__Stokes' Theorem__

**The surface-integral of the normal component of the curl of a vector field over an open surface yields the circulation of the vector field around its bounding contour.**

*Adjacent infinitesimal circulation densities cancel – leaving only the circulation components along the boundary curve, yielding the total circulation around that curve.*

- This article aims to explain the following ideas:
**(1)**\(\vplup\)How does the surface integral of the curl of a vector field \(\b{F}\) over a region \(D\) enclosed by a simple closed curve \(C\) yield the circulation of \(\b{F}\) around \(C\)?**(2)**How may Green’s Theorem be applied to determine the area of an enclosed region? For vector fields with what properties is this possible?**(3)**Can Green’s Theorem be applied to determine the area of an enclosed region with holes? How?**(4)**Why does Green’s Theorem only apply to enclosed regions without discontinuities?**(5)**In Stokes’ Theorem, why is the circulation around a simple closed curve \(C\) independent of the open surface enclosing it?**(6)**For a simple closed surface S and a vector field \(\b{F}\) continuous over S, what is the value of\(\ds{\oint_S{(\boldsymbol{\mathrm{\nabla }}\boldsymbol{\mathrm{\times }}\boldsymbol{\mathrm{\overrightarrow{F}}})\ {\cdot }}\ d\boldsymbol{\overrightarrow{S}}?}\)

What would be some of the implications if this value were anything else?**(7)**How is the curl of the vector field \(\bvec{F}=\left[\lfrac{-y}{x^2+y^2 }\right]\boldsymbol{\widehat{i}}+\left[\lfrac{x}{x^2+y^2 }\right]\boldsymbol{\widehat{j}}\) (shown below), where it is defined, zero everywhere – but its circulation, where it is

defined, around a simple closed curve \(C\) enclosing a region containing the origin, is not? What about curves*not*enclosing the origin?\(\hspace{80px}\)**(8)**In determining conservative vector fields in 3D, a hole within the region of interest is not a problem so long as it permits for some surface to be drawn within a region whose boundary is any simple closed curve within the region (but outside the discontinuity). That is, the discontinuity does not range continuously from one point of the enclosing surface to another – because then Stokes’ Theorem can be applied to test for conservativeness. In 2D, a curve can be drawn around a hole that does not range continuously from one point of the enclosing curve to another; why, then, is such a hole a problem in 2D but not in 3D?**(9)**Why must a curve/surface be simple & closed for Green's and Stokes' theorems to apply?

Links include visualizations, animations, and intuitive breakdowns of concepts underlying Stokes' theorem and vector fields in general.

** [1]**

(Circles separated for visualization purposes; imagine they are tangent at *P*)

\(\ds{\oint_C{\bnvec{F}\mathrm{\cdot }\ d\bvec{r}}=\iint_S {(\boldsymbol{\mathrm{\nabla }}\boldsymbol{\mathrm{\times }}\bnvec{F})\cdot k}\ dA.\ [1]}\)

the circulation of \(\b{F}\) around \(C\) is defined to be positive counterclockwise. Note that from the concept of surface integrals, the right-hand integral in Eq. [1] above sums the products of infinitesimal circulation densities (call them ‘microcirculations’) and infinitesimal region area increments for the entire region \(D.\) If the circulation arrows were drawn clockwise, each of the products would need to have the opposite sign – and thus the final integral would be opposite in sign as well. Also, for a single integration, the arrow orientations cannot alternate; that is, for integrating over a region \(D\), if some sub-regions \(D_n\) had circulation arrows going clockwise while others counterclockwise, the integral would be unable to account for those differences. (Doing so would require that the infinitesimal sums alternate in sign, whereas a single integral may only assume one sign.) With multiple integrals, aside from extra work, this is not a problem.Lastly, consider a visualization of the enclosed region for the limit as the ‘microcirculation’ approximation areas approach zero:

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**zero**. If the result were anything else – that is, if a finite circulation was computed, under the Stokes’ Theorem premise that the circulation around a simple closed curve is independent of the open surface enclosing it, every possible simple closed curve

*in the entire domain of*\(\b{F}\) would have the same value (by drawing any possible surface) – which is absurd.

** [7]**

All of the above suggests that that for any simple closed curve not enclosing the origin, the circulation is indeed zero.

What about simple closed curves that do enclose the origin? Assuming C1 is a circle of radius 4, \[\oint_{C_1}{\bnvec{F}\mathrm{\cdot }d\boldsymbol{r}}=\iint_{D_1}{(\boldsymbol{\mathrm{\nabla }}\boldsymbol{\mathrm{\times }}\bnvec{F})}dA\] \[C_1\left(t\right)=\left[4{\mathrm{cos} \left(t\right)}\right]\uvec{i}+\left[4{\mathrm{sin} \left(t\right)}\right]\uvec{j},\ \ \ \frac{\pi }{2}\le t\le \pi ;\] \[D_1\left(r,\theta \right)=\left[r{\mathrm{cos} \left(\theta \right)}\right]\uvec{i}+\left[r{\mathrm{sin} \left(\theta \right)}\right]\uvec{j},\ 0\le r\le 4,\ 0\le \theta \le 2\pi \] Then, \[\iint_{D_1}{(\boldsymbol{\mathrm{\nabla }}\times \bnvec{F})}dA=\int \limits_{0}^{2\pi }\int \limits_{0}^{4}{\frac{r^2{{\mathrm{cos}}^{\mathrm{2}} \left(\theta \right)}- r^2{{\mathrm{sin}}^{\mathrm{2}} \left(\theta \right)}}{{\left(r^2{{\mathrm{sin}}^{\mathrm{2}} \left(\theta \right)}+r^2{{\mathrm{cos}}^2 \left(\theta \right)}\right)}^2}drd\theta }=\int \limits_{0}^{2\pi }\int \limits_{0}^{4}{\frac{{\mathrm{cos} \left(2\theta \right)}}{r^2}drd\theta }=\mathrm{DNC^{1}}\] [1 - Does Not Converge] Thus, Green’s Theorem may not be applied to compute the circulation of \(\b{F}\) around \(C_1\) enclosing a region \(D_1\) containing the origin. The line integral, however, will work just fine in computing it: \[\bvec{F}\cdot\bvec{r}=\left[\left(\frac{-4{\mathrm{sin} \left(t\right)}}{4^2{{\mathrm{sin}}^{\mathrm{2}} \left(t\right)}+4^2{{\mathrm{cos}}^2 \left(t\right)}}\right)i+\left(\frac{4{\mathrm{cos} \left(t\right)}}{4^2{{\mathrm{sin}}^{\mathrm{2}} \left(t\right)}+4^2{{\mathrm{cos}}^2 \left(t\right)}}\right)\right]\cdot \left[\left(-4{\mathrm{sin} \left(t\right)}\right)i+\left(4{\mathrm{cos} \left(t\right)}\right)j\right]\] \[={{\mathrm{sin}}^{\mathrm{2}} \left(t\right)+{{\mathrm{cos}}^2 \left(t\right)=1}}\Rightarrow \] \[\oint_{C_1}{\bnvec{F}\mathrm{\cdot }d\boldsymbol{r}}=\int^{2\pi }_0{dt}=2\pi \] Not only does the line integral converge, but it is finite and non-zero – as expected:

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