计算∫∫Σ xdydz + ydzdx + zdxdy,其中Σ是柱面x^2 + y^2被平面z = 0及z = 3所截部分的外则.
用高斯公式：
补面Σ1：z = 3,上侧、Σ2：z = 0,下侧
于是∫∫(Σ+Σ1+Σ2) xdydz + ydzdx + zdxdy
= 3∫∫∫Ω dV
= 3 * π * 1^2 * 3
= 9π
∫∫Σ1 xdydz + ydzdx + zdxdy
= 3∫∫D dxdy
= 3 * π * 1^2
= 3π
∫∫Σ2 xdydz + ydzdx + zdxdy
= 0
于是∫∫Σ xdydz + ydzdx + zdxdy = 9π - 3π = 6π
用基本方法.
已知∫∫Σ z dxdy = 0
考虑yoz面,x = ± √(1 - y^2)
∫∫Σ xdydz + ydzdx
= ∫∫Σ前 + ∫∫Σ后
= ∫∫D [ - y * - y/√(1 - y^2) - 0 + √(1 - y^2) ] dydz - ∫∫D [ - y * y/√(1 - y^2) - 0 - √(1 - y^2) ] dydz
= 2∫∫D 1/√(1 - y^2) dydz
= 2∫(0,3) dz ∫(- 1,1) 1/√(1 - y^2) dy
= 2 * 3 * π
= 6π