3.33pt

Curl and DivergenceMATH 311, Calculus III

J. Robert Buchanan

Department of Mathematics

Spring Summer 2019

Curl

DefinitionLet F(x , y , z) = F1(x , y , z)i + F2(x , y , z)j + F3(x , y , z)k, thenthe curl of F(x , y , z) is the vector field

curlF =

(∂F3

∂y− ∂F2

∂z

)i +

(∂F1

∂z− ∂F3

∂x

)j +

(∂F2

∂x− ∂F1

∂y

)k.

Often the curl is denoted

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

F1 F2 F3

∣∣∣∣∣∣

Curl

DefinitionLet F(x , y , z) = F1(x , y , z)i + F2(x , y , z)j + F3(x , y , z)k, thenthe curl of F(x , y , z) is the vector field

curlF =

(∂F3

∂y− ∂F2

∂z

)i +

(∂F1

∂z− ∂F3

∂x

)j +

(∂F2

∂x− ∂F1

∂y

)k.

Often the curl is denoted

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

F1 F2 F3

∣∣∣∣∣∣

Example

Compute the curl of

F(x , y , z) = xzi + xyzj− y2k.

∇× F = 〈−2y − xy , x , yz〉

Example

Compute the curl of

F(x , y , z) = xzi + xyzj− y2k.

∇× F = 〈−2y − xy , x , yz〉

Graphical Interpretation

Suppose F(x , y , z) = 〈x , y ,0〉, then the vector field resembles:

. -1.0 -0.5 0.0 0.5 1.0-1.0-0.50.00.51.0

x

y

∇× F = 〈0,0,0〉

When ∇× F = 0 we say the vector field is irrotational.

Right Hand Rule

Suppose F(x , y , z) = 〈−y , x ,0〉, then the vector fieldresembles:

. -1.0 -0.5 0.0 0.5 1.0-1.0-0.50.00.51.0

x

y

∇× F = 〈0,0,2〉

∇ × F at a point is a vector parallel to the axis of rotation of theflow lines.

Divergence

DefinitionLet F(x , y , z) = F1(x , y , z)i + F2(x , y , z)j + F3(x , y , z)k, thenthe divergence of F(x , y , z) is the scalar function

divF =∂F1

∂x+

∂F2

∂y+

∂F3

∂z.

Often the divergence is denoted ∇ · F.

Divergence

DefinitionLet F(x , y , z) = F1(x , y , z)i + F2(x , y , z)j + F3(x , y , z)k, thenthe divergence of F(x , y , z) is the scalar function

divF =∂F1

∂x+

∂F2

∂y+

∂F3

∂z.

Often the divergence is denoted ∇ · F.

Example

Find the divergence of

F(x , y , z) = 〈ex sin y ,ex cos y , z〉

∇ · F = 1

Example

Find the divergence of

F(x , y , z) = 〈ex sin y ,ex cos y , z〉

∇ · F = 1

Graphical Interpretation

Suppose F(x , y , z) = 〈x , y ,0〉, then the vector field resembles:

. -1.0 -0.5 0.0 0.5 1.0-1.0-0.50.00.51.0

x

y

∇ · F = 2

When ∇ · F > 0 at a point we say the point is a source.

Incompressible Flow

Suppose F(x , y , z) = 〈−y , x ,0〉, then the vector fieldresembles:

. -1.0 -0.5 0.0 0.5 1.0-1.0-0.50.00.51.0

x

y

∇ · F = 0

When ∇ · F = 0 we say the vector field is incompressible ordivergence free.

Sink

Suppose F(x , y , z) = 〈−x ,−y ,0〉, then the vector fieldresembles:

. -1.0 -0.5 0.0 0.5 1.0-1.0-0.50.00.51.0

x

y

∇ · F = −2

When ∇ · F < 0 at a point we say the point is a sink.

Divergence, Gradient, and Curl

ExampleSuppose F(x , y , z) is a vector field and f (x , y , z) is a scalarfunction. Determine whether the following are vector fields,scalar functions, or undefined operations.

1. ∇× (∇f )

vector field

2. ∇× (∇ · F)

undefined operation

3. ∇ · (∇f )

scalar function

Divergence, Gradient, and Curl

ExampleSuppose F(x , y , z) is a vector field and f (x , y , z) is a scalarfunction. Determine whether the following are vector fields,scalar functions, or undefined operations.

1. ∇× (∇f ) vector field2. ∇× (∇ · F) undefined operation3. ∇ · (∇f ) scalar function

Remarks

Remarks:I If f (x , y , z) is a scalar function∇ · (∇f ) = ∇2f = ∆f = fxx + fyy + fzz is called theLaplacian of f .

I If f (x , y , z) is a scalar function then

∇×(∇f ) =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

fx fy fz

∣∣∣∣∣∣ = 〈fzy−fyz , fxz−fzx , fyx−fxy 〉 = 0

i.e. the curl of the gradient is always the zero vector.

Remarks

Remarks:I If f (x , y , z) is a scalar function∇ · (∇f ) = ∇2f = ∆f = fxx + fyy + fzz is called theLaplacian of f .

I If f (x , y , z) is a scalar function then

∇×(∇f ) =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

fx fy fz

∣∣∣∣∣∣ = 〈fzy−fyz , fxz−fzx , fyx−fxy 〉 = 0

i.e. the curl of the gradient is always the zero vector.

Three-dimensional Conservative Vector Fields

TheoremSuppose that F(x , y , z) = 〈F1(x , y , z),F2(x , y , z),F3(x , y , z)〉 isa vector field whose component functions have continuous firstpartial derivatives in an open region D ⊂ R3. If F isconservative then ∇× F = 0.

Note: the converse of this theorem is not true. If ∇× F = 0 itdoes not necessarily mean that F is conservative.

Three-dimensional Conservative Vector Fields

TheoremSuppose that F(x , y , z) = 〈F1(x , y , z),F2(x , y , z),F3(x , y , z)〉 isa vector field whose component functions have continuous firstpartial derivatives in an open region D ⊂ R3. If F isconservative then ∇× F = 0.

Note: the converse of this theorem is not true. If ∇× F = 0 itdoes not necessarily mean that F is conservative.

Example (1 of 2)

Determine whether the following vector field is conservative.

F(x , y , z) = 〈xz, xyz,−y2〉

Since ∇× F = 〈−(2 + x)y , x , yz〉 6= 0 then we know F is notconservative.

Example (1 of 2)

Determine whether the following vector field is conservative.

F(x , y , z) = 〈xz, xyz,−y2〉

Since ∇× F = 〈−(2 + x)y , x , yz〉 6= 0 then we know F is notconservative.

Example (2 of 2)

Determine whether the following vector field is conservative.

F(x , y , z) = 〈2xy , x2 − 3y2z2,1− 2y3z〉

We can see that ∇× F = 0, but this does not prove F isconservative.

If f (x , y , z) = x2 − y3z2 + z then we see that∇f (x , y , z) = F(x , y , z) which does show that F is conservative.

Example (2 of 2)

Determine whether the following vector field is conservative.

F(x , y , z) = 〈2xy , x2 − 3y2z2,1− 2y3z〉

We can see that ∇× F = 0, but this does not prove F isconservative.

If f (x , y , z) = x2 − y3z2 + z then we see that∇f (x , y , z) = F(x , y , z) which does show that F is conservative.

Conservative Vector Fields Revisited

TheoremSuppose that F(x , y , z) = 〈F1(x , y , z),F2(x , y , z),F3(x , y , z)〉 isa vector field whose component functions have continuous firstpartial derivatives in all of R3. Then F is conservative if and onlyif ∇× F = 0.

Conservative Vector Fields

Summary: If F(x , y , z) = 〈F1(x , y , z),F2(x , y , z),F3(x , y , z)〉 isa vector field whose component functions have continuous firstpartial derivatives throughout R3 then the following statementsare equivalent.

1. F(x , y , z) is conservative.2.∫

C F · dr is independent of path.3.∫

C F · dr = 0 for every piecewise-smooth closed curve C.4. ∇× F = 0.5. F = ∇f for some potential function f .

Connection with Green’s Theorem

Green’s Theorem:∮C

M(x , y) dx + N(x , y) dy =

∫∫R

(∂N∂x− ∂M

∂y

)dA

Suppose F(x , y , z) = 〈M(x , y),N(x , y),0〉, then

∇× F =

(∂N∂x− ∂M

∂y

)k which implies

(∇× F) · k =

(∂N∂x− ∂M

∂y

)k · k =

∂N∂x− ∂M

∂y.

Hence the vector form of Green’s Theorem can be written as∮C

F · dr =

∫∫R

(∇× F) · k dA.

Connection with Green’s Theorem

Green’s Theorem:∮C

M(x , y) dx + N(x , y) dy =

∫∫R

(∂N∂x− ∂M

∂y

)dA

Suppose F(x , y , z) = 〈M(x , y),N(x , y),0〉, then

∇× F =

(∂N∂x− ∂M

∂y

)k which implies

(∇× F) · k =

(∂N∂x− ∂M

∂y

)k · k =

∂N∂x− ∂M

∂y.

Hence the vector form of Green’s Theorem can be written as∮C

F · dr =

∫∫R

(∇× F) · k dA.

Yet Another Version

Suppose simple closed curve C is parameterized byr(t) = 〈x(t), y(t)〉 for a ≤ t ≤ b, then the outward unit normalvector is

n(t) =1

‖r′(t)‖〈y ′(t),−x ′(t)〉

and∮C

F · nds =

∫ b

a

[M(x(t), y(t))y ′(t)

‖r′(t)‖− N(x(t), y(t))x ′(t)

‖r′(t)‖

]‖r′(t)‖dt

=

∮C

M(x , y) dy − N(x , y) dx

=

∫∫R

(∂M∂x

+∂N∂y

)dA

=

∫∫R∇ · F(x , y) dA

Homework

I Read Section 14.5.I Exercises: 1–59 odd