- #1

Master1022

- 611

- 116

**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.]**

Last edited by a moderator: