计算∫∫(x^2+y^2)dzdx+zdxdy,其中∑是锥面z=√x^2+y^2被平面z=1所截下的在第一卦限的下侧

用两类积分的转换：
∫∫Σ (x^2 + y^2)dzdx + zdxdy
= ∫∫Σ [ (x^2 + y^2) * |cosβ|/|cosα| + z ] dxdy
= - ∫∫D [ (x^2 + y^2) * - y/√(x^2 + y^2) + √(x^2 + y^) ] dxdy
= ∫∫D [ (x^2 + y^2)y - (x^2 + y^2) ]/√(x^2 + y^2) dxdy
= ∫∫(0,π/2) dθ ∫(0,1) (r^3 * sinθ - r^2)/r * r dr
= 1/4 - π/6
用向量点积法：
∫∫Σ (x^2 + y^2)dzdx + zdxdy
= - ∫∫D [ - 0 - (x^2 + y^2) * y/√(x^2 + y^2) + √(x^2 + y^2) ] dxdy
= ∫∫D [ (x^2 + y^2)y - (x^2 + y^2) ]/√(x^2 + y^2) dxdy
= ∫(0,π/2) dθ ∫(0,1) (r^3 * sinθ - r^2)/r * r dr
= 1/4 - π/6
本题不太推荐高斯公式法,但可以作为对比：
取Σ：z = √(x² + y²)的下侧
取Σ1：z = 1的上侧
取Σ2：x = 0的左侧
取Σ3：y = 0的后侧
∫∫(Σ+Σ1+Σ2+Σ3) (x^2 + y^2)dzdx + zdxdy
= ∫∫∫Ω [ 0 + 2y + 1 ] dxdydz
= ∫∫∫Ω (2y + 1) dxdydz
= ∫(0,π/2) dθ ∫(0,1) r dr ∫(r,1) (2 * rsinθ + 1) dz
= ∫(0,π/2) dθ ∫(0,1) r * (2 * rsinθ + 1) * (1 - r) dr
= (π + 2)/12
∫∫Σ1 (x^2 + y^2)dzdx + zdxdy
= ∫∫D dxdy
= 1/4π * 1² = π/4
∫∫Σ2 (x^2 + y^2)dzdx + zdxdy
= 0
∫∫Σ3 (x^2 + y^2)dzdx + zdxdy
= ∫∫Σ3 x^2 dzdx
= - ∫∫D x^2 dzdx、D：z = x
= - ∫(0,1) dx ∫(x,1) x^2 dz
= - 1/12
于是∫∫Σ + ∫∫Σ1 + ∫∫Σ2 + ∫∫Σ3 = ∫∫(Σ+Σ1+Σ2+Σ3)
从而∫∫Σ = π/12 + 1/6 - π/4 - (- 1/12)
= 1/4 - π/6