Mathematical Physics Vol 1

4.6 Examples

177

Given that

i

j

k

∂ ∂ x ∂ ∂ z 2 x − y − yz 2 − y 2 z ∂ ∂ y

∇ × A =

= k ,

it follows that

I = x S

( ∇ × A ) · n d S = x S

k · n d S = x R

d x d y .

Namely, k · n d S = d x d y , R is the projection of S on the xy plane. By introducing these relations in the previous integral we obtain

√ 1

√ 1

− x 2 Z − x 2

− x 2 Z 0

1 Z

1 Z 0

I =

d x

d y =

d y = 4

d x

√ 1

− 1

−

1 Z 0 p

1 − x 2 d x = π .

= 4

Exercise 119 Prove that the necessary and sufficient condition for I C A · d r = 0 , along any closed curve C , is ∇ × A = 0.

Proof The condition is sufficient. In this case ∇ × A = 0 is valid. Then (Stokes’ theorem 102, relation (4.92)) I C A · d r = x S ( ∇ × A ) · n d S = 0 . The condition is necessary. In this case

I C

A · d r = 0 ,

(4.204)

for any closed curve C .

Further, according to Stokes’ theorem, 0 = I C A · d r = Z S

( ∇ × A ) · n d a