# Stokes' Theorem Example Question

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 realise 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.

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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.

• 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
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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.

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• 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
Staff Emeritus
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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##.

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