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=

2024-12-02 13:13:53
推荐回答(2个)
回答1:

注意:
用格林公式后是
I
=
-∫∫(2x+2)dxdy
引入广义极坐标
x
=
rcost,
y
=
(1/3)rsint,

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

回答2:

注意:
用格林公式后是
i
=
-∫∫(2x+2)dxdy
引入广义极坐标
x
=
rcost,
y
=
(1/3)rsint,

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