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", $d\vec{\sigma}= \partial G/\partial u X \partial G/\partial v du dv$.

 

[tacman]

for your question!

In Stoke's theorem, we are computing the line integral of a vector field over a closed curve C, which is equal to the surface integral of the curl of the vector field over the surface bounded by C. In order to calculate ndA, we first need to compute the curl of the vector field in x-y-z coordinates. However, if the surface is parametrized in cylindrical or spherical coordinates, we need to transform the curl of the vector field using the appropriate transformation equations. This means taking the cross product of the partial derivatives in the new coordinate system and multiplying it by the appropriate scaling factor (such as du dt or dudv). This will give us the correct value for ndA in the new coordinate system.

In cases where the vector field and n are easy to compute in x-y-z coordinates, but the surface is described more easily in spherical coordinates, we can still apply Stoke's theorem by transforming the vector field and n into the new coordinate system. We do this by first normalizing n in x-y-z coordinates, and then using the appropriate transformation equations to calculate dA as the magnitude of the cross product of the partial derivatives of the surface in the new coordinate system. This will give us the correct value for ndA in the new coordinate system.

In summary, when applying Stoke's theorem, we need to make sure that the vector field and n are in the same coordinate system. If the surface is described in a different coordinate system, we need to transform the vector field and n accordingly using the appropriate transformation equations. This will ensure that we are correctly computing ndA and can accurately apply Stoke's theorem.