∫（y+1)dx+(z+2)dy+(x+3)dz,L是球面x2+y2+z2=a2与平面x+y+z=0的交线,从x抽正向看去,L的方向是逆时针可以由stokes公式求解,学到这一章

取Σ为x + y + z = 0的上侧
Σ的单位法向量n = (i + j + k)/√3
取A = (y + 1)i + (z + 2)j + (x + 3)k
rot(A) =
[ - ∂/∂z (z + 2) ] i + [ - ∂/∂x (x + 3) ] j + [ - ∂/∂y (y + 1) ] k
= - (i + j + k)
D为x^2 + y^2 + (x + y)^2 = a^2 ==> 2(x^2 + y^2) + 2xy = a^2
2r^2 + 2r^2sinθcosθ = a^2 ==> r^2(2 + sin2θ) = a^2 ==> r = a/√(2 + sin2θ)
∮L (y + 1)dx + (z + 2)dy + (x + 3)dz
= ∫∫Σ rot(A) * n dS
= ∫∫Σ - (i + j + k) * (i + j + k)/√3 dS
= - ∫∫Σ (1 + 1 + 1)/√3 dS,Σ为z = - x - y
= - √3∫∫D √[ 1 + (- 1)^2 + (- 1)^2 ] dxdy
= - √3 * √3∫∫D dxdy
= - 3∫∫D dxdy
= - 3∫(0,2π) [ ∫(0,a/√(2 + sin2θ) r dr ] dθ
= - 3∫(0,2π) [ ( r^2/2 )：(0,a/√(2 + sin2θ) ] dθ
= (- 3a^2/2)∫(0,2π) 1/(2 + sin2θ) dθ
= (- 3a^2/2)(2π/√3)
= - √3πa^2