Question

∫∫∫(x^2+z^2)du,其中ñ={(x,y,z)|x^2+y^2+(z-R)^2<=R^2}

Answer 1

令u = x、v = y、w = z - R
|J| = 1
Ω：x² + y² + (z - R)² ≤ R² ==> ω：u² + v² + w² ≤ R²
∫∫∫Ω (x² + z²) dV
= ∫∫∫ω [u² + (w + R)²] |J| dV
= ∫∫∫ω (u² + w² + 2wR + R²) dV
= ∫∫∫ω (u² + w²) dV + 0 + R²∫∫∫ω dV
= (2/3)∫∫∫ω (u² + v² + w²) dV + (R²)(4/3)(π)(R)³
= (2/3)∫(0→2π) dθ ∫(0→π) sinφ dφ ∫(0→R) r⁴ dr + (4/3)πR⁵
= (2/3)(2π)(2)(1/5)(R⁵) + (4/3)πR⁵
= (28/15)πR⁵

另解：(直接用球坐标)
{ x = rcosθsinφ
{ y = rsinθsinφ
{ z = rcosφ
x² + y² + (z - R)² ≤ R²
==> x² + y² + z² ≤ 2Rz
==> r² ≤ (2R)(rcosφ)
==> r ≤ 2Rcosφ
∫∫∫Ω (x² + z²) dV
= ∫(0→2π) dθ ∫(0→π/2) sinφ dφ ∫(0→2Rcosφ) (r²cos²θsin²φ + r²cos²φ)r² dr
= (32/5)(R⁵)∫(0→2π) dθ ∫(0→π/2) sinφ(cos²θsin²φcos⁵φ + cos⁷φ) dφ
= (32/5)(R⁵)(1/48)∫(0→2π) (7+ cos2θ) dθ
= (2/15)(R⁵)(14π)
= (28/15)πR⁵