Calculating flux through a surface area

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To calculate the flux of the vector field F through the surface defined by the curve r(u,v), the user first computed the partial derivatives dr/du and dr/dv. They then used the cross product to find the normal vector N, resulting in a lengthy expression. The user expressed concern about the complexity of their calculations and sought a simpler method. It was suggested that the complexity of the surface could affect the calculations, and the use of the divergence theorem or Stokes' theorem was recommended as potential alternatives. Simplifying the approach may depend on the specific form of r(u,v).
dan38
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S is a portion of a curve with r(u,v)
where 0 < u < 2 and 0 < v < 2pi

I'm meant to calculate Flux of the vector field F

My Calculations
First found dr/du
then dr/dv

Using the cross product, I found N = (- u cos (v) + 5 sin (v), -5 cos (v) - u sin(v), u)
Then I dot product with the given F (after putting it in terms of "u" and "v")
Then I do the relevant integration

I think my method is correct, however since my cross product yields an extremely long vector it makes all my following calculations really tedious.
So just want to check if there's a easy way to do this.
 
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There is no way to answer that without knowing what r(u,v) is. Some surfaces have "simple" differentials of surface area, some complicated ones. (What you give does NOT seem "extremely long" to me!)
 
Use the divergence theorem or Stokes' theorem.

BiP
 
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