# Applying Stoke's Theorem: A Hint

• BloonAinte
In summary: 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.
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
Orodruin said:
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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BloonAinte said:
$$\int_S \nabla \times (a \times F) \cdot dS = \int_S dS \times \nabla \cdot (a \times F)$$

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.

Orodruin said:
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!

BloonAinte said:
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
Orodruin said:
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
Orodruin said:
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.

BloonAinte said:
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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Orodruin said:
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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BloonAinte said:
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
Orodruin said:
∇×(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).

For closure on this, I wanted to share the solution to the intended problem in Post 14.

Consider the field ##a \times F##. Applying Stoke's theorem gives

##\int_S \nabla \times (a \times \mathbf{F}) \cdot d\mathbf{S} = \int_C (a \times \mathbf{F}) \cdot d\mathbf{r}##

Then, ##(a \times \mathbf{F}) \cdot d\mathbf{r} = a \cdot (\mathbf{F} \times d\mathbf{r}) ## and ## (a \times \mathbf{F}) \cdot d\mathbf{S} = (\mathbf{F} \times d\mathbf{S}) \cdot a##. This then gives:

##a \cdot \int_S \nabla \times (\mathbf{F} \times d\mathbf{S}) = \pm a \cdot \int_C \mathbf{F} \times d\mathbf{r}##

and since ##a## is arbitrary,

##\int_S \nabla \times (\mathbf{F} \times d\mathbf{S}) = \pm \int_C \mathbf{F} \times d\mathbf{r}.##

Hence, ## \alpha = \pm 1##. We have a plus/minus sign as the orientation of C is not specified.

BloonAinte said:
and ## (a \times \mathbf{F}) \cdot d\mathbf{S} = (\mathbf{F} \times d\mathbf{S}) \cdot a##. This then gives:
This does not give:

BloonAinte said:
##a \cdot \int_S \nabla \times (\mathbf{F} \times d\mathbf{S}) = \pm a \cdot \int_C \mathbf{F} \times d\mathbf{r}##
The dot product with dS in the surface integral is with the entire expression ##\nabla \times (\vec a \times \vec F)##, not only the parenthesis. In fact, the vector structure of the entire expression would not make sense if this were not the case.

Thank you for your comment. I understand what you have said. Apologies to necrobump a thread with a wrong solution. This is actually the official solution I was given. (For future readers' notice: The correct version of this result is proved in Post #7 and and also be found in Wikipedia, under curve-surface integrals of this page. I thought that some readers may encounter the form I posted in Post 1 and in Post 25 and decided to post the solution I was given.)

I wonder if a counter-example to this can be found? The earlier example of the linear function did not work for me. I tried ##F = (x,y,z)## and C to be a circle, and it seemed to be fine, working out to (0,0,## \pm \pi##) for both sides.

## 1. What is Stoke's Theorem?

Stoke's Theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a surface to the line integral of the same vector field around the boundary of the surface.

## 2. How is Stoke's Theorem applied?

To apply Stoke's Theorem, you first need to calculate the curl of the vector field. Then, evaluate the line integral of the curl of the vector field along the boundary of the surface. This will give you the surface integral of the original vector field over the surface.

## 3. What is the significance of Stoke's Theorem?

Stoke's Theorem provides a powerful tool for simplifying calculations involving surface integrals by relating them to line integrals. It is particularly useful in physics and engineering for solving problems involving fluid flow, electromagnetism, and more.

## 4. Can Stoke's Theorem be applied to any surface?

Stoke's Theorem can be applied to any surface that is smooth and oriented with a consistent normal direction. Surfaces that have holes or self-intersections may require special considerations when applying the theorem.

## 5. Are there any limitations to Stoke's Theorem?

Stoke's Theorem is limited by the assumption that the vector field is continuously differentiable over the surface. In cases where this assumption does not hold, the theorem may not be applicable, and alternative methods may need to be used.

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