Stokes' Theorem Example Question

[Master1022]

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

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

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

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

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

Then I chose to evaluate the integral in polar coordinates: x = r\cos(\phi), y = r\sin(\phi), z = 2 - r\sin(\phi). I let dS = r dr d\phi and I used the limits r: 0 to 1 and 0 \leq \phi \leq 2\pi. However evaluating this gives me the answer \frac{\pi}{\sqrt{2}}, when the video shows that the answer is \pi. I realize that this must have to do with my evaluation of d \vec S, 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.]

 

[Orodruin]

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.

 

[Master1022]

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 x-y 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: ?
\vec S = x \hat i + y \hat j + ( 1 - x - y ) \hat k
\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}
\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

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

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

Many thanks for the help

 

[LCKurtz]

2) Do the magnitudes always cancel out for between \hat n and 'dS'? I ask this because I think it is still quicker (at least in this problem) to just get the \hat n 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.

 

[Master1022]

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

 

[Orodruin]

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 $z=2-y$.

 

[Master1022]

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 \hat n dS does yield a 'non-normalised' version of the normal vector which I calculated above.

Thanks