Mathematical Methods for Physics and Engineering : A Comprehensive Guide

11.7 INTEGRAL FORMS FOR grad, div AND curl

to the surface integral from these two faces is then

[(φ+∆φ)−φ]∆y∆zi=

(

φ+

∂φ

∂x

∆x−φ

)

∆y∆zi

=

∂φ

∂x

∆x∆y∆zi.

The surface integral over the pairs of faces withy=constantandz= constant respectively
may be found in a similar way, and we obtain
∮

S

φdS=

(

∂φ

∂x

i+

∂φ

∂y

j+

∂φ

∂z

k

)

∆x∆y∆z.

Therefore∇φat the pointPis given by

∇φ=∆x,∆limy,∆z→ 0

[

1

∆x∆y∆z

(

∂φ

∂x

i+

∂φ

∂y

j+

∂φ

∂z

k

)

∆x∆y∆z

]

=

∂φ

∂x

i+

∂φ

∂y

j+

∂φ

∂z

k.

We now turn to (11.15) and (11.17). These geometrical definitions may be

shown straightforwardly to lead to the usual expressions for div and curl in

orthogonal curvilinear coordinates.

By considering the infinitesimal volume elementdV=h 1 h 2 h 3 ∆u 1 ∆u 2 ∆u 3 shown in fig-

ure 11.10, show that (11.15) leads to the usual expression for∇·ain orthogonal curvilinear

coordinates.

Let us write the vector field in terms of its components with respect to the basis vectors
of the curvilinear coordinate system asa=a 1 ˆe 1 +a 2 ˆe 2 +a 3 eˆ 3. We consider first the
contribution to the RHS of (11.15) from the two faces withu 1 =constant,i.e.PQRS
and the face opposite it (see figure 11.10). Now, the volume element is formed from the
orthogonal vectorsh 1 ∆u 1 ˆe 1 ,h 2 ∆u 2 ˆe 2 andh 3 ∆u 3 eˆ 3 at the pointPandsoforPQRSwe
have

∆S=h 2 h 3 ∆u 2 ∆u 3 ˆe 3 ׈e 2 =−h 2 h 3 ∆u 2 ∆u 3 ˆe 1.

Reasoning along the same lines as in the previous example, we conclude that the contri-
bution to the surface integral ofa·dSoverPQRSand its opposite face taken together is
given by

∂

∂u 1

(a·∆S)∆u 1 =

∂

∂u 1

(a 1 h 2 h 3 )∆u 1 ∆u 2 ∆u 3.

The surface integrals over the pairs of faces withu 2 =constantandu 3 =constant
respectively may be found in a similar way, and we obtain
∮

S

a·dS=

[

∂

∂u 1

(a 1 h 2 h 3 )+

∂

∂u 2

(a 2 h 3 h 1 )+

∂

∂u 3

(a 3 h 1 h 2 )

]

∆u 1 ∆u 2 ∆u 3.

Therefore∇·aat the pointPis given by

∇·a= lim

∆u 1 ,∆u 2 ,∆u 3 → 0

[

1

h 1 h 2 h 3 ∆u 1 ∆u 2 ∆u 3

∮

S

a·dS

]

=

1

h 1 h 2 h 3

[

∂

∂u 1

(a 1 h 2 h 3 )+

∂

∂u 2

(a 2 h 3 h 1 )+

∂

∂u 3

(a 3 h 1 h 2 )

]

.