计算曲面积,I=?(x3+z2)dydz+(y3+x2)dzdx+(z3+y2)dxdy,其中∑为上半球面z=1?x2?y2的上侧

2025-06-21 15:22:40
推荐回答(1个)
回答1:

添加曲面S1

x2+y2≤1
z=0
,方向为z轴的负方向.
令∑和S1所围成的空间区域为Ω,根据高斯公式可得,
I1=
?
∑+S1
(x3+z2)dydz+(y3+x2)dzdx+(z3+y2)dxdy

=
?
Ω
3(x2+y2+z2)dxdy

=3
∫ 
sinφdφ
ρ4

=
6
5
π

又因为
I2=
?
S1
(x3+z2)dydz+(y3+x2)dzdx+(z3+y2)dxdy

=
?
S1
y2dxdy

=?
?
x2+y2≤1
y2dxdy

=-
sin2θdθ
r3dr

=-
π
4

所以,
I=I1-I2
=
6
5
π
+
π
4

=
29
20
π