Surface Integrals

2011-07-20 9:56 pm
http://i707.photobucket.com/albums/ww74/stevieg90/ScreenHunter_03Jul192329.gif
更新1:

To Lotus: 我是用vector的做法去做的...例如3a: 我是let ψ=9x²-y²+9z², δψ=18x[i]-2y[j]+18z[k], ψ_z=18z 然後用project S onto xz-plane的計法,請問這方法要如何做?

更新2:

例如http://i707.photobucket.com/albums/ww74/stevieg90/ScreenHunter_03Jul212314.gif的做法,你看一看,THX!

更新3:

明白了,LOTUS,感激萬分!

回答 (1)

2011-07-22 4:39 am
✔ 最佳答案
The surface y =3√(x^2+z^2) , y=3~9 can be parametrized as
R=(x,y,z)=(rcosθ, 3r, rsinθ), r=1~3, θ=0~2π
dS=| ∂R/∂r x ∂R/∂θ| dr dθ = √10 r dr dθ
∫∫_S y dS
=∫[0~2π] ∫[1~3] 3r*√10 r dr dθ
=52π√10

2011-07-21 22:49:22 補充:
(b)integrate directly.
R=(x,y,z)=(rcost, rsint, 3-rcost-rsint), r=0~6, t=0~2π
dS= ∂R/dr x ∂R/dt dr dt= (r, r, r) dr dt
∫∫_s F dot dS
=∫[0~2π] ∫[0~6] [3r^2 cost+ r - 2(3-rcost-rsint) r ] dr dt
=∫[0~2π] [216 cost- 90+ 144(sint+cost) ] dt
= -180π

2011-07-21 23:41:23 補充:
ydS=y* dx dz/|cosβ| = y dx dz/(2y) *| grad(9x^2-y^2+9z^2) |
= √(81x^2+y^2+81z^2) dx dz=3√10 *√(x^2+z^2) dx dz
∫∫s ydS=∫∫s_zx 3√10 *√(x^2+z^2) dx dz
=∫[0~2π]∫[1~3] 3√10 *r * rdr dt
=52√10*π

2011-07-21 23:59:32 補充:
Projection 的方法沒有比較好


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