只有第四第五题要补面,第六题可挖去奇点后用高斯公式.
曲面积分的符号上有圈圈的是封闭区域,可以直接用高斯公式,不是封闭的就要补面.
1.不用补面
∫∫Σ x²dydz + y²dzdx + z²dxdy
= ∫∫∫Ω (2x + 2y + 2z) dV
= ∫(0→a) dx ∫(0→a) dy ∫(0→a) 2(x + y + z) dz
= 3a⁴
2.不用补面
∫∫Σ xdydz + ydzdx + zdxdy
= 3∫∫∫Ω dV
= 3 * π * 9 * 3
= 81π
3.不用补面
∫∫Σ xzdydz + x²ydzdx + y²zdxdy
= ∫∫∫Ω (z + x² + y²) dV
= ∫(0→π/2) dθ ∫(0→1) r dr ∫(0→r²) (z + r²) dz
= π/8
4.补面Σ1：z = 0取下侧
∫∫(Σ+Σ1) (x³ + az²)dydz + (y³ + ax²)dzdx + (z³ + ay²)dxdy
= ∫∫∫Ω (3x² + 3y² + 3z²) dV
= 3∫(0→2π) dθ ∫(0→π/2) sinφ dφ ∫(0→a) r⁴ dr
= 3 * 2π * 1 * a⁵/5
= (6/5)πa⁵
∫∫Σ1 (x³ + az²)dydz + (y³ + ax²)dzdx + (z³ + ay²)dxdy
= a∫∫Σ1 y² dxdy
= (- a/2)∫∫D (x² + y²) dxdy
= (- a/2)∫(0→2π) dθ ∫(0→a) r³ dr
= (- a/2)(2π)(a⁴/4)
= - πa⁵/4
∴I = (6/5)πa⁵ + πa⁵/4 = (29/20)πa⁵
5.法向量与z轴正向夹角为锐角 ==> 曲面内测
补面Σ1：z = 1取下测
∫∫(Σ+Σ1) (2x + z)dydz + zdxdy
= - ∫∫∫Ω (2 + 1) dV
= - 3∫(0→2π) dθ ∫(0→1) r dr ∫(r²→1) dz
= - 3 * 2π * ∫(0→1) (r - r³) dr
= - 6π * (r²/2 - r⁴/4)：(0→1)
= - 6π * (1/2 - 1/4)
= (- 3/2)π
∫∫Σ1 (2x + z)dydz + zdxdy
= ∫∫Σ1 dxdy
= - ∫∫D dxdy
= - π
∴I = - 3π/2 + π = - π/2
或者直接用曲面积分的方法：
∫∫Σ (2x + z)dydz + zdxdy
= ∫∫D [- P * ∂z/∂x - Q * ∂z/∂y + R] dxdy
= ∫∫D [- (2x + x² + y²)(2x) + x² + y²] dxdy
= ∫∫D (- 4x² - 2x³ - 2xy² + x² + y²) dxdy
= ∫∫D (- 3x² + y²) dxdy
= - ∫∫D (x² + y²) dxdy
= - ∫(0→2π) dθ ∫(0→1) r³ dr
= (- 2π)(1/4) = - π/2
6.不用补面,但是要避开奇点(0,0,0)
设Σ1：x² + y² + z² = r²,半径r趋向0,取内测
∫∫(Σ+Σ1) (xdydz + ydzdx + zdxdy)/(x² + y² + z²)^(3/2) = 0
∫∫Σ1 (xdydz + ydzdx + zdxdy)/(x² + y² + z²)^(3/2)
= 1/(r²)^(3/2) * ∫∫Σ1(内测) xdydz + ydzdx + zdxdy
= (1/r³)[∫∫Σ1 x dydz + ∫∫Σ1 y dzdx + ∫∫Σ1 z dxdy]
∫∫Σ1 x dydz
= ∫∫[x = - √(r² - y² - z²)前侧] x dydz + ∫∫[x = √(r² - y² - z²)后侧] x dydz
= ∫∫D - √(r² - y² - z²) dydz - ∫∫ √(r² - y² - z²) dydz
= - 2∫(0→2π) dθ ∫(0→r) √(r² - ρ²)ρ dρ
= - 4π * (- 1/2) * (2/3)(r² - ρ²)^(3/2)：(0→r)
= - 4π * (- 1/3)(- r³)
= - 4π/3 * r³
三个加起就是- 4πr³
于是I = ∫∫(Σ+Σ1) - ∫∫Σ1 = 0 - (1/r³)(- 4πr³) = 4π
PS：三组偏导数都相等,曲面积分的结果与曲面无关,因此过程中没用过曲面的方程Σ
而选用了小圆的方程Σ1