(t) = t,
,
2
t ∈ [0, 2]
⇒
r0 (t) = h1, ti.
Line integrals (16.1).
Example
Integrate the function f (x, y ) = x 3 /y along the plane curve C
given by y = x 2 /2 for x ∈ [0, 2], from the point (0, 0) to (2, 2).
D t2 E
Solution: r(t) = t,
for t ∈ [0, 2], and r0 (t) = h1, ti.
2
Z 2 3 p
Z
Z t1
0
t
1 + t 2 dt,
f ds =
f x(t), y (t) |r (t)| dt =
2 /2
t
0
C
t0
Z
Z
f ds =
2
2t
p
Z
5
1 + t 2 dt,
u = 1 + t 2,
du = 2t dt.
0
C
Z
2 3/2 5 2 3/2
f ds =
u du = u = 5 − 1 .
3
3
1
C
1
Z
2 √
We conclude that
f ds = 5 5 − 1 .
3
C
1/2
C
Review for Exam 4.
I
(16.1) Line integrals.
I
(16.2) Vector fields, work, circulation, flux (plane).
I
(16.3) Conservative fields, potential functions.
I
(16.4) The Green Theorem in a plane.
I
(16.5) Surface area, surface integrals.
I
(16.7) The Stokes Theorem.
I
(16.8) The Divergence Theorem.
Vector fields, work, circulation, flux (plane) (16.2).
Example
Find the work done by the force F = hyz, zx, −xy i in a moving
particle along the curve r(t) = ht 3 , t 2 , ti for t ∈ [0, 2].
Solution: The formula for the work done by a force on a particle
moving along C given by r(t) for t ∈ [t0 , t1 ] is
Z
Z t1
W = F · dr =
F(t) · r0 (t) dt.
C
t0
In this case r0 (t) = h3t 2 , 2t, 1i for t ∈ [0, 2]. We now need to
evaluate F along the curve, that is,
F(t) = F x(t), y (t), z(t) = h(t 2 )t, t(t 3 ), −(t 3 )t 2 i
We obtain F(t) = ht 3 , t 4 , −t 5 i.
Vector fields, work, circulation, flux (plane) (16.2).
Example
Find the work done by the force F = hyz, zx, −xy i in a moving
particle along the curve r(t) = ht 3 , t 2 , ti for t ∈ [0, 2].
Solution: F(t) = ht 3 , t 4 , −t 5 i and r0 (t) = h3t 2 , 2t, 1i for t ∈ [0, 2].
The Work done by the force on the particle is
Z 2
Z t1
ht 3 , t 4 , −t 5 i · h3t 2 , 2t, 1i dt
W =
F(t) · r0 (t) dt =
0
t0
Z
2
5
5
3t + 2t − t
W =
5
2
Z
dt =
0
4 6 2 2 6
4t dt = t = 2 .
6 0 3
5
0
We conclude that W = 27 /3.
Vector fields, work, circulation, flux (plane) (16.2).
Example
Find the flow of the velocity field F = hxy , y 2 , −yzi from the point
(0, 0, 0) to the point (1, 1, 1) along the curve of intersection of the
cylinder y = x 2 with the plane z = x.
Solution: The flow (also called circulation) of the field F along a
curve C parametrized by r(t) for t ∈ [t0 , t1 ] is given by
Z
Z t1
F · dr =
F(t) · r0 (t) dt.
C
t0
We use t = x as the parameter of the curve r, so we obtain
r(t) = ht, t 2 , ti,
t ∈ [0, 1]
F(t) = ht(t 2 ), (t 2 )2 , −t 2 (t)i
⇒
⇒
r0 (t) = h1, 2t, 1i.
F(t) = ht 3 , t 4 , −t 3 i.
Vector fields, work, circulation, flux (plane) (16.2).
Example
Find the flow of the velocity field F = hxy , y 2 , −yzi from the point
(0, 0, 0) to the point (1, 1, 1) along the curve of intersection of the
cylinder y = x 2 with the plane z = x.
Solution: r0 (t) = h1, 2t, 1i for t ∈ [0, 1] and F(t) = ht 3 , t 4 , −t 3 i.
Z
Z t1
Z 1
F · dr =
F(t) · r0 (t) dt =
ht 3 , t 4 , −t 3 i · h1, 2t, 1i dt,
t0
C
0
Z
Z
1
3
5
t + 2t − t
F · dr =
0
C
3
Z
dt =
0
1
2 6 1
2t dt = t .
6 0
5
Z
We conclude that
1
F · dr = .
3
C
C
Vector fields, work, circulation, flux (plane) (16.2).
Example
Find the flux of the field F = h−x, (x − y )i across loop C given by
the circle r(t) = ha cos(t), a sin(t)i for t ∈ [0, 2π].
Solution: The flux (also normal flow) of the field F = hFx , Fy i
across a loop C parametrized by r(t) = hx(t), y (t)i for t ∈ [t0 , t1 ]
is given by
I
Z t1
F · n ds =
Fx y 0 (t) − Fy x 0 (t) dt.
t0
C
Recall that n =
1
hy 0 (y ), −x 0 (t)i and ds = |r0 (t)| dt, therefore
0
|r (t)|
1
0
0
F · n ds = hFx , Fy i · 0
hy (y ), −x (t)i |r0 (t)| dt,
|r (t)|
so we obtain F · n ds = Fx y 0 (t) − Fy x 0 (t) dt.
Vector fields, work, circulation, flux (plane) (16.2).
Example
Find the flux of the field F = h−x, (x − y )i across loop C given by
the circle r(t) = ha cos(t), a sin(t)i for t ∈ [0, 2π].
I
Z t1
Solution:
F · n ds =
Fx y 0 (t) − Fy x 0 (t) dt.
t0
C
We evaluate F along the loop,
F(t) = h−a cos(t), a cos(t) − sin(t) i,
and compute r0 (t) = h−a sin(t), a cos(t)i. Therefore,
I
2π Z
−a cos(t)a cos(t) − a cos(t) − sin(t) (−a) sin(t) dt
F · n ds =
0
C
I
2π Z
−a2 cos2 (t) + a2 sin(t) cos(t) − a2 sin2 (t) dt
F · n ds =
0
C
Vector fields, work, circulation, flux (plane) (16.2).
Example
Find the flux of the field F = h−x, (x − y )i across loop C given by
the circle r(t) = ha cos(t), a sin(t)i for t ∈ [0, 2π].
Solution:
I
Z
F · n ds =
2π −a2 cos2 (t) + a2 sin(t) cos(t) − a2 sin2 (t) dt.
0
C
I
F · n ds = a
2
−1 + sin(t) cos(t) dt,
0
C
I
2π Z
Z
2π h
i
1
F · n ds = a
−1 + sin(2t) dt.
2
C
0
Z 2π
I
Since
sin(2t) dt = 0, we obtain
F · n ds = −2πa2 .
0
2
C
C
Review for Exam 4.
I
(16.1) Line integrals.
I
(16.2) Vector fields, work, circulation, flux (plane).
I
(16.3) Conservative fields, potential functions.
I
(16.4) The Green Theorem in a plane.
I
(16.5) Surface area, surface integrals.
I
(16.7) The Stokes Theorem.
I
(16.8) The Divergence Theorem.
Conservative fields, potential functions (16.3).
Example
Is the field F = hy sin(z), x sin(z), xy cos(z)i conservative?
If “yes”, then find the potential function.
Solution: We need to check the equations
∂y Fz = ∂z Fy ,
∂x Fz = ∂z Fx ,
∂x Fy = ∂y Fx .
∂y Fz = x cos(z) = ∂z Fy ,
∂x Fz = y cos(z) = ∂z Fx ,
∂x Fy = sin(z) = ∂y Fx .
Therefore, F is a conservative field, that means there exists a
scalar field f such that F = ∇f . The equations for f are
∂x f = y sin(z),
∂y f = x sin(z),
∂z f = xy cos(z).
Conservative fields, potential functions (16.3).
Example
Is the field F = hy sin(z), x sin(z), xy cos(z)i conservative?
If “yes”, then find the potential function.
Solution: ∂x f = y sin(z), ∂y f = x sin(z), ∂z f = xy cos(z).
Integrating in x the first equation we get
f (x, y , z) = xy sin(z) + g (y , z).
Introduce this expression in the second equation above,
∂y f = x sin(z) + ∂y g = x sin(z)
⇒
∂y g (y , z) = 0,
so g (y , z) = h(z). That is, f (x, y , z) = xy sin(z) + h(z).
Introduce this expression into the last equation above,
∂z f = xy cos(z) + h0 (z) = xy cos(z) ⇒ h0 (z) = 0 ⇒ h(z) = c.
We conclude that f (x, y , z) = xy sin(z) + c.
C
Conservative fields, potential functions (16.3).
Example
Z
Compute I =
y sin(z) dx + x sin(z) dy + xy cos(z) dz, where C
C
given by r(t) = hcos(2πt), 1 + t 5 , cos2 (2πt)π/2i for t ∈ [0, 1].
Solution: We know that the field F = hy sin(z), x sin(z), xy cos(z)i
conservative, so there exists f such that F = ∇f , or equivalently
df = y sin(z) dx + x sin(z) dy + xy cos(z) dz.
We have computed f already, f = xy sin(z) + c.
Since F is conservative, the integral I is path independent, and
Z
I =
(1,2,π/2) y sin(z) dx + x sin(z) dy + xy cos(z) dz
(1,1,π/2)
I = f (1, 2, π/2) − f (1, 1, π/2) = 2 sin(π/2) − sin(π/2) ⇒ I = 1.