# Applying Stoke's Theorem: A Hint

• BloonAinte
I understand. Thank you for all your help! I very much appreciated your hints and how you helped me reach...In summary, you obtained that the surface integral of a vector field over a closed surface is the sum of the integrals over the boundary of the surface.f

#### BloonAinte

Homework Statement
Prove ##\int F \times dr## = ##\alpha \int (\nabla \times F) \times dS##
where ##\alpha## is a constant to be found. Here $F$ is a vector field, S is an open surface and C is the boundary of S.
Relevant Equations
Stokes' theorem: ##\int_C F \cdot dr = \int_S \nabla \times F \cdot dS##
I considered the vector field ##a \times F##, and applied Stoke's theorem. I obtained that $$\int_C (a \times F) \cdot dr = \int_C (F \times dr ) \cdot a.$$

Now, $$\nabla \times (a \times F) = a (\nabla \cdot F) - (a \cdot \nabla) F.$$

Using Stoke's theorem for the vector field ##a \times F##, I obtain $$\int_C (F \times dr ) \cdot a = \int_C (a \times F) \cdot dr = \int_S \nabla \times (a \times F) \cdot dS = \int_S [a (\nabla \cdot F) - (a \cdot \nabla) F] \cdot dS$$

But then, I cannot proceed. Can someone give me a hint? I want to get the surface integral into the form ##a \cdot \int_S (\nabla \times F) \times dS##

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Use the cyclicity of the triple product before trying to simplify the right hand side. You want to have both cross products so getting rid of them is not the way to go.

• BloonAinte
Use the cyclicity of the triple product before trying to simplify the right hand side. You want to have both cross products so getting rid of them is not the way to go.
Hi Orodruin, thank you for your message. I have tried to use your hint as follows. I am not sure how the curl behaves when I do a cyclic permutation, but I tried the following.

$$\int_S \nabla \times (a \times F) \cdot dS = \int_S dS \times \nabla \cdot (a \times F)$$

I am quite unsure as I don't think the curl should change to a divergence operator, but I don't see another way to apply cyclicity. If this is right, I obtain $$\nabla \cdot (a \times F) = F \cdot (\nabla \times a) - a \cdot (\nabla \times F) = -a \cdot (\nabla \times F)$$

Plugging this back in, I obtain $$\int_S dS \times \nabla \cdot (a \times F) = \int_S dS \times -a \cdot (\nabla \times F) = -a \cdot \int_S (\nabla \times F) \times dS$$

I am somewhat uncertain though. Could you let me know if it is okay so far? According to this, ##\alpha=-1##. Thank you for your help!

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$$\int_S \nabla \times (a \times F) \cdot dS = \int_S dS \times \nabla \cdot (a \times F)$$

I am quite unsure as I don't think the curl should change to a divergence operator,
It doesn’t. You need to be a bit more careful about the structure of your expression. Add a parenthesis or two.

It doesn’t. You need to be a bit more careful about the structure of your expression. Add a parenthesis or two.
Thank you for your comment. I am not sure about how to use the cyclicity. I am trying to use that $$\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})= \mathbf{b}\cdot(\mathbf{c}\times \mathbf{a}) = \mathbf{c}\cdot(\mathbf{a}\times \mathbf{b})$$ for numerical vectors ##a,b,c##. This formula is taken from here.

I gave it another go:

$$\int_S [\nabla \times (\mathbf{a} \times \mathbf{F})] \cdot d\mathbf{S} = \int_S [d\mathbf{S} \times \nabla] \cdot (\mathbf{a} \times \mathbf{F})$$

which is unfortunately not very different. Could you let me know what I am doing wrong? Thank you!

which is unfortunately not very different
This is not true, it is quite different. It has separated the two ##\times##. Now get them back together again by using a different cyclicity and making sure yoy get the inner product of ##a## with something.

• BloonAinte
This is not true, it is quite different. It has separated the two ##\times##. Now get them back together again by using a different cyclicity and making sure yoy get the inner product of ##a## with something.
Aha! Thank you for this comment! I got the Eureka moment.

I consider the integrand $$(dS \times \nabla) \cdot (a \times F) = a \cdot F \times (dS \times \nabla) = -a \cdot (dS \times \nabla) \times F,$$ where I use cyclicity and then antisymmetry ##a \times b = -b \times a##.

Overall, $$\int_C F \times dr = - \int_S (dS \times \nabla) \times F.$$ However, this doesn't seem to relate to the given identity. How can I reconcile this? Have I made a mistake?

You cannot reconcile it because what you are trying to prove is false.

As an easy counter example, pick ##\vec F = \vec x## and look at a circle in the x-y-plane for which ##\vec x \times d\vec x = R^2 \vec e_z d\theta## and ##\nabla \times \vec x = 0##.

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• BloonAinte
You cannot reconcile it because what you are trying to prove is false.

As an easy counter example, pick ##\vec F = \vec x## and look at a circle in the x-y-plane for which ##\vec x \times d\vec x = R^2 \vec e_z d\theta## and ##\nabla \times \vec x = 0##.
I understand. Thank you for all your help! I very much appreciated your hints and how you helped me reach the answer.

I understand. Thank you for all your help! I very much appreciated your hints and how you helped me reach the answer.
It is not my first time proving this theorem. Also, the more general statement is
$$\oint_\Gamma (d\vec x \ldots) = \int_S ([d\vec S \times \vec \nabla]\ldots)$$
where ##\ldots## may be replaced with any sort of vector structure expression - including different products. For ##\ldots = \cdot \vec F## we recover the original curl theorem. For ##\ldots = \times \vec F## you get your result from #7.

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Also, the more general statement is
$$\int_\Gamma (d\vec x \ldots) = \oint_S ([d\vec S \times \vec \nabla]\ldots)$$
where ##\ldots## may be replaced with any sort of vector structure expression - including different products. For ##\ldots = \cdot \vec F## we recover the original curl theorem. For ##\ldots = \times \vec F## you get your result from #7.
Wow, that's really cool! Do you have a reference for this? I'd like to learn more about this. Does this relate to the study of differential forms/generalized Stokes?

This should be discussed in most intermediate level textbooks on vector analysis. For example in this one • PhDeezNutz
Hi, I've been exploring this a bit more. Would the identity work if I write ##\int_S \nabla \times (F \times dS)##? I considered the circle of radius r in the xy plane, and I got ##\nabla \times (F \times dS) = (0,0,-4r \cos \theta)dr d\theta## which integrates to ##(0,0,0)## unlike the ##(0,0,2 \pi r^2)## meant to be obtained (from post 8). So I concluded that it didn't work. Would that be right? Thanks again

What works is the general expression in #11. Anything that cannot be simplified to that will generally be false.

Thank you. I did a misread and in fact it is the expression in #14 which was in the question.

Hi! I just wanted to write a quick correction for future readers. Note that what I did in post 14 is wrong. The mistake arose from wrongly using the cross product in cylindrical coordinates. To remedy this, instead use Cartesian coordinates.

We write ##F(x,y,z) = (x,y,0)## and consider a circle of radius 1. Use the parametrization ##r(x,y) = (x,y,0), -1<x<1, -\sqrt{1-x^2}<y<\sqrt{(1-x^2}##. The normal is ##(0,0,1)##. You get ##\nabla \times (F \times dS) = (0,0,-2)##, and then you get ##(0,0,-4\pi)## as the answer. Meanwhile, using ##r(\theta) = (\cos \theta, \sin \theta, 0)##, we have ##r \times dr = (0,0,1) d \theta##, giving the answer as ##(0,0,2\pi)##. For equality we need ##\alpha = -2##.

I thought that the formula was wrong, and I tried a different curve.

Let ##C## be the unit square in the first quadrant of the x-y plane. Let ##F(x,y,z) = (x^2,0,0)##. Let's find the left hand side first. We again have ##dS = (0,0,1)##. We have ##\nabla \times (F \times dS) = (0,0,-2x)##. We integrate over ##[0,1] \times [0,1]##, obtaining ##(0,0,-2)##.

Let's find the right hand side. The unit square is made up of 4 curves.

##y = 0, 0<x<1##: Parametrise using ##r(x) = (x,0,0)##. This is parallel to ##F## so ##F \times dr = 0##.
##x = 1, 0<y<1##: Parametrise using ##r(y) = (1,y,0)##. Note ##dr = (0,1,0) dy## and ##F = (1,0,0)##. So ##F \times dr = (0,0,1) dy##. The integral over this part gives ##(0,0,1)##
##y = 1, 0<x<1##: Parametrise using ##r(x) = (x,1,0)##. Note ##dr = (1,0,0) dx## which is parallel to ##F##, so the integral is 0.
##x = 0, 0<y<1##: Pametrise using ##r(y) = (0,y,0)##. Here ##F = (0,0,0)## so the integral is 0.

For equality we would need ##\alpha = -2## again. So is the corrected formula ##\int_S \nabla \times (F \times dS) = -2 \int_C F \times dr## correct?

Would appreciate if someone could check this! Thank you

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For equality we would need ##\alpha = -2## again. So is the corrected formula ##\int_S \nabla \times (F \times dS) = -2 \int_C F \times dr## correct?
No. It only "works" for the particular case of ##\vec F = \vec x## because it just so happens that ##\nabla \times (\vec x \times \vec a) = (\vec a \times \nabla) \times \vec x##. You will therefore need another field ##\vec F## to construct a counter example.

Thank you, but I tried again using ##F(x,y,z) = (x^2,0,0)## in the second part of the post, which still gave the answer. Do you have a particular suggestion?

You need to find a field for which
∇×(x→×a→)=(a→×∇)×x→.
does not hold. Then pick a suitable surface.

Thank you for your hint. Apologies to be a bother but I wasn't able to obtain the required field.

For ##F = (x^2 ,0,0)## it does not hold with a = (1,1,1) but the integrals agreed

For ##F = (e^x, e^y, 0)##, it does not hold with a = (1,1,1) but I obtain 2e - 2 for both integrals when considering C to be the unit square in the first quadrant of the xy plane. I have also tried changing it to the unit circle, obtaining 4 pi * I_1(1) = 7.102 for both where I_1(1) was a function given by WolframAlpha.

Could you let me know what I am doing wrong? Is the surface I'm picking not good? Can you please suggest a surface? Thank you and I apologise for taking up your time.

Pick ##\vec F = (\vec b \cdot \vec x) \vec c## with ##\vec b## and ##\vec c## carefully chosen constant vectors.

Thank you! I chose a surface with ##dS = (0,0,1)##, ##b=(1,2,0), c=(1,2,0)## so that ##F=(1(x+2y), 2(x+2y),0)##. Then ##\nabla \times (F \times dS) = (0,0,-5).## However, ##(dS \times \nabla) \times F = (0,0,0)##. Here ##dS \times \nabla = (- \partial_y, \partial_x, 0)## and thus ##(dS \times \nabla) \times F = (0,0,0)##.

Thanks a lot for this. Could you let me know how you found it, just for reference? Thanks again, so very much. This problem was pretty painful but I'm glad I finally managed to do it

##(\vec b \cdot \vec x) \vec c## is a quite general form of a linear vector valued function of ##\vec x## (although not the most general). I wanted a linear function to make the derivatives constant, which makes the surface integrals trivial (as long as the surface is flat).