∭Ωsin（x2+y2+z2）32dv，其中Ω是由曲面z=3(x2+y2)与z=R2-x2-y2（R＞0且R为常数）所围成．

Question

∭

Ω

sin（x²+y²+z²） (3)/(2)dv，其中Ω是由曲面z=√(3(x2+y2))与z=√(R2-x2-y2)（R＞0且R为常数）所围成．

Answer 1

由题意，知Ω={(x，y，z)|√(3(x2+y2))≤z≤√(R2-x2-y2)}
={(r，θ，φ)|0≤θ≤2π，0≤φ≤(π)/(6)，0≤r≤R}，故

∭

Ω

sin（x²+y²+z²） (3)/(2)dv=

∫

2π 0

dθ

∫

(π)/(6) 0

sinφdφ

∫

R 0

(sinr3)r2dr
=2π(1-(√(3))/(2)).(1)/(3)(1-cosR3)=(π)/(3)(2-√(3))(1-cosR3)．