Question

I=∮L(x+y)dx?(x?y)dyx2+y2，L为|x|+|y|=1的顺时针方向，则I=（　　）A．0B．2πC．-2πD．

I=∮L(x+y)dx?(x?y)dyx2+y2，L为|x|+|y|=1的顺时针方向，则I=（　　）A．0B．2πC．-2πD．π

Answer 1

取曲线C为圆周x2+y2＝

1

4

的逆时针方向，并设L+C所围区域为D．
利用格林公式可得，
 

∮

L+C

(x+y)dx?(x?y)dy

x2+y2

 
=

?

D

(

?(? x?y x2+y2 )

?x

?

?( x+y x2+y2 )

?y

)dxdy
=

?

D

(

x2?2xy?y2

(x2+y2)2

?

x2?2xy?y2

(x2+y2)2

)dxdy
=0．
又因为
 

∮

C

(x+y)dx+(x?y)dy

x2+y2

=4

?

C

(x+y)dx+(x?y)dy
=4

?

x2+y2≤ 1 4

(?1?1)dxdy
=?8 

?

x2+y2≤ 1 4

dxdy
=(?8)×

π

4

=-2π，
故
 

∮

(x+y)dx+(x?y)dy

x2+y2

=-

∮

C

(x+y)dx+(x?y)dy

x2+y2

=2π．
故选：B．