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Math 21a Curl and Divergence Spring, 2009
Define the operator ∇ (pronounced “del”) by
∇ = i
∂x
+ j
∂y
+ k
∂z
Notice that the gradient ∇f (or also grad f ) is just ∇ applied to f.
(a) We define the divergence of a vector field F, written div F or ∇ · F, as the dot product of
del with F. So if F = 〈P, Q, R〉, then
div F = ∇ · F =
∂x
∂y
∂z
· 〈P, Q, R〉 =
∂P
∂x
∂Q
∂y
∂R
∂z
Notice that div F is a scalar.
Find div F for each of the following vector fields:
(i) F = 〈xy, yz, xz〉 (ii) F = 〈yz, xz, xy〉
(iii) F =
x
r
y
r
z
r
where r =
x
+ y
+ z
(iv) F = grad f , where f is a function with
continuous second derivatives
(b) We define the curl of a vector field F, written curl F or ∇ × F, as the cross product of del
with F. So if F = 〈P, Q, R〉, then
curl F = ∇ × F =
∂x
∂y
∂z
× 〈P, Q, R〉 =
i j k
∂x
∂y
∂z
P Q R
∂R
∂y
∂Q
∂z
i +
∂P
∂z
∂R
∂x
j +
∂Q
∂x
∂P
∂y
k.
Notice that curl F is a vector.
Find curl F for each of the following vector fields:
(i) F = 〈xy, yz, xz〉 (ii) F = 〈yz, xz, xy〉
(iii) F =
x
r
y
r
z
r
where r =
x
+ y
+ z
(iv) F = grad f , where f is a function with
continuous second derivatives
2
Check the appropriate box (“Vector”, “Scalar”, or “Nonsense”) for each quantity.
Quantity Vector Scalar Nonsense
curl(∇f )
∇ · (∇ × F)
div(curl f )
curl(curl F)
∇(∇ · F)
Quantity Vector Scalar Nonsense
curl(curl f )
grad(div f )
div(grad F)
∇ · (∇f )
div(div F)
3
(a) Suppose we have a vector 〈P, Q〉 that we extend to a vector in space: F = 〈P (x, y), Q(x, y), 0 〉.
Find curl F.
(b) When is the vector field F in part (a) conservative? Use the curl in your answer.
(c) If F = ∇f = 〈f
x
, f
y
, f
z
〉 is conservative, then is curl F = 0? (See Problem 1(b)(iv).)
In fact, we can identify conservative vector fields with the curl:
Theorem: Let F be a vector field defined on all of R
. If the component
functions of F all have continuous derivatives and curl F = 0 , then F is a
conservative vector field.
4
Show that, if F is a vector field on R
with components that have continuous second-order
derivatives, then div curl F = 0. (This is less useful for us right now than curl grad f = 0 , but
it’s not a difficult computation.)
Curl and Divergence – Answers and Solutions
(a) (i) div F = x + y + z
(ii) div F = 0
(iii) div F =
y
+z
r
x
+z
r
x
+y
r
2(x
+y
+z
r
r
(iv) div F = div(grad f ) = f
xx
+ f
yy
+ f
zz
. This is the Laplace operator applied to f.
(b) (i) curl F = 〈−y, −z, −x〉
(ii) curl F = 0
(iii) curl F = 0
(iv) curl F = curl(grad f ) = 0.
Two quick comments:
• Notice that the vector from part (ii) is actually an example of this, since F =
grad(xyz) = 〈yz, xz, xy〉. The vector for part (iii) is F = grad(
x
+ y
+ z
), so
it is another example as well.
• We’ll see later on in this worksheet that curl F = 0 precisely when F = grad f (when
we make some continuity assumptions about the derivatives of the components of
F.
Here are some answers, although half are blank as they match questions from the homework:
Quantity Vector Scalar Nonsense
curl(∇f ) x
∇ · (∇ × F) x
div(curl f ) x
curl(curl F)
∇(∇ · F)
Quantity Vector Scalar Nonsense
curl(curl f ) x
grad(div f )
div(grad F) x
∇ · (∇f )
div(div F)
(a) curl F =
∂Q
∂x
∂P
∂y
= 〈 0 , 0 , Q
x
− P
y
(b) A planar vector field F = 〈P, Q〉 is conservative when Q
x
− P
y
= 0. This means there’s a
function f (x, y) with P = f
x
and Q = f
y
. We also get f
z
= 0 since f is a function of only
x and y, so F = 〈P (x, y), Q(x, y), 0 〉 is conservative under this same condition. From part
(a), this means that F is conservative when curl F = 0.
(c) Yep, we’ve already seen that curl grad f = 0.
Here’s a quick detailed computation:
div curl F = div (∇ × 〈P, Q, R〉)
[(
∂R
∂y
∂Q
∂z
i +
∂P
∂z
∂R
∂x
j +
∂Q
∂x
∂P
∂y
k
]
∂x
∂R
∂y
∂Q
∂z
∂y
∂P
∂z
∂R
∂x
∂z
∂Q
∂x
∂P
∂y
R
∂x ∂y
Q
∂x ∂z
P
∂y ∂z
R
∂y ∂x
Q
∂z ∂x
P
∂z ∂y
= (R
yx
− Q
zx
) + (P
zy
− R
xy
) + (Q
xz
− P
yz
5
(a) We do this both ways, of course. First let’s parameterize C by r(t) = 〈x, y〉 = 〈cos(t), sin(t)〉,
so dx = − sin(t) dt and dy = cos(t) dt. Thus
∮
C
F · dr =
∫
〈− sin(t), cos(t)〉 · 〈− sin(t), cos(t)〉 dt =
∫
1 dt = 2π.
The double integral is also straightforward: writing F = 〈−y, x, 0 〉 as a vector field in space,
we get curl F = 〈 0 , 0 , 2 〉. Thus, since k = 〈 0 , 0 , 1 〉, we get
∫ ∫
D
(curl F) · k dA =
∫ ∫
D
2 dA = 2 · Area(D) = 2π.
The normal component version requires us to have a unit outward-pointing normal n. For
the unit circle, the unit normal at the point (x, y) is simply n = 〈x, y〉. (One way to see
this is to parameterize the circle as r(t) = 〈x(t), y(t)〉 = 〈cos(t), sin(t)〉, so by our formula n
is the unit vector in the direction 〈y
(t), −x
(t)〉 = 〈cos(t), sin(t)〉. But this is a unit vector
already, and in fact it is the vector 〈x, y〉.) Thus F · n = 〈−y, x〉 · 〈x, y〉 = 0, so the line
integral is zero. Note also that div F =
∂x
(−y) +
∂y
(x) = 0, so the double integral is also
zero.
(b) Again we do this both ways. We’ll parameterize C in four parts (all with 0 ≤ t ≤ 1):
C
: r(t) = 〈x, y〉 = 〈t, 0 〉 C
: r(t) = 〈x, y〉 = 〈 1 , t〉
C
: r(t) = 〈x, y〉 = 〈 1 − t, 1 〉 C
: r(t) = 〈x, y〉 = 〈 0 , 1 − t〉
From this we get four integrals:
∫
C
F · dr =
∫
C
〈 2 x, 2 y〉 · 〈dx, dy〉 =
∫
〈 2 t, 0 〉 · 〈dt, 0 〉 =
∫
2 t dt = 1
∫
C
F · dr =
∫
C
〈 2 x, 2 y〉 · 〈dx, dy〉 =
∫
〈 2 , 2 t〉 · 〈 0 , dt〉 =
∫
2 t dt = 1
∫
C
F · dr =
∫
C
〈 2 x, 2 y〉 · 〈dx, dy〉 =
∫
〈2(1 − t), 2 〉 · 〈−dt, 0 〉 =
∫
2(t − 1) dt = − 1
∫
C
F · dr =
∫
C
〈 2 x, 2 y〉 · 〈dx, dy〉 =
∫
〈 0 , 2(1 − t)〉 · 〈 0 , −dt〉 =
∫
2(t − 1) dt = − 1.
Thus ∫
C
F · dr =
∫
C
∫
C
∫
C
∫
C
= 1 + 1 − 1 − 1 = 0.
The double integral is considerably easier: writing F = 〈 2 x, 2 y, 0 〉 as a vector field in space,
we compute curl F = 0 , so the integrand in the double integral is zero. Thus the double
integral is zero as well.
The normal component version requires us to have a unit outward-pointing normal n. These
are simple to find for each component (using the above parameterization):
C
: n = −j = 〈 0 , − 1 〉 C
: n = i = 〈 1 , 0 〉,
C
: n = j = 〈 0 , 1 〉, C
: n = −i = 〈− 1 , 0 〉.
In the case of the first sketch, it’s clear that F is mostly parallel to the tangent vector to the
small circle C, so F · T > 0. Thus
∮
C
F · T ds > 0, so by Green’s theorem,
(curl F(0, 0 , 0)) · k > 0.
The moral: the curl of F is non-zero means that there is some kind of rotation in the vector
field. We’ll see much more of this later.
We could do something similar with the divergence. Let’s cut straight to the chase: the (outward-
pointing) normal n produces a positive divergence at the origin. This gives us what we call a
source (when div F > 0) at the origin; if the vector fields were pointing in we’d get a sink (and
div F < 0).
