Stokes Theorum Surface integral

[wil3]

Let's assume that I have a surface defined parametrically by a vector \mathbf{\<br /> r}(r,\theta)

Is it acceptable to simplify the Stokes theorum surface integral to:

\iint\limits_D\,\nabla \times f \cdot\!(r_r\times\!r_\theta) \,\, \!r \mathrm{d}r\,\mathrm{d}\theta

Where r_r and r_theta are the derivatives of the parametrized vector with respect to r and theta. In other words, I canceled out the magnitude of the normal vector with the 3D jacobian that turns the flat area element into a 3D area element. It seems to me like this should work, but I got the wrong answer to a problem, and I couldn't find an error in my work, leading me to suggest I was doing this wrong.

Thank you very much for any advice. Happy Christmas.

 

[RadioactivMan]

Maybe, r_rXr_thetha is not the right normal vector it could be r_thetaXr_r.

 

[wil3]

I was thinking that, too, but I'm pretty sure that all that does is flip the sign. My answer is numerically off, but my convention for the outward normal gives the correct sign. Thank you very much for your guess, though.

 

[billturner90]

i just finished up my multivariable class so hopefully i can be of help. I think that stokes theorem relates the line integral to the curl of the Force field. I think the proper set up would be Del x F r dr dtheta.

 

[wil3]

So, it turns out that a parametrized surface requires the analogue of the Jacobian to be the absolute value of(!r_r\times\!r_\theta), and this "3D" Jacobian swallows the "r" that is the Jacobian for polar corrdinates. What this means is that, when I simpliy the integral formula, I should the "r" into the integral, because we are no longer using a generic area element, like dxdy=rdrd0, and instead are just using the two parametric differentials of the surface, drd0

The correct equation is:

<br /> \iint\limits_D\,\nabla \times f \cdot\!(r_r\times\!r_\theta) \,\, \mathrm{d}r\,\mathrm{d}\theta <br />

Long story short, I put an "r" in that shouldn't have been there. Stoke's theorum for parametric surfaces does not involve the differential area element unless we are dealing with a Cartesian-like coordinate system, because the Jacobian when we convert among coordinate systems is already accounted for.

Thanks very much for your replies.

 

[7thSon]

wil,

I actually had a similar problem and just figured it out before visiting this thread :).

I had another question though, which highlights my lack of knowledge about the change of variables theorem. I am wondering why, in general, the inner product in this surface integral stays a Euclidian inner product and not the derived inner product for the curvilinear coordinate system (also suppose that this coordinate system could be non-orthogonal).

It seems that once there is a change in variables for the vectors in the vector surface integral, there would need to be a change in the inner product operator as well.

Can anyone shed light on this?

 

[7thSon]

so i just got a partial answer to my own question...

it turns out that in orthogonal [CORRECTION: orthonormal] curvilinear coordinates (such as polar), the inner product and cross-product are unchanged because the relation between the curvilinear bases is still given by the Kronecker delta.

So that's enlightening, but now I still don't know the answer for non-orthogonal coordinates!