I'm having difficulties deciding the truths of the two following statements. The first one I believe is false, but I'm not entirely sure and the second one I don't know how to make heads or tales of. Any help would be much appreciated. Thank you.

First: The flux of $curl(F)$ through every closed, piecewise-smooth, oriented surface is zero when $F$ is a $C^2$ vector field.

Second: From the divergence theorem one can conclude that the flux of $$F(x,y,z)=<x^2,y+e^z,y-2xz>$$ through every closed, piecewise-smooth, oriented surface $S$ is equal to the surface area of the solid $W$ enclosed by $S$.

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    $\begingroup$ Why do you think the first statement is false? In this forum it's important that you give reasons, not just assertions. (Hint for the first problem: think about the boundary of your surface.) $\endgroup$ – symplectomorphic Apr 29 '14 at 1:47
  • $\begingroup$ Ah okay the hint helped! So it is indeed true. Thank you. Also, do you have any advice for the second? $\endgroup$ – Valentino Apr 29 '14 at 1:52

The best way to understand Stokes's theorem in its many forms (the divergence theorem, Green's theorem, etc.) is to understand what those various forms allow you to do:

  • The divergence theorem allows you to turn the flux of a vector field $\mathbf{F}$ through a closed surface $S$ into a volume integral of $\text{div}\,\mathbf{F}=\nabla\cdot\mathbf{F}$ over the volume enclosed by the surface. (The surface has to be closed, i.e. have no boundary, in order for "the volume enclosed by the surface" to make sense.) In short, the divergence theorem allows the transformation $$\left\{\text{flux integral of a vector field}\right\}\leftrightarrow\left\{\text{volume integral of the divergence}\right\}$$ So you should reach for this theorem only if you want to transform a flux integral into a volume integral or a volume integral into a flux integral.
  • Stokes's theorem (which, in the form I'm about to give it, is really only a special case of a more general theorem of the same name) allows you to the flux of $\text{curl}\,\mathbf{F}=\nabla\times\mathbf{F}$ through a surface $S$ with boundary $\partial S$ into a line integral of $\mathbf{F}$ over the boundary $\partial S$. In short, Stokes's theorem allows the transformation $$\left\{\text{flux integral of the curl}\right\}\leftrightarrow\left\{\text{line integral of the vector field}\right\}$$ So you should only reach for this theorem if you want to transform the flux integral of a curl into a line integral. Notice that this theorem doesn't require the surface to be closed: your surface can have a boundary; indeed the boundary is the "line" you integrate over in the line integral.

These heuristics should immediately help you understand the first problem: you're asked about the flux of a curl through a closed surface... but closed surfaces don't have boundaries...

In both of these examples, the trick to remembering them is that the "derivative operator" changes places from one side of the equation to the other. In its most general form, Stokes's theorem is a statement about a kind of "duality" between derivatives of vector fields (divergences and curls are just different notions of the "derivative" of a vector field) and "derivatives" of manifolds (like curves and surfaces and volumes), where taking the "derivative" of a manifold means taking its boundary. So think of Stokes's theorem in its most general form as the statement $$\boxed{\int_{\text{some manifold}} (\text{derivative of vector field})=\int_{\text{boundary of the manifold}}(\text{the vector field})}$$

The major point is that on one side the derivative is upstairs (in the integrand) and on the other side the derivative, aka the boundary, is downstairs (in the region you're integrating over). (If you know anything about "adjoints," what Stokes's theorem is really saying is that the "exterior derivative" of things like vector fields -- I say "things like" because in truth the theorem applies to objects called "differential forms" -- and the boundary operator are adjoints.)

Thinking of the theorem in this way allows you to remember all the various specialized versions dealing with curls, divergences, etc. All you have to do is move the "derivative" upstairs and downstairs depending on the context.


Both would be true if the second said the volume of the solid $W$.

  • $\begingroup$ Thank you very much Dr. Shifrin! $\endgroup$ – Valentino Apr 29 '14 at 1:57

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