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Surface Integral of Vector Field |
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| Nov24-08, 12:23 PM | #1 |
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Surface Integral of Vector Field
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. |
| Nov24-08, 01:47 PM | #2 |
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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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| integral, parametrization, surface |
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