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!