Curl and Divergence
Definition
~ = (F1 , F2 , F3 ) be a vector field. The curl of F
~ is the vector
Let F
field defined by
~) =
curl(F
δF3 δF2 δF1 δF3 δF2 δF1
−
,
−
,
−
.
δy
δz δz
δx δx
δy
Curl and Divergence
Definition
~ = (F1 , F2 , F3 ) be a vector field. The curl of F
~ is the vector
Let F
field defined by
~) =
curl(F
δF3 δF2 δF1 δF3 δF2 δF1
−
,
−
,
−
.
δy
δz δz
δx δx
δy
~ as the
As a mnemonic device, one can think of the curl of F
symbolic cross product:
~) = ∇×F
~ = ( δ , δ , δ ) × (F1 , F2 , F3 ).
curl(F
δx δy δz
Curl and Divergence, contd.
Definition
~ = (F1 , F2 , F3 ) be a vector field. The divergence of F
~
Again let F
is the real-valued function in three variables defined by
~) =
div(F
δF1 δF2 δF3
+
+
.
δx
δy
δz
Curl and Divergence, contd.
Definition
~ = (F1 , F2 , F3 ) be a vector field. The divergence of F
~
Again let F
is the real-valued function in three variables defined by
~) =
div(F
δF1 δF2 δF3
+
+
.
δx
δy
δz
For a mnemonic device, we can think of the divergence as the
symbolic dot product:
~) = ∇·F
~ = ( δ , δ , δ ) · (F1 , F2 , F3 ).
div(F
δx δy δz
Physical Significance
The physical applications of the notions of curl and divergence of a vector field are
impossible to fully capture within the scope of this class (and this slide!). However, we
can give some terse indications in the context of fluid dynamics.
~ as representing the velocity field of a three-dimensional body of liquid in
Think of F
motion.
Physical Significance
The physical applications of the notions of curl and divergence of a vector field are
impossible to fully capture within the scope of this class (and this slide!). However, we
can give some terse indications in the context of fluid dynamics.
~ as representing the velocity field of a three-dimensional body of liquid in
Think of F
motion.
Imagine taking a paddle-wheel (which can spin in any direction) and fixing it at a point
~ at (x, y , z) may be imagined as the axis on which
(x, y , z). Then the curl vector of F
the fluid makes the wheel spin according to the right-hand rule: that is, if you stick
your right thumb up in the direction of the curl, the wheel will spin in the direction
that your fingers curl. The magnitude of the curl vector is how fast the wheel rotates.
Physical Significance
The physical applications of the notions of curl and divergence of a vector field are
impossible to fully capture within the scope of this class (and this slide!). However, we
can give some terse indications in the context of fluid dynamics.
~ as representing the velocity field of a three-dimensional body of liquid in
Think of F
motion.
Imagine taking a paddle-wheel (which can spin in any direction) and fixing it at a point
~ at (x, y , z) may be imagined as the axis on which
(x, y , z). Then the curl vector of F
the fluid makes the wheel spin according to the right-hand rule: that is, if you stick
your right thumb up in the direction of the curl, the wheel will spin in the direction
that your fingers curl. The magnitude of the curl vector is how fast the wheel rotates.
~ represents the expansion/compression of the fluid at a given
The divergence of F
point (x, y , z). A positive divergence corresponds to fluid expansion, i.e. the fluid is
generally moving away from the point, while a negative divergence corresponds to fluid
compression, i.e. the fluid is generally moving toward the point.
Linearity of Curl and Divergence
~, G
~ are any two vectors fields then
If F
~ +G
~ ) = curl(F
~ ) + curl(G
~ ) and
curl(F
~ +G
~ ) = div(F
~ ) + div(G
~ ),
div(F
and if c is any constant then
~ ) = c curl(F
~ ) and div(c F
~ ) = c div(F
~ ).
curl(c F
In other words curl and div are linear transformations.
Boundary Orientations
Let G (u, v ) be a smooth one-to-one parametrization with domain
D of a two-dimensional surface S in R3 . Define the boundary of
S, denoted δS, to be the image of δD under G .
We informally fix a boundary orientation on S as follows: if you
are a normal vector standing on the surface S walking along the
boundary with the correct orientation, then the surface is on your
left and the void is on your right. (See picture.)
Stokes’ Theorem
Theorem (Stokes’ Theorem)
Let S be a surface parametrized by a smooth one-to-one function
G (u, v ) with domain D, where δD is comprised of simple closed
curves. Then
I
Z Z
~
~ ) · dS.
F · ds =
curl(F
δS
S
Example
~ = (−y , 2x, x + z) and the
Verify Stokes’ theorem for the field F
upper unit hemisphere S, with outward-pointing normal vectors.
Solution
~ = (−y , 2x, x + z)
Known: F
First we will just compute the line integral about the boundary δS, which is the unit
circle in the xy -plane oriented counterclockwise. We can parametrize δS in the usual
way:
~c (t) = (cos t, sin t, 0) on the domain [0, 2π].
Solution
~ = (−y , 2x, x + z)
Known: F
First we will just compute the line integral about the boundary δS, which is the unit
circle in the xy -plane oriented counterclockwise. We can parametrize δS in the usual
way:
~c (t) = (cos t, sin t, 0) on the domain [0, 2π].
Now compute:
I
~ · ds =
F
δS
Z
2π
~ (~c (t)) · ~c 0 (t)dt
F
0
Z
2π
(− sin t, 2 cos t, cos t) · (− sin t, cos t, 0)dt
=
0
Z
2π
=
(sin2 t + 2 cos2 t)dt
0
Z
2π
=
(1 + cos2 t)dt
0
2π
3
1
+ cos 2t dt
2
2
0
2π
3
1
=
t + sin 2t
= 3π.
2
4
0
Z
=
Solution, contd.
~ = (−y , 2x, x + z)
Known: F
H
~ · ds = 3π
F
δS
Now we will compute the flux integral over S. First we parametrize the upper unit
hemisphere in the usual way:
G (θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ) on the domain [0, 2π] × [0,
π
].
2
Solution, contd.
~ = (−y , 2x, x + z)
Known: F
H
~ · ds = 3π
F
δS
Now we will compute the flux integral over S. First we parametrize the upper unit
hemisphere in the usual way:
G (θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ) on the domain [0, 2π] × [0,
π
].
2
Now compute the normal:
~ θ = (− sin θ sin φ, cos θ sin φ, 0) and T
~ φ = (cos θ cos φ, sin θ cos φ, − sin φ);
T
Solution, contd.
~ = (−y , 2x, x + z)
Known: F
H
~ · ds = 3π
F
δS
Now we will compute the flux integral over S. First we parametrize the upper unit
hemisphere in the usual way:
G (θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ) on the domain [0, 2π] × [0,
π
].
2
Now compute the normal:
~ θ = (− sin θ sin φ, cos θ sin φ, 0) and T
~ φ = (cos θ cos φ, sin θ cos φ, − sin φ);
T
~θ × T
~ φ = (cos θ sin2 φ, sin θ sin2 φ, sin φ cos φ).
~n = T
Solution, contd.
~ = (−y , 2x, x + z)
Known: F
H
~ · ds = 3π
F
δS
Now we will compute the flux integral over S. First we parametrize the upper unit
hemisphere in the usual way:
G (θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ) on the domain [0, 2π] × [0,
π
].
2
Now compute the normal:
~ θ = (− sin θ sin φ, cos θ sin φ, 0) and T
~ φ = (cos θ cos φ, sin θ cos φ, − sin φ);
T
~θ × T
~ φ = (cos θ sin2 φ, sin θ sin2 φ, sin φ cos φ).
~n = T
At this moment we verify mentally or graphically that we have chosen the correct
orientation for our normal vector ~n. (If we had accidentally parametrized S in such a
way that ~n faced the wrong direction, then we could fix the problem by just flipping
the sign on ~n.)
Solution, contd.
~ = (−y , 2x, x + z)
Known: F
H
~ · ds = 3π
F
δS
~n = (cos θ sin2 φ, sin θ sin2 φ, sin φ cos φ)
Solution, contd.
~ = (−y , 2x, x + z)
Known: F
H
~ · ds = 3π
F
δS
~n = (cos θ sin2 φ, sin θ sin2 φ, sin φ cos φ)
~ ):
Next compute curl(F
 ~
i
~ ) = det 
curl(F
=
δ
δx
~j
~k

δ
δy
δ
δz

−y 2x x + z
δ
δ
δ
δ
δ
δ
(x + z) −
(2x), − (x + z) +
(−y ),
(2x) −
(−y )
δy
δz
δx
δz
δx
δy
= (0, −1, 3).
Solution, contd.
~ = (−y , 2x, x + z)
Known: F
H
~ · ds = 3π
F
δS
~n = (cos θ sin2 φ, sin θ sin2 φ, sin φ cos φ)
~ ) = (0, −1, 3)
curl(F
Solution, contd.
~ = (−y , 2x, x + z)
Known: F
H
~ · ds = 3π
F
δS
~n = (cos θ sin2 φ, sin θ sin2 φ, sin φ cos φ)
~ ) = (0, −1, 3)
curl(F
~ ) through S and show that it equals
Finally we are ready to compute the flux of curl(F
3π as per Stokes’ theorem.
Z Z
~ ) · dS =
curl(F
π/2
Z
S
π/2
2π
Z
=
0
(0, −1, 3) · (cos θ sin2 φ, sin θ sin2 φ, sin φ cos φ)dθdφ
0
π/2
Z
~ )(G (θ, φ)) · ~n(θ, φ)dθdφ
curl(F
0
0
Z
2π
Z
Z
=
2π
(− sin θ sin2 φ + 3 sin φ cos φ)dθdφ
0
Z π/2
0
= 3 · 2π
sin φ cos φdφ
0
= 6π ·
1
sin2 φ
2
π/2
0
= 3π[1 − 0] = 3π.
Higher-Dimensional Boundary Orientations
Now let W denote a closed bounded region in R3 enclosed by a
smooth parametrized surface S = δW, the boundary of W.
We orient δW by requiring that the normal vectors ~n point
outward, away from W.
Divergence Theorem
Theorem (Divergence Theorem)
Let S = δW be a smooth parametrized surface enclosing a region
~ be a vector field defined on W. Then
W in R3 . Let F
Z Z
Z Z Z
~
~ )d(x, y , z).
F · dS =
div(F
δW
W
Example
~ = (x 2 , z 4 , e z ) and let S be the boundary of the box [0, 2] × [0, 3] × [0, 1]
Let F
~ through S.
in R3 . Use the divergence theorem to calculate the flux of F
Example
~ = (x 2 , z 4 , e z ) and let S be the boundary of the box [0, 2] × [0, 3] × [0, 1]
Let F
~ through S.
in R3 . Use the divergence theorem to calculate the flux of F
Solution.
~:
First compute the divergence of F
~) =
div(F
δ
(x 2 )
δx
+
δ
δy
(z 4 ) +
δ
(e z )
δz
= 2x + e z .
Example
~ = (x 2 , z 4 , e z ) and let S be the boundary of the box [0, 2] × [0, 3] × [0, 1]
Let F
~ through S.
in R3 . Use the divergence theorem to calculate the flux of F
Solution.
~:
First compute the divergence of F
~) =
div(F
δ
(x 2 )
δx
+
δ
δy
(z 4 ) +
δ
(e z )
δz
= 2x + e z .
Now apply the divergence theorem to convert the flux integral to a volume
integral:
Z Z
~ · dS =
F
1
Z
S
0
0
1
Z
3
Z
(2x + e z )dxdydz
0
3
Z
[x 2 + xe z ]2x=0 dydx
=
0
0
1
Z
2
Z
3
Z
(4 + 2e z )dydz
=
0
0
1
Z
(12 + 6e z )dz
=
0
= 12 + 6e − 6 = 6(1 + e).
Consequences of Stokes’ and Divergence Theorems
Fact
Conservative vector fields have zero curl. That is:
~ = ∇V , then curl(F
~ ) = ~0.
If F
Consequences of Stokes’ and Divergence Theorems
Fact
Conservative vector fields have zero curl. That is:
~ = ∇V , then curl(F
~ ) = ~0.
If F
Proof.
Cross-partials property!
Consequences of Stokes’ and Divergence Theorems
Fact
Conservative vector fields have zero curl. That is:
~ = ∇V , then curl(F
~ ) = ~0.
If F
Proof.
Cross-partials property!
Corollary
~ is a conservative vector field, then the circulation of F
~ about any simple
If F
closed curve C is 0.
Consequences of Stokes’ and Divergence Theorems
Fact
Conservative vector fields have zero curl. That is:
~ = ∇V , then curl(F
~ ) = ~0.
If F
Proof.
Cross-partials property!
Corollary
~ is a conservative vector field, then the circulation of F
~ about any simple
If F
closed curve C is 0.
Proof.
H
~ · ds is equal to some surface
Stokes’ theorem implies that the circulation C F
~
integral of the field curl(F ) = ~0, which is always 0.
(We also already know this from the fundamental theorem for conservative
vector fields.)
Consequences of Stokes’ and Divergence Theorems, contd.
Fact
Curl fields have zero divergence. That is:
~ = curl(A),
~ then div(F
~ ) = 0.
If F
Consequences of Stokes’ and Divergence Theorems, contd.
Fact
Curl fields have zero divergence. That is:
~ = curl(A),
~ then div(F
~ ) = 0.
If F
Proof.
A consequence of Clairaut’s theorem! Remember this theorem says mixed
second-order partial derivatives are equal for continuously differentiable
functions. Try working through this on your own.
Consequences of Stokes’ and Divergence Theorems, contd.
Fact
Curl fields have zero divergence. That is:
~ = curl(A),
~ then div(F
~ ) = 0.
If F
Proof.
A consequence of Clairaut’s theorem! Remember this theorem says mixed
second-order partial derivatives are equal for continuously differentiable
functions. Try working through this on your own.
Corollary
~ is the curl field of some vector field A,
~ then the flux of F
~ through any
If F
closed surface is 0.
Consequences of Stokes’ and Divergence Theorems, contd.
Fact
Curl fields have zero divergence. That is:
~ = curl(A),
~ then div(F
~ ) = 0.
If F
Proof.
A consequence of Clairaut’s theorem! Remember this theorem says mixed
second-order partial derivatives are equal for continuously differentiable
functions. Try working through this on your own.
Corollary
~ is the curl field of some vector field A,
~ then the flux of F
~ through any
If F
closed surface is 0.
Proof.
~ is equal to a volume integral of
The divergence theorem says that the flux of F
~
the function div(F ), again a 0 function which gives a 0 integral.
Summary of Vector Field Operations
f
real-valued function
∇
→
~
F
vector field
curl
→
~
G
vector field
div
→
g
real-valued function
Doing any two consecutive operations in a row results in zero, i.e. curl(∇f ) = ~0 and
~ )) = 0.
div(curl(F
Thanks
Thanks for a great semester!