Stokes' theorem

This is derived in a similar fashion to Gauss' theorem, except now, instead of considering a volume and taking a surface integral, we consider a surface and take a line integral around the edge of the surface. In this case the surface does not enclose a volume. A good picture that describes what is being done is figure 2.23 in Berkeley Physics Course Volume 2. (Unfortunately it is copyrighted, and so it can not be shown on the web - If somebody reading this can provide some suitable drawings I would incorporate them with an acknowledgment). The following figures are not as good as in the book, but will have to do for now. We have some surface with a vector field passing through it:

We can take the line integral around the edge of the surface and evaluate that. We can also slice the surface into two parts The line integral along the common edge will have opposite signs for each half and so the sum of the two individual line integrals will equal the line integral of the complete surface. We can subdivide as much as we want and we will always have: ∮ C F ⃗ ⋅ d s ⃗ = ∑ i = 1 N ∮ C i F ⃗ ⋅ d s i ⃗ . ∮ C F ⃗ ⋅ d s ⃗ = ∑ i = 1 N ∮ C i F ⃗ ⋅ d s i ⃗ . Since we can subdivide as much as we want we can break the surface down into a collection of tiny squares, each of which lies in the { x , y } { x , y } , { y , z } { y , z } or { x , z } { x , z } planes.

Consider asquare in the x y plane and lets find the line integral Let us find the line integral of F ⃗ F ⃗ . First consider the x component at the center of the bottom of the square F x = F x ( x , y ) + Δ x 2 ∂ F x ∂ x F x = F x ( x , y ) + Δ x 2 ∂ F x ∂ x and then at the center of the top F x = F x ( x , y ) + Δ x 2 ∂ F x ∂ x + Δ y ∂ F x ∂ y F x = F x ( x , y ) + Δ x 2 ∂ F x ∂ x + Δ y ∂ F x ∂ y So multiplying by Δ x Δ x and subtracting the top and bottom to get the F x F x contribution to the line integral gives − Δ x Δ y ∂ F x ∂ y − Δ x Δ y ∂ F x ∂ y Minus sign comes about from the integration direction (if at the top F x F x is more positive this results in a negative contribution)Similarly in y y we get the contribution Δ x Δ y ∂ F y ∂ x Δ x Δ y ∂ F y ∂ x because F y F y is more positive in the right you get a positive contribution

So for this simple example ∮ C i F ⃗ ⋅ d s i ⃗ = Δ x Δ y [ ∂ F y ∂ x − ∂ F x ∂ y ] ∮ C i F ⃗ ⋅ d s i ⃗ = Δ x Δ y [ ∂ F y ∂ x − ∂ F x ∂ y ] which is just ( ∇ ⃗ × F ⃗ ) z ( ∇ ⃗ × F ⃗ ) z or if we define Δ A ⃗ Δ A ⃗ such that it has the area Δ x Δ y Δ x Δ y and a direction perpendicular to the x y x y plane it is ∮ C i F ⃗ ⋅ d s i ⃗ = Δ A ⃗ ⋅ ( ∇ ⃗ × F ⃗ ) ∮ C i F ⃗ ⋅ d s i ⃗ = Δ A ⃗ ⋅ ( ∇ ⃗ × F ⃗ ) We have shown the above is true for a square in the { x , y } { x , y } plane. Similarly we would look at the other possible planes and in { x z } { x z } plane would get Δ x Δ z [ ∂ F x ∂ z − ∂ F z ∂ x ] Δ x Δ z [ ∂ F x ∂ z − ∂ F z ∂ x ] which is ( ∇ ⃗ × F ⃗ ) y ( ∇ ⃗ × F ⃗ ) y . In the { y z } { y z } plane we get Δ y Δ z [ ∂ F z ∂ y − ∂ F y ∂ z ] Δ y Δ z [ ∂ F z ∂ y − ∂ F y ∂ z ] which is ( ∇ ⃗ × F ⃗ ) x ( ∇ ⃗ × F ⃗ ) x

So in the 3d case we see that Now ∮ C F ⃗ ⋅ d s ⃗ = ∑ i = 1 N ∮ C i F ⃗ ⋅ d s i ⃗ ∮ C F ⃗ ⋅ d s ⃗ = ∑ i = 1 N ∮ C i F ⃗ ⋅ d s i ⃗ becomes ∮ C F ⃗ ⋅ d s ⃗ = ∑ i = 1 N Δ A ⃗ i ⋅ ∇ ⃗ × F ⃗ ∮ C F ⃗ ⋅ d s ⃗ = ∑ i = 1 N Δ A ⃗ i ⋅ ∇ ⃗ × F ⃗ which in the infinitesimal limit is ∮ C F ⃗ ⋅ d s ⃗ = ∫ S d A ⃗ ⋅ ∇ ⃗ × F ⃗ . ∮ C F ⃗ ⋅ d s ⃗ = ∫ S d A ⃗ ⋅ ∇ ⃗ × F ⃗ . This is the fundamental definition of curl. We have shown that the del operator "crossed" into the vector field captures the definition of the operation.

As a side note, Physicists uniformly refer to the above as Stoke's theorem but in fact that is a more general theorem and the above can correctly be called the "curl theorem". This is a physics course though, so we will call it Stoke's theorem.