Results 1 to 6 of 6
Like Tree4Thanks
  • 2 Post By HallsofIvy
  • 2 Post By HallsofIvy

Math Help - Stoke's Theorem Question

  1. #1
    Newbie
    Joined
    May 2013
    From
    Australia
    Posts
    4

    Stoke's Theorem Question

    Evaluate \iint_S \nabla \times \mathbf{F} \cdot dS Where S consists of the top and four sides of the cube with vertices (\pm1,\pm1,\pm1) oriented outwards.


     \mathbf{F}(x,y,z) = xyz\ \mathbf{i} + xy\ \mathbf{j} + x^2yz\ \mathbf{k}


    pls help, never worked with cubes before.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,595
    Thanks
    1420

    Re: Stoke's Theorem Question

    Do the six faces separately. On the face x= 1, d\vec{S}= dydz\vec{i} oriented outward. y goes from -1 to 1, z goes from -1 to 1. \nabla\times F(1, y, z)= z\vec{i}+ (-2yz)\vec{j}+ (y- z)\vec{k} so that \nabla F(1, y, z)\cdot d\vec{S}= z dydz and the integral over that face is \int_{y= -1}^1\int_{z= -1}^1 z dydz.

    Do the same for the other 5 faces, x= -1, y= 1, y= -1, z= 1, z= -1.
    Last edited by HallsofIvy; May 3rd 2013 at 07:32 AM.
    Thanks from usagi_killer and PathlessGiant
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2013
    From
    Australia
    Posts
    4

    Re: Stoke's Theorem Question

    wow i get it now thanks alot
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2013
    From
    Australia
    Posts
    4

    Re: Stoke's Theorem Question

    some of the terminology doesnt sit well with me.

    when i did my curl i got x^2z\mathbf{i} - (2xyz - xy)\mathbf{j} + (y - xz)\mathbf{k} which at \mathbf{F}(1,y,z) is different to what you have, plus i dont understand where \nabla\mathbf{F}(1,y,z) = zdydz comes from
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,595
    Thanks
    1420

    Re: Stoke's Theorem Question

    You are right. I accidently dropped the "-xy" term in \vec{i}i. \nabla\times F= x^2z\vec{i}- (2xyz- xy)\vec{j}+ (y- xz)\vec{k} which, with x= 1 is z\vec{i}- (2yz- y)\vec{j}+ (y- z)\vec{k}

    But it is not \nabla\times F that is zdydz, it is \nabla\times F \cdot d\vec{S}. On the plane x= 1, d\vec{S}= dydz\vec{i} and the dot product of that with z\vec{i}- (2yz- y)\vec{j}+ (y- z)\vec{k} is zdydz.
    Thanks from usagi_killer and PathlessGiant
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2013
    From
    Australia
    Posts
    4

    Re: Stoke's Theorem Question

    ahhh i see now, thankyou alot
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stoke's Theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 12th 2010, 09:04 PM
  2. Stoke Theorem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 30th 2009, 07:41 PM
  3. Stoke's Theorem
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 23rd 2009, 04:33 AM
  4. Stoke's theorem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 12th 2009, 05:00 PM
  5. Stoke's Theorem Question
    Posted in the Calculus Forum
    Replies: 7
    Last Post: October 5th 2008, 06:45 PM

Search Tags


/mathhelpforum @mathhelpforum