Stokes theorem

  • Thread starter Telemachus
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  • #1
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Homework Statement


Hi there. I was trying to solve this problem, from the book. The problem statement says:
Integrate [tex]\nabla \times{F},F=(3y,-xz,-yz^2)[/tex] over the portion of the surface [tex]2z=x^2+y^2[/tex] under the plane z=2, directly and using Stokes theorem.

So I started solving the integral directly. For the rotational I got:

[tex]\nabla \times{F}=\left|\begin{matrix}i&j&k\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\3y&-xz&-yz^{2}\end{matrix}\right|=(x-z^{2},0,-z-3)[/tex]

Then I parametrized the surface.:

[itex]\begin{matrix}x=x=r\cos \theta\\ y=r \sin \theta \\z=\displaystyle\frac{r^2}{2}\end{matrix}, \theta[0,2\pi] ,r[0,2][/itex]

Then I did: [tex]T_r\times{T_{\theta}}[/tex]

[tex]T_r=(\cos \theta,\sin \theta,r),T_{\theta}=(-r\sin \theta,r\cos \theta,0)[/tex]

For the cross product I got:
[tex]T_r\times{T_{\theta}}=(r^2\cos \theta,-r^2\sin \theta,r)[/tex]

And then I computed the integral

[tex]\displaystyle\int_{0}^{2}\int_{0}^{2\pi}(r\cos \theta-\displaystyle\frac{r^4}{4},0,\displaystyle\frac{-r^2}{2}-3)\cdot{(-r^2\cos \theta,-r^2\sin \theta,r)}d\theta dr=-12\pi[/tex]

The result given by the book is: [tex]20\pi[/tex].

I don't know what I did wrong, and is one of the first exercises that I've solved for the stokes theorem, so maybe I could get some advices and corrections from you :)

Thank you in advance. Bye.
 
Last edited:

Answers and Replies

  • #2
833
30
I've corrected some typos. Is there anybody out there? :P
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
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Homework Statement


Hi there. I was trying to solve this problem, from the book. The problem statement says:
Integrate [tex]\nabla \times{F},F=(3y,-xz,-yz^2)[/tex] over the portion of the surface [tex]2z=x^2+y^2[/tex] under the plane z=2, directly and using Stokes theorem.

So I started solving the integral directly. For the rotational I got:

[tex]\nabla \times{F}=\left|\begin{matrix}i&j&k\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\3y&-xz&-yz^{2}\end{matrix}\right|=(x-z^{2},0,-z-3)[/tex]

Then I parametrized the surface.:

[itex]\begin{matrix}x=x=r\cos \theta\\ y=r \sin \theta \\z=\displaystyle\frac{r^2}{2}\end{matrix}, \theta[0,2\pi] ,r[0,2][/itex]

Then I did: [tex]T_r\times{T_{\theta}}[/tex]

[tex]T_r=(\cos \theta,\sin \theta,r),T_{\theta}=(-r\sin \theta,r\cos \theta,0)[/tex]

For the cross product I got:
[tex]T_r\times{T_{\theta}}=(r^2\cos \theta,-r^2\sin \theta,r)[/tex]
This is wrong. It should be [itex]\left<-r^2cos(\theta), -r^2sin(\theta), r\right>[/itex] (oriented "upward").

And then I computed the integral

[tex]\displaystyle\int_{0}^{2}\int_{0}^{2\pi}(r\cos \theta-\displaystyle\frac{r^4}{4},0,\displaystyle\frac{-r^2}{2}-3)\cdot{(-r^2\cos \theta,-r^2\sin \theta,r)}d\theta dr=-12\pi[/tex]

The result given by the book is: [tex]20\pi[/tex].

I don't know what I did wrong, and is one of the first exercises that I've solved for the stokes theorem, so maybe I could get some advices and corrections from you :)

Thank you in advance. Bye.
 
  • #4
833
30
Thank you very much HallsofIvy. How did you realize that the orientation was negative? I can identify a positive oriented normal in the Cartesian coordinates, but I don't know how to do it in cylindrical or spherical coordinates.
 

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