Results 1 to 6 of 6

Math Help - Stokes' Theorem

  1. #1
    Newbie
    Joined
    Jun 2011
    Posts
    23

    Stokes' Theorem

    Consider the cylindrical can of radius 1 with a
    differentiable board in the counter-clockwise's direction. Calculate:

    \int\limits_C \vec{F} .d\vec{r} :

    \vec{F} (x, y, z) = ( y(x-2), x^2y, z)

    OBS.: there is the figure of the cylinder and the curve, but it just says that the curve is closed and impossible to be parametrized.

    A.: 3pi
    _____________________________________________

    Attempt:

    Parametrization: S --> (cosv, senv, u)
    n = (-cosv, sinv, 0)

    \nabla\vec{F} = (0, 0, 6x^2 + 6y^2)

    \nabla\vec{F}\vec{r}(u,v)\vec{n} = 0

    Where is the mistake?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    May 2010
    From
    Los Angeles, California
    Posts
    274
    Thanks
    1

    Re: Stokes' Theorem

    Your question is unclear. In order to apply Stokes's theorem, you need a surface whose boundary is the curve C. You seem to be saying that the surface is the open cylinder but then the boundary for this is composed of two circles, not some closed curve you seem to be suggesting. A figure will help a lot.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2011
    Posts
    23

    Re: Stokes' Theorem

    Figure:

    Stokes' Theorem-digitalizar0001.jpg
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    May 2010
    From
    Los Angeles, California
    Posts
    274
    Thanks
    1

    Re: Stokes' Theorem

    OK, that makes things clearer. Using Stokes's theorem, we can write the line integral as a surface integral. Now split the surface integral up into two parts; one over the curved lateral side of the cylinder and the other over the bottom of the cylinder. Since we don't know what the lateral surface is exactly, we expect this integral to be zero. That is, check that \text{ curl }\mathbf{F\cdot n}=0 here. The bottom one you can do directly since you know the equation for this surface.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    May 2010
    From
    Los Angeles, California
    Posts
    274
    Thanks
    1

    Re: Stokes' Theorem

    Quote Originally Posted by PedroMinsk View Post
    Attempt:

    Parametrization: S --> (cosv, senv, u)
    n = (-cosv, sinv, 0)

    \nabla\vec{F} = (0, 0, 6x^2 + 6y^2)

    \nabla\vec{F}\vec{r}(u,v)\vec{n} = 0

    Where is the mistake?
    Do you mean \nabla \times \vec{F} here?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jun 2011
    Posts
    23

    Re: Stokes' Theorem

    Yes, it was what I meant.

    I forgot to calcalute the bottom, that was the problem.

    Thanks a lot for your help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stokes theorem
    Posted in the Calculus Forum
    Replies: 8
    Last Post: August 5th 2011, 11:50 AM
  2. Replies: 3
    Last Post: May 14th 2010, 11:04 PM
  3. stokes theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 21st 2010, 05:20 AM
  4. Replies: 2
    Last Post: April 3rd 2010, 05:41 PM
  5. Help with Stokes Theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 5th 2009, 06:01 AM

Search Tags


/mathhelpforum @mathhelpforum