Download The Divergence Theorem. (Sect. 16.8) The divergence of a vector

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 Z Zrelates line integrals with surface integrals,
that is,
E·r=
(∇ × E) dσ.
C
S
Using the Divergence Theorem we obtain the differential form of
Gauss’ law,
∂t B = −∇ × E.