Results 1 to 4 of 4

Math Help - stokes

  1. #1
    Super Member
    Joined
    Aug 2009
    Posts
    639

    stokes

    May I know what is wrong with my workings?


    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,356
    Thanks
    36
    Quote Originally Posted by alexandrabel90 View Post
    May I know what is wrong with my workings?


    Yes - two things. First {\bf n} = <1,1,1> and second, it has to be a unit normal so you have to divide each component by \sqrt{3} .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2009
    Posts
    639
    must it always be a normal vector?

    because i thought i can actually use N(x,y) = (-fx, -fy,1) dA where dA is the area on the xy plane.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,537
    Thanks
    1392
    There are two ways of thinking about this. A vector integral over a surface is usually written " \int\int \vec{f}\cdot\vec{n}dS" where n is a unit normal vector to the surface and dS is the "differential of surface area".

    Here, the surface is the plane x+ y+ z= 5. The way I prefer to calculate dS is to write z= 5- x- y so that the "position vector" of any point in the plane is \vec{X}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (5- x- y)\vec{k} in terms of the two parameters x and y. Now, the derivative vectors \vec{X}_x= \vec{i}- \vec{k} and \vec{X}_y= \vec{j}- \vec{k} are in the tangent plane and their cross product, \vec{X}_x\times\vec{X}_y= \vec{i}+ \vec{j}+ \vec{k} is normal to the surface (you could have gotten that from the fact that A\vec{i}+ B\vec{j}+ C\vec{k} is normal to Ax+ By+ Cz= D) and has length \sqrt{3} so that \vec{n}= \frac{1}{\sqrt{3}}\vec{i}+ \frac{1}{\sqrt{3}}\vec{j}+ \frac{1}{\sqrt{3}}\vec{k} is the unit normal and the "differential of surface area" is the length of that vector times dxdy or \sqrt{3}dxdy.

    That is, here \vec{n}dS= (\frac{1}{\sqrt{3}}\vec{i}+ \frac{1}{\sqrt{3}}\vec{j}+ \frac{1}{\sqrt{3}}\vec{k})\sqrt{3}dxdy= (\vec{i}+ \vec{j}+ \vec{k})dxdy. The two "lengths", \sqrt{3}, cancel!

    That is why I prefer to write " d\vec{S}" and write the integral of a vector function over a surface as \int\int \vec{f}\cdot d\vec{S} to begin with.

    Here, \nabla\times\vec{f}= \vec{i}+ \vec{j}+ \vec{k} and d\vec{S}= (\vec{i}+ \vec{j}+ \vec{k})dxdy so that \nabla\times\vec{f}\cdot d\vec{S}= 3dxdy so that the integral is just 3 times the area of the surface- which is a "square of side length 2".

    Note that \vec{X}_x\times\vec{X}_y= -\vec{X}_y\times\vec{X}_x so that you have to choose in which order you will take the cross product. I took \vec{X}_x\times\vec{X}_y= \vec{i}+ \vec{j}+ \vec{k} rather than \vec{X}_y\times\vec{X}_y= -\vec{i}- \vec{j}- \vec{k}.

    That is equivalent to choosing an orientation for the surface and for its boundary. Here we are told "oriented clockwise as seen from the origin". Given that orientation for the boundary, a person "walking around" the boundary, with left side toward the interior, would have his/her head in the positive z direction. That means that we must choose the sign so that the z-component is positive.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stokes' Theorem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 28th 2011, 06:02 AM
  2. stokes
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 8th 2010, 03:30 AM
  3. stokes thm
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 24th 2009, 04:26 PM
  4. Stokes theorem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: August 16th 2009, 05:49 PM
  5. Stokes
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 3rd 2006, 02:28 AM

Search Tags


/mathhelpforum @mathhelpforum