Mathematical Physics Vol 1

4.6 Examples

181

It can be seen from Figure 4.33 theta x = 2cos θ , y = 2sin θ , d S 3 = 2d θ d z , and it follows that

Figure 4.33

2 π Z θ = 0 2 π Z θ = 0 2 π Z θ = 0

3 Z z = 0

I 3 = x S 3

[ 2 ( 2cos θ ) 2 − ( 2sin θ ) 3 )] 2d z d θ =

A · n d S 3 =

( 48cos 2 θ − 48sin 3 θ ) d θ =

=

48cos 2 θ d θ = 48 π .

=

The surface integral over the entire surface is the sum of these three integrals, namely I = I 1 + I 2 + I 3 = 0 + 36 π + 48 π = 84 π . Thus, the obtained result is the same as the one obtained by the volume integral, which confirms the divergence theorem.

Exercise 123 If div A represents the divergence of a vector field A at point P , showthat

s ∆ S

A · n d S ∆ V

div A = lim ∆ V → 0

,

where ∆ V is the volume bounded by the closed contour of the surface ∆ S , and the limit value is obtained by " shrinking" ∆ V around point P .