Calculating Line Integrals Using Stokes' Theorem

Niles
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[SOLVED] Calculus - Stokes' theorem

Homework Statement


I have F in Cartesian coordinates (F is a vector): F = (y , x , x*z) and a curve C given by the quarter-circle in the z-plane z=1 (so t : (cos(t) , sin(t) , 1) for t between 0 and Pi/4).

I have found the line integral, and it equals 1/2.

For fun I wanted to find the same line-integral using Stokes' theorem, so I find the curl of F to be (0 , -z , 0) and dS I find by finding the normalvector, which is the cross-product between n_r and n_t. This gives a z-component (of course) with magnitude r - but then the surface integral is zero?

Where is my error?
 
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You can only use Stokes theorem if you have a closed contour. Try doing the contour integral around the whole boundary of the quarter circle. Then you should get zero.
 
Ah man, of course; hadn't thought of that.

You saved me again, thanks.
 
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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