Mathematical Physics Vol 1

Chapter 4. Field theory

172

Solution It can be shown that M ( x , y ) d x + N ( x , y ) d x is a total differential if ∂ M ∂ x = ∂ N ∂ y 23 . In this Example M = 10 x 4 − 2 xy 3 and N = − 3 x 2 y 2 , which satisfies the condition ∂ M ∂ x = ∂ N ∂ y , and it follows that ( 10 x 4 − 2 xy 3 ) d x − 3 x 2 y 2 d y is the total differential of the function ϕ = 2 x 5 − x 2 y 3 . From there, it follows that ( 2 , 1 ) Z ( 0 , 0 ) ( 10 x 4 − 2 xy 3 ) d x − 3 x 2 y 2 d y = ( 2 , 1 ) Z ( 0 , 0 ) d ϕ = = ϕ = 60 .

( 2 , 1 ) ( 0 , 0 )

( 2 , 1 ) ( 0 , 0 )

= 2 x 5 − x 2 y 3

Exercise 113 Show that the area bounded by a simple curved line C is given by

1 2 I C

x d y − y d x .

Solution According to Stokes’ theorem (relation (4.90), p. 102) I C x d y − y d x = x R ∂ ∂ x ( x ) − ∂ ∂ y

( − y ) d x d y

= 2 x R

d x d y = 2 A ,

where A is the required area. From here, it follows that

1 2 I C

A =

x d y − y d x .

23

∂ f ∂ x

∂ f ∂ y

d f =

d x +

d y = M ( x , y ) d x + N ( x , y ) d y ,

∂ 2 f ∂ x ∂ y

∂ 2 f ∂ y ∂ x

∂ f ∂ x

∂ f ∂ y ⇒

∂ N ∂ y

∂ M ∂ x

N =

M =

i

⇒

,

=

=

∂ N ∂ y

∂ M ∂ x

=

.