Surface Integral of Vector Field

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SUMMARY

The discussion focuses on calculating the surface integral of the vector field F(x, y, z) = xi + x²j - yzk over the surface S defined by z=0 and the bounds 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. The parametrization of the surface is established as x = u, y = v, z = 0. The key equations used include the transformation equations for the surface and the integral formula, specifically \int\int_{\Phi} F dS = \int\int_{D} F * (T_u \times T_v) du dv. The orientation of the normal vector is discussed, although it is noted that it does not affect the outcome of this particular integral.

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  • Understanding of surface integrals in vector calculus
  • Familiarity with parametrization of surfaces
  • Knowledge of vector fields and their operations
  • Proficiency in using the cross product in three-dimensional space
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  • Study the concept of surface integrals in vector calculus
  • Learn about parametrization techniques for various surfaces
  • Explore the properties of vector fields and their integrals
  • Review the application of the cross product in calculating normal vectors
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Homework Statement


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.


Homework 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

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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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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