Michael Corral: Vector Calculus

4.6 Gradient, Divergence, Curl and Laplacian 179
Solution: (a)∇‖r‖ 2 = 2xi+2yj+2zk=2r
(b)∇·r= ∂ ∂x (x)+ ∂ ∂y (y)+ ∂ ∂z (z)=1+1+1=3
(c)
i j k
∇×r=
∂ ∂ ∂
∂x ∂y ∂z
= (0−0)i−(0−0)j+(0−0)k=0
∣
x y z∣
(d)∆‖r‖ 2 = ∂2
∂x 2 (x 2 +y 2 +z 2 )+ ∂2
∂y 2 (x 2 +y 2 +z 2 )+ ∂2
∂z 2 (x 2 +y 2 +z 2 )=2+2+2=6
Notethatwecouldhavecalculated∆‖r‖ 2 anotherway,usingthe∇notationalongwith
parts (a) and (b):
∆‖r‖ 2 =∇·∇‖r‖ 2 =∇·2r=2∇·r=2(3)=6
Notice that in Example 4.17 if we take the curl of the gradient of‖r‖ 2 we get
∇×(∇‖r‖ 2 )=∇×2r=2∇×r=20=0.
The following theorem shows that this will be the case in general:
Theorem 4.15. For any smooth real-valued function f(x,y,z),∇×(∇f)=0.
Proof: We see by the smoothness of f that
i j k
∂ ∂ ∂
∇×(∇f)=
∂x ∂y ∂z
∂f ∂f ∂f
∣
∂x ∂y ∂z
∣
( ∂ 2 )
f
=
∂y∂z − ∂2 f
∂z∂y
( ∂ 2 ) (
f
i−
∂x∂z − ∂2 f ∂ 2 )
f
j+
∂z∂x ∂x∂y − ∂2 f
k=0,
∂y∂x
since the mixed partial derivatives in each component are equal.
QED
Corollary 4.16. If a vector field f(x,y,z) has a potential, then curl f=0.
Another way of stating Theorem 4.15 is that gradients are irrotational. Also, notice
that in Example 4.17 if we take the divergence of the curl of r we trivially get
∇·(∇×r)=∇·0=0.
The following theorem shows that this will be the case in general:
Theorem 4.17. For any smooth vector field f(x,y,z),∇·(∇×f)=0.
The proof is straightforward and left as an exercise for the reader.