Surface Integral of Vector Field


by ma3088
Tags: integral, parametrization, surface
ma3088
ma3088 is offline
#1
Nov24-08, 12:23 PM
P: 4
1. The problem statement, all variables and given/known data
Find [tex]\int\int_{S}[/tex] F dS where S is determined by z=0, 0[tex]\leq[/tex]x[tex]\leq[/tex]1, 0[tex]\leq[/tex]y[tex]\leq[/tex]1 and F (x,y,z) = xi+x2j-yzk.


2. Relevant equations
Tu=[tex]\frac{\partial(x)}{\partial(u)}[/tex](u,v)i+[tex]\frac{\partial(y)}{\partial(u)}[/tex](u,v)j+[tex]\frac{\partial(z)}{\partial(u)}[/tex](u,v)k

Tv=[tex]\frac{\partial(x)}{\partial(v)}[/tex](u,v)i+[tex]\frac{\partial(y)}{\partial(v)}[/tex](u,v)j+[tex]\frac{\partial(z)}{\partial(v)}[/tex](u,v)k

[tex]\int\int_{\Phi}[/tex] F dS = [tex]\int\int_{D}[/tex] F * (TuxTv) du dv

3. The attempt at a solution
To start off, I'm not sure how to parametrize the surface S. Any help is appreciated.
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HallsofIvy
HallsofIvy is offline
#2
Nov24-08, 01:47 PM
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PF Gold
P: 38,904
Since you are just talking about a portion of the xy-plane, x= u, y= v, z= 0. Oh, and the order of multiplication in [itex]T_u\times T_v[/itex] is important. What is the orientation of the surface? (Which way is the normal vector pointing?)

(Actually that last point doesn't matter because this integral is so trivial.)


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