Question

计算∫(2xy3-y2cosx)dx+(1-2ysinx+3x2y2)dy,其中L为在抛物线2x=πy2上由(0,0)到(

Answer 1

简单计算一下即可，答案如图所示

Answer 2

其中L是在抛物线2x
=
πy^2上由点(0，0)到(π/2，1)的一段弧。
——————————————————————————————————————————
补线：
L1：x
=
π/2、逆时针方向、dx
=
0、由y
=
0变化到y
=
1
L2：y
=
0、逆时针方向、dy
=
0、由x
=
0变化到x
=
π/2
由于L是顺时针方向，现在设L⁻是L的逆时针方向
∮(L⁻+L1+L2)
(2xy^3
-
y^2cosx)
dx
+
(1
-
2ysinx
+
3x^2y^2)
dy
=
∫∫D
[∂/∂x
(1
-
2ysinx
+
3x^2y^2)
-
∂/∂y
(2xy^3
-
y^2cosx)]
dxdy、用Green公式
=
∫∫D
[(-
2ycosx
+
6xy^2)
-
(6xy^2
-
2ycosx)]
dxdy
=
∫∫D
(-
2ycosx
+
6xy^2
-
6xy^2
+
2ycosx)
dxdy
=
0
而∫(L1)
(2xy^3
-
y^2cosx)
dx
+
(1
-
2ysinx
+
3x^2y^2)
dy
=
∫(0→1)
[0
+
1
-
2y
+
3(π/2)^2y^2]
dy
=
∫(0→1)
[1
-
2y
+
(3/4)π^2
*
y^2]
dy
=
y
-
y^2
+
(3/4)π^2
*
(1/3)y^3：(0→1)
=
1
-
1
+
(3/4)π^2
*
1/3
=
(1/4)π^2
而∫(L2)
(2xy^3
-
y^2cosx)
dx
+
(1
-
2ysinx
+
3x^2y^2)
dy
=
∫(L2)
0
dx
=
0
于是∫(L⁻)
+
∫(L1)
+
∫(L2)
=
∮(L⁻+L1+L2)
∫(L⁻)
+
(1/4)π^2
+
0
=
0
∫(L⁻)
=
-
(1/4)π^2
∫(L)
=
(1/4)π^2
即原式∫(L)
(2xy^3
-
y^2cosx)
dx
+
(1
-
2ysinx
+
3x^2y^2)
dy
=
(1/4)π^2