Section 14.5: Divergence and Curl
Divergence and curl are two operations on vector fields that are used frequently in the
study of fluid flow.
Definition: Consider a vector field
F~ (x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i.
The divergence of F~ is given by
divF~ =
∂Q ∂R
∂P
+
+
,
∂x
∂y
∂z
and the curl of F~ is given by
curlF~ =
∂R ∂Q ∂P
∂R ∂Q ∂P
−
,
−
,
−
∂y
∂z ∂z
∂x ∂x
∂y
.
Note: The divergence of a vector field is a scalar function and the curl is a vector field. These
operations can be expressed in terms of the del operator
∂ ∂ ∂
∇=
, ,
.
∂x ∂y ∂z
In particular,
∂
∂
∂
∂P
∂Q ∂R
divF~ = ∇ · F~ =
, ,
· hP, Q, Ri =
+
+
.
∂x ∂y ∂z
∂x
∂y
∂z
~k
~i
~j
∂
∂
∂
.
curlF~ = ∇ × F~ = ∂x
∂y
∂z
P (x, y, z) Q(x, y, z) R(x, y, z) Example: Find the divergence and curl of F~ (x, y, z) = hx2 yz, xy 2 z, xyz 2 i.
The divergence of F~ is
divF~ = 2xyz + 2xyz + 2xyz = 6xyz.
The curl of F~ is
~i
~k
~j
∂
∂
∂
curlF~ = ∂y
∂z
∂x
x2 yz xy 2 z xyz 2
= hxz 2 − xy 2 , x2 y − yz 2 , y 2 z − x2 zi.
Example: Find the divergence and curl of F~ = hexz , −2eyz , 3xey i.
The divergence of F~ is
divF~ = zexz − 2zeyz .
The curl of F~ is
curlF~ = ~k
~i
~j
∂
∂
∂
∂x
∂y
∂z
exz −2eyz 3xey
= h3xey + 2yeyz , xexz − 3ey , 0i.
Definition: A vector field F~ is called conservative if it is the gradient of some scalar function.
That is, if there exists a function f such that ∇f = F~ . In this case, the function f is called
a potential function or scalar potential for F~ .
Definition: If any two points A and B in a plane region D can be joined by a piecewise-smooth
curve that lies entirely in D, then the region D is called connected. A simply connected
plane region D is a region that is connected and every closed curve in D encloses only points
that are also in D. That is, a simply connected region has no holes.
Theorem: (Cross-Partials Test for a Conservative Vector Field)
Consider the vector field F~ (x, y) = hP (x, y), Q(x, y)i, where P and Q have continuous firstorder partial derivatives in the open, simply connected plane region D. Then F~ (x, y) is
conservative in D if and only if
∂Q
∂P
=
.
∂y
∂x
Example: Show that the vector field F~ (x, y) = hex sin y − y, ex cos y − x − 2i is conservative
and find a scalar potential for F~ .
The partial derivatives of the component functions are
∂P
∂Q
= ex cos y − 1 =
.
∂y
∂x
Thus, F~ is a conservative vector field. Note that if f is a scalar potential of F~ , then
fx (x, y) = ex sin y − y and fy (x, y) = ex cos y − x − 2. Then
Z
f (x, y) =
fx (x, y)dx
Z
=
ex sin y − ydx
= ex sin y − xy + K(y).
Now let fy (x, y) = ex cos y − x − 2. That is,
ex cos y − x + K 0 (y) = ex cos y − x − 2
K 0 (y) = −2
K(y) = −2y + C.
Thus, a scalar potential for F~ is
f (x, y) = ex sin y − xy − 2y.
Theorem: (The Curl Criterion for a Conservative Vector Field)
Suppose the vector fields F~ and curlF~ are continuous in R3 . Then F~ is conservative if and
only if curlF~ = ~0.
Example: Show that the vector field F~ (x, y, z) = h20x3 z + 2y 2 , 4xy, 5x4 + 3z 2 i is conservative
and find a scalar potential for F~ .
The curl of F~ is
~k
~i
~j
∂
∂
∂
curlF~ = ∂x
∂y
∂z
20x3 z + 2y 2 4xy 5x4 + 3z 2
= ~0.
Thus, F~ is a conservative vector field. Note that if f is a scalar potential of F~ , then
fx = 20x3 z + 2y 2
fy = 4xy
fz = 5x4 + 3z 2 .
Then,
Z
f (x, y, z) =
20x3 z + 2y 2 dx = 5x4 z + 2xy 2 + K1 (y, z).
Now let fy (x, y, z) = 4xy. That is
∂K1
= 4xy
∂y
∂K1
= 0
∂y
K1 (y, z) = K2 (z).
4xy +
Now let fz (x, y, z) = 5x4 + 3z 2 . That is
5x4 + K20 (z) = 5x4 + 3z 2
K20 (z) = 3z 2
K2 (z) = z 3 + C.
Thus, a scalar potential for F~ is
f (x, y, z) = 5x4 z + 2xy 2 + z 3 .