# Stokes' Theorem and vector integrand in curvilinear coordinates

I want to compute the following vector integral:

$$\iint_\Omega d\Omega \{\nabla_s\times \mathbf{F}\} \tag{1}\label{eq1}$$

Note that $$d\Omega$$ is a scalar surface element, the result should therefore be vectorial. Further we can not apply Stoke's Theorem, as it would require an oriented surface element $$\mathbf{d\Omega}$$ such that the integrand becomes a scalar.

In the particular case I'm studying $$\mathbf{F}$$ is tangential to the surface $$\Omega$$, so in curvilinear coordinates, using Dupin's orthogonal system $$\hat{\mathbf{v}}_1, \hat{\mathbf{v}}_2, \hat{\mathbf{n}}$$:

$$F_n=0 \implies \mathbf{F} = F_1 \hat{\mathbf{v}}_1+ F_2 \hat{\mathbf{v}}_2,$$

with $$\hat{\mathbf{v}}_1, \hat{\mathbf{v}}_2$$ being aligned to the lines of curvature.

The surface curl $$\nabla_s\times$$ is defined as

$$\nabla_s \times \mathbf{G} = (\nabla_s G_n)\times\hat{\mathbf{n}}+\frac{G_2}{R_2}\hat{\mathbf{v}_1}-\frac{G_1}{R_1} \hat{\mathbf{v}_2} + \hat{\mathbf{n}} \nabla_s\cdot(\mathbf{G}\times \hat{\mathbf{z}})$$

This definition was found in "J. van Bladel, Electromagnetic fields, 2nd ed., 2007" and has the advantage that there is no normal derivative applied to the tangential components, which is important when dealing with surface vectors that are non-zero only on the surface $$\Omega$$. The quantities $$R_1, R_2$$ denote the radii of curvature.

Now, considering first a plane in Cartesian coordinates, we have $$R_1=R_2=\infty$$ and a constant normal which is invariant to the location, e.g., $$\hat{\mathbf{n}}=\hat{\mathbf{z}},\hat{\mathbf{v}_1}=\hat{\mathbf{x}},\hat{\mathbf{v}_2}=\hat{\mathbf{y}}$$, and therefore

$$\nabla_s \times \mathbf{F} = \hat{\mathbf{z}} \nabla_s\cdot(\mathbf{F}\times \hat{\mathbf{z}}) \\ \iint_\Omega d\Omega \{\nabla_s\times \mathbf{F}\} = \hat{\mathbf{z}} \iint_\Omega d\Omega \{ \nabla_s\cdot(\mathbf{F}\times \hat{\mathbf{z}}) \}.$$

In the last equation it was possible to move $$\hat{\mathbf{z}}$$ in front of the integral because $$\hat{\mathbf{z}}$$ does not vary over $$(x,y)$$. The integrand is now a scalar, and thus the (Surface) Divergence Theorem can be applied which yields

$$\hat{\mathbf{z}} \iint_\Omega d\Omega \{ \nabla_s\cdot(\mathbf{F}\times \hat{\mathbf{z}}) \} = \hat{\mathbf{z}} \int_{\partial\Omega} dc \{ (\mathbf{F}\times \hat{\mathbf{z}})\cdot \hat{\mathbf{m}} \} = \hat{\mathbf{z}} \int_{\partial\Omega} dc \{ \mathbf{F} \cdot \hat{\mathbf{t}}\},$$

where $$\hat{\mathbf{t}}$$ is the unit vector aligned to the boundary $$\partial\Omega$$ and $$\hat{\mathbf{m}}$$ is defined such that $$\hat{\mathbf{n}}\times\hat{\mathbf{m}}=\hat{\mathbf{t}}$$. In other words, a transformation of $$\eqref{eq1}$$ into a boundary line integral is possible for a plane in Cartesian coordinates by using Stokes' or the Divergence Theorem, because the normal is invariant. As an application, e.g., take the case that $$\mathbf{F}$$ vanishes outside a circular region, centered at the origin. If the boundary $$\partial\Omega$$ encloses this region, we find

$$\iint_\Omega d\Omega \{\nabla_s\times \mathbf{F}\} =0,$$

because $$\mathbf{F} \cdot \hat{\mathbf{t}}=0$$ on the entire boundary $$\partial\Omega$$.

Now I would like to check if the same applies for curved surfaces. More importantly, I'm wondering if the surface $$\Omega$$ is closed (consider e.g. a sphere), that is it has no Stokes Boundary $$\partial\Omega$$, is it true that

$$\iint_\Omega d\Omega \{\nabla_s\times \mathbf{F}\} = 0 ?$$

I was not able to find any hint in my favorite text books ... so I'm happy for any advice. Thank you.

• If I'm understanding correctly and $\nabla_s = P(\nabla) = P(e^i)\partial_i$ where $P$ is the orthogonal projection onto the tangent planes of $\Omega$ and $e^i$ is (say) the standard basis, then I can use some geometric calculus to get you something like... Commented Jan 6, 2023 at 0:21
• $$\int_\Omega\mathrm d\Omega\,\nabla_s\times\mathbf F = \int_\Omega\mathrm d\Omega\,(\nabla\times\mathbf\omega)\times(\mathbf F\times\mathbf\omega) - \int_{\partial\Omega}\mathrm d\mathbf l\times(\mathbf F\times\mathbf\omega)$$ where $\mathbf\omega$ is the outward unit normal of $\Omega$ and $\nabla\times$ is the usual unrestricted curl. Would this be helpful? Commented Jan 6, 2023 at 0:21
• Dear Nicholas, thank you for the reply. Because I was not familiar with geometric algebra/calculus, it took me some time to get a grasp of it. I finally arrived at a similar expression, but I suspect there is a typo (missing "=") in your solution? I will add an explanation below (as answer, to have some more space).
– 3ak
Commented Feb 15, 2023 at 12:04

$$\renewcommand\vec\mathbf \newcommand\vecg\boldsymbol \newcommand\R{\mathbb R} \newcommand\G{\mathcal G} \newcommand\dd{\mathrm d} \newcommand\grade[1]{\left\langle#1\right\rangle} \newcommand\lcontr{\mathbin\rfloor} \newcommand\rcontr{\mathbin\lfloor}$$Let $$\vec F, \Omega$$ be as defined in the quesion. We work the in the Euclidean geometric algebra $$\G_3(\R)$$ over $$\R^3$$. Let $$I$$ be the unique right-handed unit pseudo scalar, i.e. $$I^2 = -1$$ and the cross product of vectors $$\vec a, \vec b$$ is given by $$\vec a\times\vec b = -I\,\vec a\wedge\vec b.$$ Note that $$I$$ commutes with the whole algebra. The surface $$\Omega$$ at point $$\vec x \in \Omega$$ has a unique tangent unit bivector $$\vecg\Omega(\vec x)$$ defined such that $$\vecg\Omega(\vec x)\vec n(\vec x) = I$$ when $$\vec n(\vec x)$$ is the outward-pointing unit vector normal to $$\Omega$$ at $$\vec x$$. We write $$\dd^k\vec x$$ for the $$k$$-vector-valued measure over a $$k$$-surface with $$\vec x$$ the vector variable of integration; for example, integrating over $$\Omega$$ we have $$\dd^2\vec x = |\dd^2\vec x|\vecg\Omega(\vec x)$$ where $$|\dd^2\vec x|$$ is the usual unoriented scalar surface measure (denoted $$\dd\Omega$$ in the question). We will omit the $$\vec x$$-dependence from all quantities from here out.

The vector derivative splits into terms parallel and orthogonal to $$\Omega$$, namely $$\nabla = \partial + \partial_\perp$$. Here, $$\partial$$ is the same as the $$\nabla_s$$ used in the question. We can now begin to consider the integral in question, first converting the cross product to an exterior product: $$\int_\Omega|\dd^2\vec x|\partial\times\vec F = -I\int_\Omega|\dd^2\vec x|\partial\wedge\vec F.$$ We then use the grade projection operator $$\grade\cdot_2$$ that takes the bivector part of an expression, and we insert a factor of $$\vecg\Omega\vecg\Omega^{-1}$$: $$-I\grade{\int_\Omega\dd^2\vec x\,\vecg\Omega^{-1}\dot\partial\dot{\vec F}}_2.$$ Here I've used overdots to make it clear what $$\partial$$ is applying to. Because $$\partial$$ is parallel to the plane $$\vecg\Omega$$ they anticommute. We then use the "generalized product rule" to get $$I\grade{\int_\Omega\dd^2\vec x\,\dot\partial\dot{\vecg\Omega}^{-1}\dot{\vec F} - \int_\Omega\dd^2\vec x\,\dot\partial\dot{\vecg\Omega}^{-1}\vec F}_2. \tag{*}$$ We now focus on the first term. This is now susceptible to the Fundamental Theorem of Geometric Calculus; if $$\Omega$$ is closed then this term is zero, and if not we get $$I\grade{\int_\Omega\dd^1\vec x\,\vecg\Omega^{-1}\vec F}_2.$$ $$\vecg F$$ is parallel to $$\vecg\Omega$$, so they anticommute. Then using the identity $$\dd^1\vec x\,\vec F = \dd^1\vec x\cdot\vec F + \dd^1\vec x\wedge\vec F$$ and applying the bivector projection we get $$-I\int_{\partial\Omega}\dd^1\vec x\cdot\vec F\,\vecg\Omega^{-1} - I\int_{\partial\Omega}(\dd^1\vec x\wedge\vec F)\times\vecg\Omega^{-1}$$ where here $$\times$$ is the commutator product $$A\times B = \frac12(AB - BA)$$. Since both $$\dd^1\vec x$$ and $$\vec F$$ are parallel to $$\vecg\Omega$$, their exterior product commutes with $$\vecg\Omega$$ making the second term zero. $$I\vecg\Omega^{-1} = \vecg\Omega^{-1}I = \vec n$$, so finally $$-\int_{\partial\Omega}\dd^1\vecg x\cdot\vec F\,\vec n.$$

We now work on the second term of ($$*$$). We have $$-I\grade{\int_\Omega\dd^2\vec x\,\dot\partial\lcontr\dot{\vecg\Omega}^{-1}\vec F}_2.$$ $$\partial\lcontr\vecg\Omega^{-1}$$ is a vector parallel to $$\vecg\Omega$$, so it anticommutes with $$\dd^2\vec x$$. Since $$\vec F$$ is also parallel, $$\dd^2\vec x\,\vec F = \dd^2\vec x\rcontr\vec F$$, so $$I\int_\Omega\grade{\dot\partial\lcontr\dot{\vecg\Omega}^{-1}\,\dd^2\vec x\rcontr\vec F}_2.$$ We now pull the scalar measure out of $$\dd^2\vec x = |\dd^2\vec x|\vecg\Omega$$ and write $$\vecg\Omega = I\vec n$$ (where $$\vec n^{-1} = \vec n$$ since $$\vec n$$ is a unit vector). This yields \begin{aligned} I\int_\Omega|\dd^2\vec x|\grade{\dot\partial\lcontr(\dot{\vec n}I^{-1})\,(I\vec n)\rcontr\vec F}_2 &= I\int_\Omega|\dd^2\vec x|\grade{\dot\partial\wedge\dot{\vec n}\,I^{-1}I\,\vec n\wedge\vec F}_2 \\ &= -I\int_\Omega|\dd^2\vec x|(\dot\partial\times\dot{\vec n})\wedge(\vec n\times\vec F) \end{aligned} using the fact that $$I^{-1} = -I$$ in the last equality. Here $$\times$$ is the vector cross product. Finally we may write $$\int_\Omega|\dd^2\vec x|(\partial\times\vec n)\times(\vec n\times\vec F)$$ where of course $$\partial$$ is acting on the adjacent $$\vec n$$.

All together, we have found that $$\int_\Omega|\dd^2\vec x|\partial\times\vec F = -\int_{\partial\Omega}\dd^1\vecg x\cdot\vec F\,\vec n + \int_\Omega|\dd^2\vec x|(\partial\times\vec n)\times(\vec n\times\vec F).$$ Of course, the first term vanishes when $$\Omega$$ is a closed surface.

• The overall result seems to be correct. In my opinion, the first equation should read $\mathbf{a}\times\mathbf{b}=−I\mathbf{a}\wedge\mathbf{b}$ instead of $\mathbf{a}\times\mathbf{b}=−I\mathbf{a}\times\mathbf{b}$, because the dual of the exterior product is the cross product.
– 3ak
Commented Feb 23, 2023 at 11:43
• Now I am wondering howto interpret the curl of the normal $\nabla\times\mathbf{n}$, which occurs in the last equation. The normal $\mathbf{n}$ is defined only on the surface $\Omega$, so I have difficulties applying the normal derivative $\partial^\perp \mathbf{n}$. A partial answer maybe found here: math.stackexchange.com/questions/1574529/… But the result seems not hold in general.
– 3ak
Commented Feb 23, 2023 at 11:48
• @3ak You're right, that $\times$ is a typo, I will fix it. Concerning $\nabla\times\mathbf n$, looking now I think I was overzealous in writing this. You are right that it doesn't make much sense since $\mathbf n$ is only defined on $\Omega$. I should really just leave it as $\partial\times\mathbf n$. I will change this as well. Commented Feb 23, 2023 at 17:29
• Nicely done! One last thought ... I suspect that $\partial\times\mathbf{n}=0$. For a unit sphere this can be easily shown. I have not found a way yet to show it in general. But lets assume that the vector curl of the surface normal actually vanishes everywhere on the surface, then the RHS is zero if $\Omega$ is closed. I guess this could be shown via differential geometry, I will try to dive into it if time permits.
– 3ak
Commented Feb 24, 2023 at 16:28

I have removed my answer, since a correction was needed which finally resulted in the expression proposed by Nicholas Todoroff.

• I think there are some subtle errors here. I'll go over it more thoroughly and post my own derivation when I have the time. Commented Feb 15, 2023 at 16:24
• Your expression for $\partial\times\mathbf f$ is wrong. Cross products are only defined in 3D, and it should be $\partial\times\mathbf f = -I\,\partial\wedge\mathbf f$ where $I$ is the constant pseudoscalar of the ambient 3D space. You expression $\mathbf I_m\,\partial\wedge\mathbf F$ can't make sense because $\partial\wedge\mathbf F$ is a trivector and an $m$-vector times a trivector can't even have a vector component when $m$ is odd. What follows is also wrong because you use $\mathbf I_2$ instead of the ambient pseudoscalar $I = e_1e_2e_3$ where $e_1, e_2, e_3$ is the standard basis. Commented Feb 16, 2023 at 4:22
• You are right! It should have been $-I_n \partial\wedge\mathbf{f}$ instead of $I_m \partial\wedge(-(\mathbf{f}\wedge\mathbf{n}))$.
– 3ak
Commented Feb 23, 2023 at 11:41