Mathematical Physics Vol 1

4.6 Examples

183

Exercise 125

Prove that

y V

∇ φ d V = x S

φ n d S .

Solution Let A = φ C , where C is an arbitrary constant vector, and φ a scalar function. According to the divergence theorem y V ∇ ( φ C ) d V = x S φ C · n d S . As (see Example 51b on p. 129) ∇ · ( φ C )=( ∇ φ ) · C = C · ( ∇ φ ) i φ C · n = C · ( φ n ) , it follows that y V C · ∇ φ d V = x S C · ( φ n ) d S , that is C · y V ∇ φ d V = C · x S φ n d S . From here, given that C is an arbitrary vector, we finally obtain y V ∇ φ d V = x S φ n d S .

Exercise 126

Prove that

y V

∇ × B d V = x S

n × B d S .

Solution Let A = B × C ,where C is an arbitrary constant vector. According to the divergence theorem (Gauss’s theorem (4.92)) y V ∇ · ( B × C ) d V = x S ( B × C ) · n d S . (4.209) Based on the properties of the delta operator (see p. 88) and the properties of divergence (div C =0), we obtain ∇ · ( B × C )= C · ( ∇ × B ) . (4.210) On the other hand (see properties of the mixed product on p.26) ( B × C ) · n = B · ( C × n )=( C × n ) · B = C · ( n × B ) . (4.211)