Surface Integral of Vector Field

ma3088
Messages
4
Reaction score
0

Homework Statement


Find \int\int_{S} F dS where S is determined by z=0, 0\leqx\leq1, 0\leqy\leq1 and F (x,y,z) = xi+x2j-yzk.


Homework Equations


Tu=\frac{\partial(x)}{\partial(u)}(u,v)i+\frac{\partial(y)}{\partial(u)}(u,v)j+\frac{\partial(z)}{\partial(u)}(u,v)k

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

\int\int_{\Phi} F dS = \int\int_{D} 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.
 
Physics news on Phys.org
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 T_u\times T_v 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.)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top