Computing ndA (dS) in Stoke's theorem

[chy1013m1]

in one side of Stoke's theorem we compute curl(F ) . ndA .
When we have computed curl(F ) in x-y-z coordinate, but have parametrized the surface in cylindrical / spherical coordinates, then in computing ndA, we do the cross product of the partials then times that by du dt (or somethin else) . Can we then transform curl(F ) using the transformation of the surface and dot the 2 quantities together ?

also, there are cases where curl(F ) and n are easy to compute in the x-y-z coordinates, but the surface can be described easily in spherical coordinate. In that case how do we correctly proceed ? do we have to make sure the vector n is normalized (in x-y-z) then compute dA as |dG/du x dG/dv|dudv ?

thanks

 

[HallsofIvy]

in one side of Stoke's theorem we compute curl(F ) . ndA .
When we have computed curl(F ) in x-y-z coordinate, but have parametrized the surface in cylindrical / spherical coordinates, then in computing ndA, we do the cross product of the partials then times that by du dt (or somethin else) . Can we then transform curl(F ) using the transformation of the surface and dot the 2 quantities together ?

That's awkwardly phrased but yes, you convert curl(F) to your parameters and then multiply.

also, there are cases where curl(F ) and n are easy to compute in the x-y-z coordinates, but the surface can be described easily in spherical coordinate. In that case how do we correctly proceed ? do we have to make sure the vector n is normalized (in x-y-z) then compute dA as |dG/du x dG/dv|dudv ?

? Normalized is normalized! n must have length 1. It doesn't matter what coordinates it is written in- the length of a vector is independent of the coordinate system.

I personally dislke the notation $\vec{n}\cdot d\sigma$ for exactly this reason. If you take it literally, you would do exactly what you say here: calculate $\vec{n}$ by finding the "fundamental vector product", $\partial G/\partial u X \partial G/\partial v$ and dividing by its length. But then you find $d\sigma$ by multiplying dudv by that length! The two cancel out! Much better is to think of the "vector differential of surface area", $d\vec{\sigma}= \partial G/\partial u X \partial G/\partial v du dv$.