Mathematical Physics Vol 1

4.6 Examples

173

Exercise 114

Calculate I C

( y − sin x ) d x + cos x d y ,

where C is a triangle, depicted in Figure 4.31 a) directly, b) by applying Stokes’ theorem in the plane.

Figure 4.31: Path of integration.

Solution a) Along OA , y = 0, d y = 0, and the integral is π / 2 Z 0 ( 0 − sin x ) d x + cos x ( 0 )= π / 2 Z 0 Along AB , x = π / 2, d x = 0, and the integral is 1 Z 0

π / 2

− sin x d x = cos x

= − 1 .

0

( y − 1 )( 0 )+ 0d y = 0 .

2 x π

2 π d x , and the integral is

Along BO , y =

, d y =

0 Z π / 2

sin x d x +

cos x d x =

sin x

0

x 2 π

π 4 −

2 x π −

2 π

2 π

2 π

+ cos x +

= 1 −

.

π / 2

Consequently, the integral along the curve C is = − 1 + 0 + 1 − π 4 − 2 π = − π 4 − 2 π . b) According to Stokes’ theorem (relation (4.90), p. 102) I c M ( x , y ) d x + N ( x , y ) d y = x R ∂ N ∂ x − ∂ M ∂ y d x d y . As in our case M = y − sin x , N = cos x , ∂ M ∂ y = 1, ∂ N ∂ x = − sin x , we obtain I C M d x + N d y = x R ∂ N ∂ x − ∂ M ∂ y d x d y = x R ( − sin x − 1 ) d x d y =

 

( − sin x − 1 ) d y 

π / 2 Z

2 x / π Z

π / 2 Z

2 x / π

 d x =

( − y sin x − y )

d x

=

0

x = 0

y = 0

x = 0

π / 2 Z x = 0

π / 2

2 x π

x 2 π

π 4 −

2 x π

2 π ( − x cos x + sin x ) −

2 π

sin x −

d x = −

−

= −

=

.

0