6.7 Stokes’ theorem  (Page 2/10)

To approximate the flux over the entire surface, we add the values of the flux on the small squares approximating small pieces of the surface ( [link] ). By Green’s theorem, the flux across each approximating square is a line integral over its boundary. Let F be an approximating square with an orientation inherited from S and with a right side $E_l$ (so F is to the left of E ). Let $F_r$ denote the right side of $F$ ; then, $E_l=\text{−}{F}_r.$ In other words, the right side of $F$ is the same curve as the left side of E , just oriented in the opposite direction. Therefore,

As we add up all the fluxes over all the squares approximating surface S , line integrals ${\displaystyle {\int }_{E_l}\text{F}·d\text{r}}$ and ${\displaystyle {\int }_{F_r}\text{F}·d\text{r}}$ cancel each other out. The same goes for the line integrals over the other three sides of E . These three line integrals cancel out with the line integral of the lower side of the square above E , the line integral over the left side of the square to the right of E , and the line integral over the upper side of the square below E ( [link] ). After all this cancelation occurs over all the approximating squares, the only line integrals that survive are the line integrals over sides approximating the boundary of S . Therefore, the sum of all the fluxes (which, by Green’s theorem, is the sum of all the line integrals around the boundaries of approximating squares) can be approximated by a line integral over the boundary of S . In the limit, as the areas of the approximating squares go to zero, this approximation gets arbitrarily close to the flux.

Let’s now look at a rigorous proof of the theorem in the special case that S is the graph of function $z=f\left(x,y\right),$ where x and y vary over a bounded, simply connected region D of finite area ( [link] ). Furthermore, assume that $f$ has continuous second-order partial derivatives. Let C denote the boundary of S and let C ′ denote the boundary of D . Then, D is the “shadow” of S in the plane and C ′ is the “shadow” of C . Suppose that S is oriented upward. The counterclockwise orientation of C is positive, as is the counterclockwise orientation of $C^{\prime }.$ Let $\text{F}\left(x,y,z\right)=⟨P,Q,R⟩$ be a vector field with component functions that have continuous partial derivatives.

We take the standard parameterization of $S:x=x,y=y,z=g\left(x,y\right).$ The tangent vectors are ${\text{t}}_x=⟨1,0,g_x⟩$ and ${\text{t}}_y=⟨0,1,g_y⟩,$ and therefore, ${\text{t}}_x·{\text{t}}_y=⟨\text{−}{g}_x,\text{−}{g}_y,1⟩.$ By [link] ,

where the partial derivatives are all evaluated at $\left(x,y,g\left(x,y\right)\right),$ making the integrand depend on x and y only. Suppose $⟨x\left(t\right),y\left(t\right)⟩,a\le t\le b$ is a parameterization of $C^{\prime }.$ Then, a parameterization of C is $⟨x\left(t\right),y\left(t\right),g\left(x\left(t\right),y\left(t\right)\right)⟩,a\le t\le b.$ Armed with these parameterizations, the Chain rule, and Green’s theorem, and keeping in mind that P , Q , and R are all functions of x and y , we can evaluate line integral ${\displaystyle {\int }_C\text{F}·d\text{r}}\text{:}$