Calculus - Stokes' theorem

  • Thread starter 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?
 

Answers and Replies

  • #2
Dick
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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.
 
  • #3
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Ah man, of course; hadn't thought of that.

You saved me again, thanks.
 

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