Vector Calculus 3

b(ii)題目是否有問題?

2011-07-30 23:48:34 補充：
a. vector F=(y^3-2x, e^z-y^2, 2z), ∇∙F= -2-2y+2= -2y
Let V={(x,y,z) | 0 <= √(3x^2+3y^2) <= 3 }, A be the boundary of V, then
A= S ∪ { (x,y,z) | z=3, x^2+y^2 <= 1 }= S∪S'
∫∫_A F dot dS = ∫∫∫_V ∇∙F dV (by divergence theorem)
∫∫_S F dot dS +∫∫_S' F dot dS=∫∫∫_V (-2y) dV
∫∫_S F dot dS+∫∫_S' F dot (0,0, dxdy) = 0
∫∫_S F dot dS+∫∫_S' 6 dxdy =0
so, ∫∫_S F dot dS= - 6π ( Note: S' is an unit circle with r=1)

b-1.
(1) x^2+y^2=2y then r=2sinθ, θ=0~π,
then the surface S is (x,y,z)=(r cosθ, r sinθ, 16-r^2), r=0~2sinθ, θ=0~π
n dS= ( 2r^2 cosθ, 2r^2 sinθ, r) dr dθ
(2) F=(-2xy, x, -z^2), ∇xF=(0,0,1+2x)
∫_γ F dot dr
= ∫∫_S (0,0,1+2x) dot ( 2r^2 cosθ, 2r^2 sinθ, r) dr dθ
=∫_(0,π)∫_(0, 2sinθ) r(1+2r cosθ) dr dθ
=∫_(0,π) [ 2(sinθ)^2+ (16/3) (sinθ)^3 cosθ] dθ
=π

b-2.
f F dot dr= -2xyf dx + xf dy - z^2 f dz
Let f=1, C be (x,y,z)=(0, cosθ, sinθ), θ=0~2π, then
f F dot dr= 0+0 - (sinθ)^2 cosθ dθ
∫_C f F dot dr = ∫_(0,2π) -(sinθ)^2 cosθ dθ = 0
so, the problem is wrong.

um...我不知道啊! 如果有問題可以不用做呢part!