Stokes’ theorem converts a line integral to a surface integral. Divergence theorem relates a surface integral to a triple integral. So is it possible to take the result from Stokes’ theorem, and apply the divergence theorem to it? I’m not good at using latex, else I would’ve written down what I meant.

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    $\begingroup$ No, because divergence theorem requires a closed surface. If a surface had a line boundary to begin with, it could never have been closed. $\endgroup$ Dec 29, 2020 at 3:25
  • $\begingroup$ Here is a MathJax tutorial $\endgroup$ Dec 29, 2020 at 3:29
  • $\begingroup$ @Ninad Munshi, I recommend expanding that into an official answer so this doesn't stay on the unanswered list $\endgroup$
    – Mark S.
    Dec 29, 2020 at 3:32
  • $\begingroup$ Pls note you are dealing with two different measurements of a vector field in both cases - curl vs. divergence. $\endgroup$
    – Math Lover
    Dec 29, 2020 at 6:06
  • $\begingroup$ @Math Lover, Ignoring the boundary problem Ninad brought up, that's not a problem because they're being done consecutively: you'd take the divergence of the curl of the vector field that appears in the line integral...Except that the divergence of the curl of something is zero when the region has no holes (as in the applications of the divergence theorem). This duality between "derivative of derivative is zero" and "boundary of boundary is zero" is notable in deep perspectives on this like "de Rahm cohomology". $\endgroup$
    – Mark S.
    Dec 29, 2020 at 10:10

1 Answer 1


Short Answer

Arguably "yes, you can", but certainly not in a useful way. Any application of the combination of the Divergence Theorem and (Kelvin–)Stokes' Theorem results in $0$ for a trivial reason.


Using $\mathop{\rm bd}R$ to mean "the boundary of $R$", and ignoring some of the technical conditions of the theorems, there's no immediate reason why we can't chain them together to get something like the following (notations vary a lot) as long as all the conditions are satisfied: \begin{equation}\int_{\mathop{\rm bd}(\mathop{\rm bd}R)} \mathbf F\cdot\widehat{\mathbf{r}}\,\mathrm ds=\iint_{\mathop{\rm bd}R} (\nabla\times\mathbf F)\cdot\widehat{\mathbf{n}}\,\mathrm dS=\iiint_R\nabla\cdot(\nabla\times\mathbf F)\,\mathrm dV\tag{$\star$}\end{equation}

However, calculus puts some severe restrictions on what $\nabla\cdot(\nabla\times\mathbf F)$ can be, and geometry puts some severe restrictions on what $\mathop{\rm bd}(\mathop{\rm bd}R)$ can be. Either of these restrictions alone turn out to force the triple/single integral to be $0$.

Integrand Issue

In "nice situations" where the order of partial derivatives doesn't matter, it turns out that $\nabla\cdot(\nabla\times\mathbf F)=0$. You can see an algebraic proof in this answer to Why is $\operatorname{Div}\big(\operatorname{Curl} F\big) = 0$? Intuition? and get some intuition for it among the answers to What is an intuitive explanation for $\operatorname{div} \operatorname{curl} F = 0$?.

For reasons studied in real analysis, if the triple integral makes sense, we're guaranteed to be in a "nice situation" at almost all the points we're integrating over, so the triple integral must come out to be $0$. The main details are a version of Clairaut's Theorem and a generalization of Lebesgue's criterion for Riemann integrability (see, for example, the comments on this question).

This means that if you had a triple integral that you wanted to calculate via $\star$, you'd know it was $0$ without looking at the region $R$ because the integrand would just be $0$.

Domain Issue

Just like the integrand of the triple integral forced a value of zero, it turns out that the domain of integration of the single integral forces a value of zero, too.

This means that if you had a single integral that you wanted to calculate via $\star$, you'd know it was $0$ without looking at the integrand $\mathbf F$ because of the nature of the curve(s) you're integrating over.

Geometric Intuition

If you start with a nice round 3D region $R$ for the triple integral, like a ball or the region bounded by an ellipsoid or a bumpy variant*, then $\mathop{\rm bd}R$ would be a smooth surface with no edges. In that case, $\mathop{\rm bd}\mathop{\rm bd}R$, if it exists at all, would geometrically seem to be something trivial/without length — like the empty set, or a single point. Whatever it is, it would make the single integral equal to zero.

And if we instead have a region that's not so smooth, like a solid cube, then we could fix it by pulling in the edges and rounding them slightly. This would lose an arbitrarily tiny fraction of the volume (so the triple integral is not significantly affected), and then we'd again be in a situation where the single integral is forced to be zero.

*like those displayed in Samantha Driskill's Footenote 18 for Naala Brewer's Spring 2008 section of "Math 267 Calculus for Engineers III" at Arizona State University

Careful Consideration

We can understand this a bit more precisely by thinking about the orientation relationship between the surface normal(s) we use in the double integral, and the directions of any curves used in the single integral. For a refresher, the relationship required by (Kelvin–)Stokes' is described on Paul's Online Notes and in OpenStax Calculus Volume 3.

If we have a closed surface $\mathop{\rm bd}R$ that bounds a 3d region $R$, then there must be some of the surface on two sides of any curves making up the edges (in $\mathop{\rm bd}\mathop{\rm bd}R$) of the surface. For example, one of the $12$ edges of a cube bounds two faces. The equator of a sphere bounds two hemispheres, etc.

Note that the Divergence Theorem gives the surface an outward normal everywhere. But this forces the normal vectors for the pieces of the surface on two sides of an edge to go in "roughly opposite" directions — directions that require following the edge in opposite directions in order to satisfy the orientation condition of (Kelvin–)Stokes'. For example, in the cube below, the middle horizontal edge must be followed rightwards for the top blue face, but leftwards for the front yellow face:

cube with yellow face coming out of the screen, blue face on top, and orange face to the right

What this means is that any curve pieces in the single integral must be followed in both directions (an equal number of times), so that the integrals along each directed/oriented piece of the curve cancel out because switching the direction/orientation changes the sign. Since everything cancels out because of the nature of $\mathop{\rm bd}\mathop{\rm bd}R$, the singe integral must be $0$.

Deeper Context

The equations and patterns seen here generalize to higher dimensions.

Divergence and Stokes' Theorems

These theorems (and others like the fundamental theorem for line integrals) generalize to (generalized) Stokes' Theorem and the fundamental theorem of geometric calculus.

Getting zero

Zero Integrand

That the two "derivative" operations of divergence and curl end up giving you zero is generalized in the theory of differential forms (or closely-related things like "geometric calculus"): the exterior derivative gives you $0$ when applied twice, in any dimension. This is used as the basis of things like de Rham cohomology, where we can detect topological holes in a space by noting things like "on this region, the divergence of this field is zero even though it's not the curl of another field" (see the answer to Problem understanding a solenoidal vector field that is not a curl. for an example that shows a hole at the origin of 3D space).

"Zero" curves

That the boundary of the boundary of $R$ seems to give curves that "cancel out" comes up in discussions of simplicial homology and singular homology, which can also be used to measure topological holes.

There are deep connections between these sorts of things in algebraic topology, but they're not directly related to your question.


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