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Question about Stokes theorem

By Stokes theorem, I mean the one in the blue box here.
http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx

I want to know what it means when it says that C bounds the surface. Like suppose we have the surface of a sphere centered at the origin. What would a boundary curve be? It seems that it could be a circle centered at the origin in some plane, but surely that doesn't cover all points of the surface? I mean, doesn't it only cover the points in the surface that are in the same plane as the circle?

Why is this version of Stokes theorem so different to the one I am used to?
http://en.wikipedia.org/wiki/Stokes'_theorem#General_formulation

Reply 1

thebadgeroverlord

Original post by gangsta316

I don't know how much help I can be as my recollection of some areas of vector calculus, like surface integrals, is slighty dodgy but I'll try and help.

If you take the top half of a sphere centered at the origin, then its boundary curve is just the circle of the same radius in the z = 0 plane like in the Pauls online notes. I think this is becasue the boundary curve is the curve that bounds all of the points on the surface when projected onto a z=constant plane. I think you need to take to pick a specific z plane to consider but for the ones you would be asked to calculate it should be fairly obvious. For example for a nothern hemisphere of a sphere not centred at the origin you would take the curve in the same z plane as the origin of the sphere.

Now for a sphere centred at the origin. You should probably check this is true yourself. I think it is but I'm not sure.

For Stokes theorem you have to consider the orientation of your boundary curve ie which way you go round it. The orientation is taken so that if you walk round the curve the surafce would be on your left and the outward pointing normal of the surface pointing to your right. This means for the sphere you have a problem with orientation of the curve. If you split your sphere up into northern and southern hemisphere then you apply Stokes Theorem to both terms. I think you will then be integrating the sane function around the same curve but in the opposite directions but these shpuld cancel so the integral is zero. I would check this yourself.

As for the form of Stokes Theorem on Wikipedia. Thats the general version of Stokes theorem for integration on n-dimensional objects (manifolds). Calculius on manifolds is the generalisation of vector calculus to n dimensions and is generally fourth year or graduate work but a bastardised version would appear in a General Relativity course. . The Fundamental Theorem of Calculus, Green's Theorem, Divergence Theroem and your version of Stoke's Theorem are all special cases of the general one.

Hope this helps