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Stoke's Theorem.

  1. Oct 24, 2006 #1
    Hi guys... I'm stuck in this question. Its not a homework nor coursework, just practice. The answer for both is the same, that is -1 ...I'm able to get (a) but (b) .... still wondering what went wrong. Please enlighten me. Thanks in advance.
    Last edited: Oct 24, 2006
  2. jcsd
  3. Oct 25, 2006 #2


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    That ds=dxdy/2 part.. it's not how it works. what if I wanted to integrate over a circle? would you take ds=pi (dx²+dy²)?
  4. Oct 25, 2006 #3
    Also, how can you say this:
    ds = 1/2dxdy

    You are saying that ds is a vector. Yet, 1/2dxdy is not.
  5. Oct 25, 2006 #4
    sorry... that ds suppose to have z_hat with it
  6. Oct 25, 2006 #5
    i'm not sure about the ds part as usually there is this table i would refer to, and it says ds = dxdy z_hat but ... i've also tried doing using it, i wont get the same answer.

    If need to integrate over a circle, transform the vector to cylindrical type and use cylindrical variables r phi z to get it?
  7. Oct 25, 2006 #6
    In tutorial classes, they gave a semicircle and that was quite easy to get it.... but this geometrical shape triangle.... is my first time trying it, and i got so say... its cracking my head (i know its simple stuff...but i really dont get where went wrong)
  8. Oct 25, 2006 #7


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    First, at least one of the arrows in your picture is going the wrong way. I suspect that your contour is supposed to go FROM (1, 1) T0 (0, 0), not the otherway round as you have it.

    [itex]d\vec{S}[/itex] is NOT "[itex]\frac{1}{2} dxdy \vec{z}[/itex]". The fact that you are integrating over a triangle has nothing to do with the differential of area (everything to do with limits of integration!). Since this is in the xy-plane, [itex]d\vec{S}= dxdy \vec{z}[/itex]. The integral is NOT for [itex]0\le x\le 1[/itex], [itex]0\le y\le 1[/itex]; that would be over the unit square. You want to integrate over the triangle with boundaries y= 0, x= 1, y= x.

    For part (b) break the triangle into the three sides. Write parametric equations for each.
  9. Oct 25, 2006 #8
    yup...sorry... the contour suppose to go from (1,1) to (0,0) thanks for the mistake pointed out :)
  10. Oct 25, 2006 #9
    anyway...mind reedit the latex thing... cant seem to get them =_=
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