The vector differential operator , called "del" or "nabla", is defined in three dimensions to be:

= /∂xi + /∂yj + /∂zk

- where the derivatives are partial, i.e. with respect to the x, y and z axes, and i, j, k are the unit vectors in the x, y and z directions, respectively.

If a scalar function, ƒ(x, y, z), is defined and differentiable at all points in some region, then ƒ is a differentiable scalar field. The del vector operator, , may be applied to scalar fields and the result, ƒ, is a vector field. It is called the gradient of ƒ (see the module on Gradients and Directional Derivatives).

Quiz 1: As a revision exercise, what is the correct gradient of the scalar field ƒ(x, y, z) = xy2yz?

(a) i + (2xz)jyk Incorrect - please try again!
(b) 2xyi + 2xyj + yk Incorrect - please try again!
(c) y2i − 2zjyk Incorrect - please try again!
(d) y2i + (2xyz)jyk Correct - well done!

Solution: If the scalar field is ƒ(x, y, z) = xy2yz, its gradient is:

 ∇ƒ = ∂/∂x(xy2 − yz) + ∂/∂y(xy2 − yz)j + ∂/∂z(xy2 − yz)k = y2 × ∂/∂x(x)i + [x × ∂/∂y (y2) − z × ∂/∂y (y]j + (−y) × ∂/∂z(z)k = y2i + (2xy − z)j − yk

The vector operator may also be allowed to act upon vector fields. Two different ways in which it may act, the subject of this module, are extremely important in mathematics, science and engineering. First though, we will briefly review some useful properties of vectors.

Consider the (three dimensional) vector, a = a1i + a2j + a2k, which we may also write as a = (a1, a2, a3). If we multiply it by a constant c, then every component of the vector is multiplied by c:

ca = ac = (ca1, ca2, ca3)

If we introduce a second vector, b = (b1, b2, b3), then we recall that there are two different ways of multiplying vectors together, the scalar and vector products.

The scalar product (also called dot product) is defined by:

ab = a1b1 + a2b2 + a3b3

It is a scalar (as the name 'scalar product' implies).

Quiz 2: What is the scalar product of a= (1, 2, 3) and b= (3, −2, 1)?

(a)2 Correct - well done!
(b)10 Incorrect - please try again!
(c) 3x − 4y + 3z Incorrect - please try again!
(d)4

Solution: The scalar product of the two vectors a= (1, 2, 3) and b = (3, −2, 1) is:

 a⋅b = a1b1 + a2b2 = 1 × 3 + 2 × (−2) + 3 × 1 = 3 − 4 + 3 = 2

The vector product (or cross product) is defined by:

a × b = (a2b3a3b2)i − (a1b3a3b1)j + (a1b2a2b1)k
=

 i j k a1 a2 a3 b1 b2 b3

It is a vector (as the name 'vector product' implies). Note that the second line (the determinant representation) is a useful shorthand for the first.

Quiz 3: What is the vector product of a = (1, 2, 3) and b = (3, −2, 1)?

(a) 8i − 8j − 8k Incorrect - please try again!
(b) −4i − 10j + 4k Incorrect - please try again!
(c) 8i + 8j − 8k Correct - well done!
(d) 8i − 10j − 8k

Solution: The vector product of the two vectors a= (1, 2, 3) and b= (3, −2, 1) is:

a×b =

 i j k a1 a2 a3 b1 b2 b3

=

 i j k 1 2 3 3 −2 1

= (2 × 1 − 3 × (−2))i − (1 × 1 − 3 × 3)j + (1 × (−2) − 2 × 3)k
= (2 + 6)i − (1 − 9) j + (−2 − 6)k
= 8i − (−8) j − 8k = 8i + 8 j − 8k
= 8i + 8 j − 8k

# 2. Divergence (Div)

If F(x, y) is a vector field, then its divergence is written as div F(x, y) = F(r), which in two dimensions is:

F(x, y) = F(r) = ( /∂xi + /∂yj ) (F1(x, y)i + (F2(x, y)j)
= ∂F1/∂x + ∂F2/∂y

It is obtained by taking the scalar product of the vector operator applied to the vector field F(x, y). The divergence of a vector field is a scalar field.

Example 2: The divergence of F(x, y) = 3x2i + 2yj is:

 ∇⋅F(x, y) = ∂F1/∂x + ∂F2/∂y = ∂/∂x(3x2) + ∂/∂y(2y) = 6x + 2

Quiz 4: What is the divergence of F(x, y) = x/yi + (2x − 3y)j ?

(a)1/y − 3 Correct - well done!
(b) − x/y2 + 2 Incorrect - please try again!
(c) 1/yx /y2 Incorrect - please try again!
(d) −2 Incorrect - please try again!

Solution: The vector field F(x, y) = x/yi + (2x − 3y)j has components:

F1(x, y) = x/y  and  F2(x, y) = 2x − 3y , so it's divergence is:
 ∇⋅F(x, y) = ∂F1/∂x + ∂F2/∂y = ∂/∂x (x/y) + ∂/∂y (2x − 3y) = 1/y − 3

The definition of the divergence may be directly extended to vector fields defined in three dimensions, F(x, y, z) = F1i + F2j + F3k:

F(x y, z)  =  ∂F1/∂x + ∂F2/∂y + ∂F3/∂z

Exercise 1: Calculate the divergence of the vector fields F(x, y) and G(x, y, z):

(a) F = xi + yj

Solution: The vector field F = xi + yj has components:

F1 = x , F2 = y

and its divergence is:

 ∇⋅F(x, y) = ∂F1/∂x + ∂F2/∂y = ∂/∂xx + ∂/∂yy = 1 + 1 = 2
(b) F = y2i + xyj

Solution: If the vector field is F = y2i + xyj, its components are:

F1 = y3 , F2 = xy

and its divergence is:

 ∇⋅F(x, y) = ∂F1/∂x + ∂F2/∂y = ∂/∂xy3 + ∂/∂yxy = 0 + x = x
(c) F = 3x2i − 6xyj

Solution: If the vector field is F = 3x2i − 6xyj, its components are:

F1 = 3x2 , F2 = −6xy

and its divergence is:

 ∇⋅F(x, y) = ∂F1/∂x + ∂F2/∂y = ∂/∂x3x2 + ∂/∂y(−6xy) = 6x − 6x = 0

Note: A vector field with vanishing divergence is called a solenoidal vector field.

(d) G = x2i + 2zjyk

Solution: The vector field G = x2i + 2zjyk has components:

G1 = x2 , G2 = 2z, G3 = −y

and its divergence is:

 ∇⋅G = ∂G1/∂x + ∂G2/∂y + ∂G3/∂z = ∂/∂xx2 + ∂/∂y(2z) + ∂/∂z(−y) = 2x +0 + 0 = 2x
(e) G = 4y/x2i + sin(y)j + 3k

Solution: Consider the vector field G = 4y/x2 i + sin(y)j + 3k has components:

G1 = 4y/x2 , G2 = sin(y), G3 = 3

and its divergence is:

 ∇⋅G = ∂G1/∂x + ∂G2/∂y + ∂G3/∂z = ∂/∂x ( 4y/x2 ) + ∂/∂ysin(y) + ∂/∂z3 = 4y × ∂G1/∂xx−2 + cos(y) = −8yx−3 + cos(y) = −8y/x3 + cos(y)
(f) G = exi + ln(xy)j + exyzk

Solution: Consider the vector field G = exi + ln(xy)j + exyzk. Its components are:

G1 = ex , G2 = ln(xy), G3 = exyz

and its divergence is:

 ∇⋅G = ∂G1/∂x + ∂G2/∂y + ∂G3/∂z = ∂/∂xex + ∂/∂yln(xy) + ∂/∂zexyz = ex + ∂/∂y(ln(x) + ln(y)) + exyz × ∂/∂z(xyz) = ex + 1/y + xyexyz

# 3. Curl

The curl of a vector field, F(x, y, z), in three dimensions may be written: curl F(x, y, z) = × F(x, y, z):

× F(x, y, z) = ( ∂F3/∂y∂F2/∂z )i( ∂F3/∂x∂F1/∂z )j + ( ∂F2/∂x∂F1/∂y )k
=

 i j k ∂/∂x ∂/∂y ∂/∂z F1 F2 F3

It is obtained by taking the vector product of the vector operator applied to the vector field F(x, y, z). The second line is again a formal shorthand. The curl of a vector field is also a vector field.

Note: × F is sometimes called the rotation of F and written rot F.

Example 3: The curl of F(x, y, z) = 3x2i + 2zjxk is:

× F(x, y, z) = ( ∂F3/∂y∂F2/∂z )i( ∂F3/∂x∂F1/∂z )j + ( ∂F2/∂x∂F1/∂y )k
=

 i j k ∂/∂x ∂/∂y ∂/∂z 3x2 2z −x

= ( ∂(−x)/∂y(2z)/∂z )i( ∂(−x)/∂x(3x3)/∂z )j + ( (2z)/∂x(3x2)/∂y )k
= (0 − 2)i − (−1 − 0)j + (0 − 0)k = −2i + j

Quiz 5: Which of the following is the curl of F(x, y,z) = xi + yj + zk?

(a)2i − 2j + 2k Incorrect - please try again!
(b)xi + yj + zk Incorrect - please try again!
(c)0Correct - well done!
(d)i + j + k Incorrect - please try again!

Explanation: The components of the vector field F(x, y,z) = xi + yj + zk are:

F1 = x,   F1 = y,   F1 = z

and its curl is:

 ∇ × F(x, y, z) = ( ∂F3/∂y − ∂F2/∂z )i − ( ∂F3/∂x − ∂F1/∂z )j + ( ∂F2/∂x − ∂F1/∂y )k = ( ∂(z)/∂y − ∂(y)/∂z )i − ( ∂(z)/∂x − ∂(x)/∂z )j + ( ∂(y)/∂x − ∂(x)/∂y )k = 0i − 0j + 0k

Note: A vector field with vanishing curl is called an irrotational vector field.

Exercise 2: Find the solution to each of the following equations:

(a)F = xiyj + zk

Solution: The components of the vector field F = xiyj + zk are:

F1 = x,    F3 = −y,    F3 = z

and its curl is:

 ∇ × F = ( ∂F3/∂y − ∂F2/∂z )i − ( ∂F3/∂x − ∂F1/∂z )j + ( ∂F2/∂x − ∂F1/∂y )k = ( ∂(z)/∂y − ∂(−y)/∂z )i − ( ∂(z)/∂x − ∂(x)/∂z )j + ( ∂(−y)/∂x − ∂(x)/∂y )k = 0i − 0j + 0k = 0

Therefore the vector field F = xiyj + zk is an irrotational vector field.

(b)F = y3i + xyjzk

Solution: The components of the vector field F = y3i + xyjzk are:

F1 = x,    F3 = −y,    F3 = z

and its curl is:

 ∇ × F = ( ∂F3/∂y − ∂F2/∂z ) i − ( ∂F3/∂x − ∂F1/∂z ) j + ( ∂F2/∂x − ∂F1/∂y ) k = (∂(−z)/∂y − ∂(xy)/∂z ) i − (∂(−z)/∂x − ∂(y3)/∂z) j + ( ∂(xy)/∂x − ∂(y3)/∂y ) k = 0i − 0j + (y − 3y2k = (y − 3y2k

i.e., the curl vector is in the k direction.

(c)F = xi + yj + zk / x2 + y2 + z2

Solution: The components of the vector field F = xi + yj + zk / x2 + y2 + z2 are:

F1 = x/r,    F3 = y/r,    F3 = z/r ,   where r = x2 + y2 + z2

The i component of × F is:

∂F3/∂y∂F2/∂z = /∂y ( z/r ) − /∂z ( y/r ) = z/∂y ( 1/r ) − y/∂z ( 1/r )

The derivative of 1/r with respect to y is:

/∂y ( 1/r ) = /∂y 1/ (x + y + z)1/2 = (− 1/2 ) × 2y/ (x + y + z)3/2 = − y/r3

and similarly /∂z ( 1/∂z ) = − z/r3 . Thus, the i component of the curl is (− zy/r3) − (− yz/r3) = 0 . It may be confirmed that that the j and k components of the curl also vanish.

(d) F = x2i + 2zjyk

Solution: The components of the vector field F = x2i + 2zjyk are:

F1 = x2,    F3 = 2z,    F3 = −y

and its curl is:

 ∇ × F = ( ∂F3/∂y − ∂F2/∂z ) i − ( ∂F3/∂x − ∂F1/∂z ) j + ( ∂F2/∂x − ∂F1/∂y ) k = (∂(−y)/∂y − ∂(2z)/∂z ) i − (∂(−y)/∂x − ∂(x2)/∂z) j + ( ∂(2z)/∂x − ∂(x2)/∂y ) k = (−1 − 2)i − (0 − 0)j + (0 − 0)k = − 3i

Here is a review exercise before the final quiz:

Exercise 3: Let ƒ be a scalar field and F(x, y, z) and G(x, y, z) be vector fields. What, if anything, is wrong with each of the following expressions?

(a)ƒ = x3 − 4y

Solution: The formula ƒ = x3 − 4y must be incorrect because the gradient of a scalar function is a vector field by definition, while the expression on the right hand side of this equation is a scalar.

(b)F = ix2yjzk

Solution: The equation F = ix2yjzk must be incorrect, because the divergence of a vector field must be a scalar by definition, but the right hand side of the equation is a vector.

(c) × G = F

Solution: The equation × G = F must be incorrect because its left hand side is a vector field, a curl, while its right hand side is a scalar function, a divergence.

# 5. Quiz on Div and Curl

Choose the solutions from the options given

1. Select the divergence of G(x, y, z) = 2x3i − 3xyj + 3x2zk
(a)9x2 − 3x
(b)6x2 + 3x
(c)0
(d)3x2 − 3x
2.Select the divergence of r ⁄ r3, where r = r and r = xi + yj + zk
(a) −1/r3
(b)0
(c) −2/r3
(d) 3/r3
3.Choose the curl of G(x, y, z) = x2i + xyzj zk at the point (2, 1, −2)
(a)2i + 2k
(b)−2i − 2j
(c)4i − 4j + 2k
(d)−2i − 2k
4.Select the irrotational vector field (i.e., whose curl is zero)
(a)yzi − 2xzj + xyzk
(b)yzi + xzj + xzk
(c)ziz2j + yzk
(d)yi + (xz)jyk