3.5 Stokes' theorem

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 andtake 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 doneis 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 thiscan provide some suitable drawings I would incorporate them with an acknowledgment). The following figures are not as good as in the book, butwill have to do for now. We have some surface with a vector field passing throughit:

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

So for this simple example $${\oint }_{C_i}\vec{F}\cdot d\overrightarrow{s_i}=\Delta x\Delta y\left[\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right]$$ which is just ${\left(\vec{\nabla }\times \vec{F}\right)}_z$ or if we define $\Delta \vec{A}$ such that it has the area $\Delta x\Delta y$ and a direction perpendicular to the $xy$ plane it is $${\oint }_{C_i}\vec{F}\cdot d\overrightarrow{s_i}=\Delta \vec{A}\cdot \left(\vec{\nabla }\times \vec{F}\right)$$ We have shown the above is true for a square in the $\{x,y\}$ plane. Similarly we would look at the other possible planes and in $\{xz\}$ plane would get $$\Delta x\Delta z\left[\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x}\right]$$ which is ${\left(\vec{\nabla }\times \vec{F}\right)}_y$ . In the $\{yz\}$ plane we get $$\Delta y\Delta z\left[\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\right]$$ which is ${\left(\vec{\nabla }\times \vec{F}\right)}_x$

So in the 3d case we see that $$$$ Now $${\oint }_C\vec{F}\cdot d\vec{s}=\sum _{i=1}^N{\oint }_{C_i}\vec{F}\cdot d\overrightarrow{s_i}$$ becomes $${\oint }_C\vec{F}\cdot d\vec{s}=\sum _{i=1}^N\Delta {\vec{A}}_i\cdot \vec{\nabla }\times \vec{F}$$ which in the infinitesimal limit is $${\oint }_C\vec{F}\cdot d\vec{s}={\int }_{S}d\vec{A}\cdot \vec{\nabla }\times \vec{F}\text{.}$$ This is the fundamental definition of curl. We have shown that the del operator "crossed" into the vector field captures the definition of theoperation.

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 calledthe "curl theorem". This is a physics course though, so we will call it Stoke's theorem.