Evaluating a Surface Integral: xze^y i -xze^y j +z k

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Homework Statement



Evalute the surface integral

Homework Equations



F(x,y,z)=xze^y i -xze^y j +z k for the surface is partof the plane x+y+2z=2 in the first octant and orientated downwards

The Attempt at a Solution



\displaystyle \int \int_{\sigma} F dS=\int \int_R (xze^y i -xze^y j +z k)(z_x i+ z_y j -k) dA=\int \int_R (x^2z^2e^y-xyz^2e^y-z) dA


Is this correct so far...if so have I to substitute for z and put in above integral. Looks like a difficult integral...?
 
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Something like

\displaystyle \int \int_{\sigma} F dS=\int \int_R (xze^y i -xze^y j +z k)(z_x i+ z_y j -k) dA=\int \int_R (x^2z^2e^y-xyz^2e^y-z) dA \implies

\displaystyle \int \int_{\sigma} F dS=\int \int_R (x^2(\frac{2-x-y}{2})^2 e^y-xy(\frac{2-x-y}{2})^2 e^y-(\frac{2-x-y}{2})) dA.?

Posted at this link also. Will notify both forums of any responses. Thanks
http://www.freemathhelp.com/forum/threads/73614-surface-integral
 
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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...
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