Mathematical Physics Vol 1

Chapter 4. Field theory

182

Solution According to Gauss’s theorem (see relation (4.92) p. 102) y ∆ V div A d V = x ∆ S A · n d S .

(4.207)

and the mean value theorem ((4.93), p. 102) y ∆ V div A d V = div A y ∆ V

d V = div A ∆ V ,

(4.208)

where div A is the mean value of div A within ∆ V , we obtain

1 ∆ V x ∆ S

div A = A · n d S . If we now let ∆ V → 0, so that point P remains always within ∆ V , then div A takes the value div A at point P . Consequently

s ∆ S

A · n d S ∆ V

div A = lim ∆ V → 0

.

Exercise 124 Calculate the integral

x S

r · n d V ,

where S is a closed surface.

Solution According to the divergence theorem, we have x S r · n d S = y V ∇ · r d V = = y V ∂ ∂ x i + ∂ ∂ y j + ∂ ∂ z

k · ( x i + y j + z k ) d V =

V

∂ z ∂ z

= y

d V = 3 y V

∂ x ∂ x

∂ y ∂ y

d V = 3 V ,

+

+

where V is the volume bounded by the closed surface S .