急求！！！高数题 L是顺时针方向的椭圆曲线x^2+9y^2=1,则积分∫L-2ydx+x^2dy=

Question

如题所述

Answer 1

注意：
用格林公式后是
I
=
-∫∫<D>(2x+2)dxdy
引入广义极坐标
x
=
rcost,
y
=
(1/3)rsint,
则
I
=
-∫<0,
2π>dt∫<0,
1>2(
rcost+1)(1/3)rdr
=
-(2/3)∫<0,
2π>dt∫<0,
1>(
rcost+1)rdr
=
-(2/3)∫<0,
2π>dt[(1/3)r^3cost+r^2/2]<0,1>
=
-(2/3)∫<0,
2π>[(1/3)cosr+1/2]dt
=
-(2/3)[(1/3)sint
+t/2]<0,
2π>
=
-2π/3

Answer 2

注意：
用格林公式后是
i
=
-∫∫<d>(2x+2)dxdy
引入广义极坐标
x
=
rcost,
y
=
(1/3)rsint,
则
i
=
-∫<0,
2π>dt∫<0,
1>2(
rcost+1)(1/3)rdr
=
-(2/3)∫<0,
2π>dt∫<0,
1>(
rcost+1)rdr
=
-(2/3)∫<0,
2π>dt[(1/3)r^3cost+r^2/2]<0,1>
=
-(2/3)∫<0,
2π>[(1/3)cosr+1/2]dt
=
-(2/3)[(1/3)sint
+t/2]<0,
2π>
=
-2π/3