I am working on an integration problem, where Stokes' theorem seems to be applicable.
I have found the curl of F, and the normal vector, n, to the surface. The surface that I will integrate over is a cross-section of a sphere of radius a, gotten from an intersection of a plane with this sphere. So the curve from this intersection is closed, as necessary for application of the theorem.
My questions are: what exactly does ds represent? Can I think of it as the "area element", except that it's now in space, and not on the xy-plane? I'm guessing it depends on the context — so, ds is just dxdy on the xy-plane, while ds is $rdrd\theta$, on some disk. Is my thinking correct?
And the normal vector n = (1,1,1) doesn't necessarily emanate from the origin, (0,0,0), right? I got it from computing cross-products of two vectors on the disk. But, say, the disk does not pass through the origin, and is way higher and above the xy-plane, then the normal vector, n, emanates from the disk, not from the origin. Does this violate any rules of vector calculus — without getting into differential geometry stuff. Because don't vectors, by definition, start from the origin?
Thanks,
Best Answer
"Stokes' Theorem" is a name given to a much broader theorem than you appear to be refering to here. I am assuming you are referring to the Curl Theorem (one special case of Stokes' Theorem): $$\int_S \nabla \times \mathbf{F}\cdot d\mathbf{a} = \int_{\partial S}\mathbf{F}·d\mathbf{s}$$ In this case it is $d\mathbf{a}$ that represents the area element on the surface. $d\mathbf{s}$ is the element of arclength along the curve $\partial S$ which forms the boundary of the surface $S$.