18 Chapter 1 Vector Analysis

Example 1.4. Suppose the functions in Fig. 1.18 are va = r = x x̂ + y ŷ + z ẑ,

vb = ẑ, and vc = z ẑ. Calculate their divergences.
Solution
∂ ∂ ∂
∇ · va = (x) + (y) + (z) = 1 + 1 + 1 = 3.
∂x ∂y ∂z
As anticipated, this function has a positive divergence.

∂ ∂ ∂
∇ · vb = (0) + (0) + (1) = 0 + 0 + 0 = 0,
∂x ∂y ∂z
as expected.
∂ ∂ ∂
∇ · vc = (0) + (0) + (z) = 0 + 0 + 1 = 1.
∂x ∂y ∂z

Problem 1.15 Calculate the divergence of the following vector functions:

(a) va = x 2 x̂ + 3x z 2 ŷ − 2x z ẑ.

(b) vb = x y x̂ + 2yz ŷ + 3zx ẑ.

(c) vc = y 2 x̂ + (2x y + z 2 ) ŷ + 2yz ẑ.

• Problem 1.16 Sketch the vector function

r̂
v= ,
r2
and compute its divergence. The answer may surprise you. . . can you explain it?

! Problem 1.17 In two dimensions, show that the divergence transforms as a scalar
under rotations. [Hint: Use Eq. 1.29 to determine v y and v z , and the method of
Prob. 1.14 to calculate the derivatives. Your aim is to show that ∂v y /∂ y + ∂v z /∂z =
∂v y /∂ y + ∂vz /∂z.]

1.2.5 The Curl

From the definition of ∇ we construct the curl:
 
 x̂ ŷ ẑ 

∇ × v =  ∂/∂ x ∂/∂ y ∂/∂z 
 vx vy vz 
     
∂vz ∂v y ∂vx ∂vz ∂v y ∂vx
= x̂ − + ŷ − + ẑ − . (1.41)
∂y ∂z ∂z ∂x ∂x ∂y
1.2 Differential Calculus 19

z
z

y y

x
(a) x (b)

FIGURE 1.19

Notice that the curl of a vector function7 v is, like any cross product, a vector.
Geometrical Interpretation: The name curl is also well chosen, for ∇ × v is
a measure of how much the vector v swirls around the point in question. Thus
the three functions in Fig. 1.18 all have zero curl (as you can easily check for
yourself), whereas the functions in Fig. 1.19 have a substantial curl, pointing in the
z direction, as the natural right-hand rule would suggest. Imagine (again) you are
standing at the edge of a pond. Float a small paddlewheel (a cork with toothpicks
pointing out radially would do); if it starts to rotate, then you placed it at a point
of nonzero curl. A whirlpool would be a region of large curl.

Example 1.5. Suppose the function sketched in Fig. 1.19a is va = −y x̂ + x ŷ,

and that in Fig. 1.19b is vb = x ŷ. Calculate their curls.
Solution
 
 x̂ ŷ ẑ 
 
∇ × va =  ∂/∂ x ∂/∂ y ∂/∂z  = 2ẑ,

 −y x 0 

and
 
 x̂ ŷ ẑ 
 
∇ × vb =  ∂/∂ x ∂/∂ y ∂/∂z  = ẑ.

 0 x 0 

As expected, these curls point in the +z direction. (Incidentally, they both have
zero divergence, as you might guess from the pictures: nothing is “spreading
out”. . . it just “swirls around.”)

7 There’s no such thing as the curl of a scalar.

20 Chapter 1 Vector Analysis

Problem 1.18 Calculate the curls of the vector functions in Prob. 1.15.

Problem 1.19 Draw a circle in the x y plane. At a few representative points draw
the vector v tangent to the circle, pointing in the clockwise direction. By comparing
adjacent vectors, determine the sign of ∂vx /∂ y and ∂v y /∂ x. According to Eq. 1.41,
then, what is the direction of ∇ × v? Explain how this example illustrates the geo-
metrical interpretation of the curl.

Problem 1.20 Construct a vector function that has zero divergence and zero curl
everywhere. (A constant will do the job, of course, but make it something a little
more interesting than that!)

1.2.6 Product Rules

The calculation of ordinary derivatives is facilitated by a number of rules, such as
the sum rule:
d df dg
( f + g) = + ,
dx dx dx
the rule for multiplying by a constant:
d df
(k f ) = k ,
dx dx
the product rule:
d dg df
( f g) = f +g ,
dx dx dx
and the quotient rule:

  g d f − f dg
d f
= dx 2 dx .
dx g g
Similar relations hold for the vector derivatives. Thus,
∇( f + g) = ∇ f + ∇g, ∇ · (A + B) = (∇ · A) + (∇ · B),

∇ × (A + B) = (∇ × A) + (∇ × B),
and
∇(k f ) = k∇ f, ∇ · (kA) = k(∇ · A), ∇ × (kA) = k(∇ × A),
as you can check for yourself. The product rules are not quite so simple. There
are two ways to construct a scalar as the product of two functions:
fg (product of two scalar functions),
A·B (dot product of two vector functions),