Stokes' Theorem Example Question

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Summary:: This question is about a Stokes' Theorem question that I saw on Khan Academy and I am trying to attempt to solve it a different way.

The problem is as follows:

Problem: Let [itex] \vec{F} = \begin{pmatrix} -y^2 \\ x \\ z^2 \end{pmatrix} [/itex]. Evaluate [itex] \oint \vec F \cdot d \vec {r} [/itex] over the path [itex] C [/itex] in the CCW direction where C is given by the intersection between [itex] x^2 + y^2 = 1 [/itex] and [itex] y + z = 2 [/itex].

My attempt:
So I first started by noting Stokes' Theorem: [itex] \oint_C \vec F \cdot d \vec {r} = \iint_S (\nabla \times \vec F) \cdot d \vec S [/itex].

I thought to find the normal vector [itex] \hat n [/itex]:
[tex] \hat n = \frac{\nabla (y + z - 2 = 0)}{||\nabla (y + z - 2 = 0)||} = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} [/tex]

Then I went to find the curl of the surface:
[tex] \nabla \times \vec F = \begin{pmatrix} 0 \\ 0 \\ 1 + 2y \end{pmatrix} [/tex]

and therefore: [itex] (\nabla \times \vec F) \cdot (\hat n dS) = \frac{1 + 2y}{\sqrt{2}} dS[/itex]

Then I chose to evaluate the integral in polar coordinates: [itex] x = r\cos(\phi), y = r\sin(\phi), z = 2 - r\sin(\phi) [/itex]. I let [itex] dS = r dr d\phi [/itex] and I used the limits r: 0 to 1 and [itex] 0 \leq \phi \leq 2\pi [/itex]. However evaluating this gives me the answer [itex] \frac{\pi}{\sqrt{2}} [/itex], when the video shows that the answer is [itex] \pi [/itex]. I realise that this must have to do with my evaluation of [itex] d \vec S [/itex], and is perhaps suggesting that I shouldn't normalised the normal vector.

Would anyone be able to help shed some insight on this for me? Thanks in advance.

[Moderator's note: Moved from a technical forum and thus no template.]
 
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  • #2
Orodruin
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I let dS=rdrdϕdS=rdrdϕ dS = r dr d\phi
In general, it is far easier to treat ##\hat n## and ##dS## together as
$$
d\vec S = \hat n \, dS = \left(\frac{\partial \vec x}{\partial s} \times \frac{\partial \vec x}{\partial t}\right) ds\, dt,
$$
where ##t## and ##s## represent any choice of coordinates on your surface.

Also, note that your plane is not parallel to the x-y-plane. Your assumed dS is only valid if it is.
 
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In general, it is far easier to treat ##\hat n## and ##dS## together as
$$
d\vec S = \hat n \, dS = \left(\frac{\partial \vec x}{\partial s} \times \frac{\partial \vec x}{\partial t}\right) ds\, dt,
$$
where ##t## and ##s## represent any choice of coordinates on your surface.

Also, note that your plane is not parallel to the x-y-plane. Your assumed dS is only valid if it is.
Thank you very much. I think I have skipped steps in just trying to reduce the problem to the [itex] x-y [/itex] plane.

I just have one or two quick follow up questions based on that:

1) So should I have gone about that part of the working like: ?
[tex] \vec S = x \hat i + y \hat j + ( 1 - x - y ) \hat k [/tex]
[tex] \frac{\partial \vec S}{\partial x} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} , \frac{\partial \vec S}{\partial y} = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} [/tex]
[tex] \hat n dS = \frac{\partial \vec S}{\partial x} \times \frac{\partial \vec S}{\partial y} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} dx dy [/tex]

So is this telling me that an 'elemental area' on the surface of the plane is [itex] \sqrt{3} [/itex] times larger than the area [itex] dx dy [/itex] in the [itex] x-y [/itex] plane?

2) Do the magnitudes always cancel out for between [itex] \hat n [/itex] and 'dS'? I ask this because I think it is still quicker (at least in this problem) to just get the [itex] \hat n [/itex] as I did above, but I could just ignore the normalization constant, knowing that it will cancel out.

Many thanks for the help
 
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LCKurtz
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2) Do the magnitudes always cancel out for between [itex] \hat n [/itex] and 'dS'? I ask this because I think it is still quicker (at least in this problem) to just get the [itex] \hat n [/itex] as I did above, but I could just ignore the normalization constant, knowing that it will cancel out.

Many thanks for the help
Yes, they always cancel out. If your surface is ##S: \vec R = \vec R(s,t)## then your unit normal vector will be $$\hat n = \pm \frac {\vec R_s \times \vec R_t}{|\vec R_s \times \vec R_t|}$$ and your element of surface area is ##dS = |\vec R_s \times \vec R_t|ds dt##. So when you calculate
##d\vec S = \pm \hat n dS## the ## |\vec R_s \times \vec R_t|## terms cancel out. The choice of the ##\pm## is taken so that ##\pm \vec R_s \times \vec R_t## agrees with the given orientation.
 
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Yes, they always cancel out. If your surface is ##S: \vec R = \vec R(s,t)## then your normal vector will be $$\hat n = \pm \frac {\vec R_s \times \vec R_t}{|\vec R_s \times \vec R_t|}$$ and your element of surface area is ##dS = |\vec R_s \times \vec R_t|ds dt##. So when you calculate
##d\vec S = \pm \hat n dS## the ## |\vec R_s \times \vec R_t|## terms cancel out. The choice of the ##\pm## is taken so that ##\pm \vec R_s \times \vec R_t## agrees with the given orientation.
Thank you very much for responding
 
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Orodruin
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1) So should I have gone about that part of the working like: ?

→S=x^i+y^j+(1−x−y)^k​
The z-component does not look correct to me. You want the surface given by ##z=2-y##.
 
  • #7
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The z-component does not look correct to me. You want the surface given by ##z=2-y##.
Yes, you are correct. I accidentally mixed up a similar question with this one when typing that. Then calculating [itex] \hat n dS [/itex] does yield a 'non-normalised' version of the normal vector which I calculated above.

Thanks
 

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