I've been taught Green's Theorem, Stokes' Theorem and the Divergence Theorem, but I don't understand them very well. In particular I don't understand in what circumstances I would choose to use any one of these over any one of the others. What criteria should I be looking for to help me decide? I specifically want to use the theorems to make integration as easy or trivial as possible.

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    $\begingroup$ Note that all three are the same thing! Or rather, Green's Theorem and the Divergence Theorem are both special cases of Stokes' Theorem, in 2 and 3 dimensions respectively. $\endgroup$ – user7530 Oct 22 '11 at 14:45
  • $\begingroup$ What a great question. I'm now going to read some answers, I hope at least one of them make a good case that's not only in math-speak. $\endgroup$ – Mattis Aug 12 '13 at 19:52

In simplest terms, the integral theorems can be described as follows:

Stokes' Theorem equates the single integral of a function $f$ along the boundary of a surface with the double integral of some kind of derivative of $f$ along the surface itself.

Gauss's Theorem (a.k.a. the Divergence Theorem) equates the double integral of a function along a closed surface which is the boundary of a three-dimensional region with the triple integral of some kind of derivative of $f$ along the region itself.

Thus the situation in Gauss's Theorem is "one dimension up" from the situation in Stokes's Theorem, so it should be easy to figure out which of these results applies. If you see a three dimensional region bounded by a closed surface, or if you see a triple integral, it must be Gauss's Theorem that you want. Conversely, if you see a two dimensional region bounded by a closed curve, or if you see a single integral (really a line integral), then it must be Stokes' Theorem that you want.

This is only two theorems: what about Green's Theorem? Green's Theorem is in fact the special case of Stokes's Theorem in which the surface lies entirely in the plane. Thus when you are applying Green's Theorem you are technically applying Stokes's Theorem as well, however in a case which leads to some simplifications in the formulas. Especially, when you have a vector field in the plane, the curl of the vector field is always a purely vertical vector, so it makes sense to identify this with a scalar quantity, and this scalar quantity is precisely the "derivative" appearing in the double integral in Green's Theorem.

For more on this perspective on the curl of plane vector fields, see this handout. For more on Green's Theorem (including two formulations of it, one in terms of "flux" and one in terms of "divergence" -- this seems like more than you are ready for at the moment so I didn't mention it, but perhaps later you will want/need it) see this handout. For a (possibly too brief; sorry) review of all three of these theorems, see this handout. These all came from a sophomore level course on multivariable calculus I taught to an audience of engineering students at Concordia University in 2004.

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    $\begingroup$ I never thanked you for this fantastic answer - I was too busy trying to wrap my head around it in time for the exam! I was only in first year, so this stuff was a wee bit advanced, but in the end I managed to score over 80% for the exam. Trust me, a VERY big portion of that was due to the links you posted here. Thanks a million! $\endgroup$ – Katherine Rix Feb 25 '12 at 15:31
  • $\begingroup$ The only way to improve this answer would be graphic figures that show the difference between each case. $\endgroup$ – Mattis Aug 13 '13 at 21:45

All these theorems have the same property: namely it relates the integral over some type of space to the integral over the boundary of that space. In the following examples $A$ is some kind of space (2, 3, 4, etc dimensional) and $\partial A$ represents its boundary. For example, if $A$ is a surface, then its boundary is a line. If $A$ is a volume, its boundary is a surface. If $A$ is 4-dimensional, then its boundary could be a 3-dimensional volume.

Here are some particular examples:

  • Green's Theorem (we turn the double integral over a SURFACE to a line integral around its BOUNDARY, a line): $$\int\int_A \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dxdy = \oint_{\partial A} (Ldx + Mdy)$$

  • Divergence Theorem (we turn the triple integral over a VOLUME into a surface integral around its BOUNDARY, a surface):

$$\int\int\int_V (\nabla \cdot F) dV = \int\int_{\partial V} (F \cdot n) dS$$

An instructive example follows:

  • the Fundamental theorem of calculus (here we turn an integral over an INTERVAL into an "integral" over its BOUNDARY, the endpoints (integration on the right is called "integration over $0$-forms", it's written this way to emphasize the use of the boundary of $[a,b]$ in the regular fundamental theorem of calculus): $$\int_a^b f'(x) dx = f(b) - f(a) = \int_{\{b,-a\}} f'$$

All of these theorems have the same "general form" called "the general Stokes' theorem":

$$\int_A df = \int_{\partial A} f$$

where $df$ is defined appropriately.

All of the integrands in these theorems are related to each other by what's called the "exterior derivative" of the "differential form". The "differential form" is literally your integrand (including the dx) and the calculus of differential forms explains how you perform a computation to get from one type of integrand to another (how you get from $\left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dxdy$ to $Ldx + Mdy$ in the case of Green's Theorem).

The details here aren't important, but it should illustrate to you that to use these theorems, you need to

  1. Figure out which one you are "in" by writing it in a form that matches those in the theorems.

  2. Figure out which one is related to the one you are "in", i.e. the other side of the "=" sign to the one you are "in".

  3. Rewrite what you have with respect to the one you are related to, usually by taking a derivative or finding anti-derivatives.

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    $\begingroup$ Also a really, really good answer that got me thinking about calculus and the underpinnings of it a lot. I wish I was able to accept this as an answer too! $\endgroup$ – Katherine Rix Feb 25 '12 at 15:32

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