Post on 01-Jan-2016
Curl and Divergence
The curl of the vector field is
Let be a vector field on and assume the partial derivatives of , and all exist.
Curl
curl = =x y zP Q R
i j k
F F
Let (the “del” operator)
, ,y z z x x yR Q P R Q P
Example 1: Let Find curl F.2 2 2 2 .( , , ) , 2 ,x y z xyz x yz x y z F
2 2 2 22
=
yz y z
x y zxyz x x
i j k
F
2 2 2 2 2 2 2 2( ) ( 2 ) ( ) ( ) ( 2 ) ( )x y z x yz x y z xyz x yz xyzy z x z x y
i j k
2 2 2 2y y x xy x xz i j k 0, 2 ,2xy x x xz
P Q R
Clairaut’s Theorem
Curl
Example 2: Let F. Evaluate curl
A field with is called curl free or irrotational.
=
y
x y zx z
i j k
F ( ) ( ) ( ) ( ) ( ) ( )z y z x y xy z x z x y
i j k
0,0,0
Theorem 1: A conservative vector field with continuous derivatives is irrotational.
Proof: Suppose is a conservative vector field and let be its potential function:
= , , 0,0,0zy yz xz zx yx xy
x y z
f f f f f fx y zf f f
i j k
F
Curl
Theorem 2: If is a vector field on all of with continuous partial derivatives and then is a conservative vector field.
REMARKS: 1. Theorem 2 provides a quick and easy way to check whether a 3D vector
field is conservative.
2. The 2D vector field can be written as a 3D vector field . Then . Therefore, in 2D, is equivalent to
3. The Fundamental Theorem of Line Integrals holds also in 3D for conservative vector fields.
The converse of Theorem 1 is also true if we assume the domain of the vector field is simply connected (in particular, the whole
¿ 𝑦+6 𝑧
Curl – Example 3
𝑓 (𝑥 , 𝑦 , 𝑧 )=𝑥𝑦+𝑔(𝑦 , 𝑧)Integrate w.r.t
differentiate w.r.t 𝑓 𝑦=𝑥+𝑔𝑦(𝑦 ,𝑧 )
𝑔 𝑦 (𝑦 ,𝑧 )=𝑧 g (𝑦 , 𝑧 )=𝑧𝑦+h(𝑧)
¿ 𝑥+ 𝑧
differentiate w.r.t h ′ (𝑧 )=6 𝑧 h (𝑧 )=3 𝑧 2+𝐾
a) S is conservative and find a potential function.
curl = =
6x y zy x z y z
i j k
F F 1 1 0 1 1) i j k 0
is irrotational, therefore there exists a potential function such that
Curl - Example 3 continued
b) Evaluate the line integral of along the curve comprised of the line segment from (0,0,0) to (1,1,2) and the line segment from (1,1,2) to (3,1,4)
Initial point: (0,0,0) Terminal point: (3,1,4)
Potential Function:
Fundamental Theorem for Line Integrals:
(3,1,4) (0,0,0)C
d f f F r
23(1) 4(1) 3 4 0 55
(0,0,0)
(a)
Interpreting the curl: If we think of the vector field as a velocity vector field of a fluid in motion, the curl measures rotation.
Based on the direction of the rotation and the Right Hand Rule, the curl will point in the direction.
At a given point, the curl is a vector parallel to the axis of rotation of flow lines near the point, with direction determined by the Right Hand Rule.
Example 4: Determine whether the curl of each vector field at the origin is the zero vector or points in the same direction as ± i, ± j, ± k.
Curl
(b)
Based on the direction of the rotation and the Right Hand Rule, the curl points in the direction of
(c)
The vector field is clearly irrotational, thus the curl is the zero vector: .
Curl – Example 4 continued
The divergence of the vector field is (a scalar)
Let be a differentiable vector field on a region in .
Divergence
, , , ,div = = P Q RP Q Rx y z x y z
F F
( , , ) ( , , ), ( , , ), ( , , )F x y z P x y z Q x y z R x y z
Let (the “del” operator)
Example 5: Let Find div F.
2 2 2( sin ) ( )xyze yz x x y zx y z
F sin 2xyzyze z x z
Example 6: Let Find div F.
( ) ( ) 1 1 1 3x y zx y z
F
Example 7: Determine whether the divergence of each vector field at the indicated point P is positive, negative or zero
(a) Draw a box around the point P.
Thus the divergence is negative and we say the point P is a sink.
We can clearly see from the magnitude of the vectors that the net flow out of the box is negative (more “stuff” coming in than out).
Interpreting the Divergence: The divergence at a given point measures the net flow out of a small box around the point, that is, it measures what is produced (source) or consumed (sink) at a given point in space.
Divergence
(b)
Draw a box around the point. The net flow is positive (less “stuff” coming in than out).
Thus the divergence is zero and we say that the vector field is incompressible at the point.
(c)
Draw a box around the point. We can see that the net flow out of the box is zero (same amount of “stuff” is coming in than coming out).
Thus the divergence is positive and we say that the point is a source.
Divergence – Example 7 continued