Question

对坐标的曲面积分，Σ为球面x²＋y²＋z²＝a²的外侧，则∫∫Σydxdy＝

Answer 1

解：原式=∫∫
√(a²-x²)dzdx-∫∫
[-√(a²-x²)]dzdx+∫∫
(x+1)dxdy-∫∫
(x+0)dxdy
(s1：-a≤x≤a:，0≤z≤1。s2：x²+y²≤a²)
=2∫∫
√(a²-x²)dzdx+∫∫
dxdy
=2∫<-a,a>√(a²-x²)dx∫<0,1>dz+∫<0,2π>dθ∫<0,a>rdr
(第二个积分作极坐标变换)
=2∫<-a,a>√(a²-x²)dx+πa²
=2∫<-π/2,π/2>a²cos²tdt+πa²
(作变换x=asint)
=a²∫<-π/2,π/2>[1+cos(2t)]dt+πa²
(应用倍角公式)
=a²[t+sin(2t)/2]│<-π/2,π/2>+πa²
=a²(π/2+π/2)+πa²
=2πa²。

Answer 2

D是∑在xOy平面的投影，方程为x^2+y^2=4
∫∫[∑]
x^2dxdy=∫∫[D]
x^2dxdy
由轮换对称性有∫∫[D]
x^2dxdy=∫∫[D]
y^2dxdy
所以∫∫[D]
x^2dxdy=(1/2)∫∫[D]
x^2+y^2dxdy=(1/2)∫[0->2π]∫[0->2]
r^3
drdθ=4π

Answer 3

被积曲面关于xOy对称,被积函数关于z是奇函数,根据第二类曲面积分的对称性原理
原式=2∫∫xy√1-x²-y²dxdy
(其中,被积区域为x²+y²=1,x,y≥0)
原式=2∫[0->π/2]∫[0->1]r³√1-r²drdθ=(π/2)∫[0->1]r²√1-r²dr²
=(π/2)[∫[0->1]√1-r²dr²-∫[0->1](1-r²)√1-r²dr²]
=(π/2)[(-2/3)(1-r²)^(3/2)
|
[0->1]
-
(-2/5)(1-r²)^(5/2)
|
[0->1]
]
=2π/15