Stokes' Theorem Example Question

In summary, the conversation is about a student attempting to solve a Stokes' Theorem question on Khan Academy. The problem involves evaluating an integral over a path in the CCW direction given by the intersection of two surfaces. The student's attempt involves using Stokes' Theorem and finding the normal vector and curl of the surface. However, the student's solution does not match the given answer and they seek clarification on their approach. Another user provides insight on treating the normal vector and surface area together and clarifies that the magnitudes always cancel out.
  • #1
Master1022
611
117
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 realize 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
Master1022 said:
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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  • #3
Orodruin said:
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
 
  • #4
Master1022 said:
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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  • #5
LCKurtz said:
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
 
  • #6
Master1022 said:
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
Orodruin said:
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
 

1. What is Stokes' Theorem?

Stokes' Theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a closed surface to the line integral of the same vector field along the boundary of the surface.

2. How do you use Stokes' Theorem to solve problems?

To use Stokes' Theorem, you first need to determine the surface and the boundary of the surface. Then, you can calculate the line integral around the boundary and the surface integral over the surface. Finally, you can equate the two values to solve for the unknown variables.

3. Can you give an example problem of using Stokes' Theorem?

Sure, for example, you can use Stokes' Theorem to calculate the surface integral of a vector field over a sphere and equate it to the line integral around the boundary of the sphere. This can help in solving problems involving fluid flow or electromagnetism.

4. What are the applications of Stokes' Theorem?

Stokes' Theorem has various applications in physics and engineering, such as in electromagnetism, fluid dynamics, and conservation laws. It is also used in mathematical fields like differential geometry and topology.

5. Is Stokes' Theorem similar to the Divergence Theorem?

Both Stokes' Theorem and the Divergence Theorem are fundamental theorems in vector calculus, but they are used for different purposes. Stokes' Theorem relates a surface integral to a line integral, while the Divergence Theorem relates a volume integral to a surface integral. However, both theorems are based on the concept of flux and are often used together in solving problems.

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