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Homework Help: Problem with stokes' theorem

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Let: [tex]\vec{F}(x,y,z) = (2z^{2},6x,0),[/tex] and S be the square: [tex]0\leq x\leq1, 0\leq y\leq1, z=1. [/tex]

    a) Evaluate the surface integral (directly):
    [tex]\int\int_{S}(curl \vec{F})\cdot\vec{n} dA[/tex]

    b) Apply Stokes' Theorem and determine the integral by evaluating the corresponding contour integral.

    2. Relevant equations

    [tex]\int\int_{S}(curl \vec{F})\cdot\vec{n} dA = \oint_{C}\vec{F}\cdot d\vec{r}[/tex]

    3. The attempt at a solution

    a) Basically I took the curl of F and got (0, 4z, 6) and dotted it with the normal vector which is either (0,0,+1) or (0,0,-1) as no orientation was given in the question! which gave me a value of +6 or -6.

    b) now this is the part which has confused me!! How on earth do I find the corresponding contour integral?? I have no idea how to find dr and I'm not sure what to do about parametizing the square.

    Any help would be greatly appreciated!!!

    P.S. I have a similar problem involving Gauss' Law so am hoping to kill two birds with one stone!
  2. jcsd
  3. Nov 8, 2009 #2


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    the normal to use for the first part is determeined by a right hand rule, to line up with the direction you go around the contour

    now to do the contour integral, see it as the sum of 4 line intergals, one for each side of the square
  4. Nov 8, 2009 #3


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    [itex] \vec{dr} [/itex] will be an infinitesimal in the direction of the line, be careful with direction. eg if the direction is the +ve x, it essentially reduces to [itex] \vec{dr} = (dx, 0, 0) [/itex]

    Pick a way to go around say counter clockwise form above and stick to it, setting the [itex] \vec{dr} [/itex] direction & limits accordingly
  5. Nov 8, 2009 #4
    I still don't get it!!

    if i name the corners of the square a,b,c and d for (0,0,1), (0,1,1), (1,1,1) and (1,0,1) respectively, then will dr be:
    a->b: (0,dy,0)
    b->c: (dx,0,0)
    c->d: (0,-dy,0)
    d->a: (-dx,0,0)?

    But then when I dot product with f=(2z^2,6x,0) and integrate each seperately, I get:
    6x + 2z^2 - 6x - 2z^2 = 0

    Forgive me if I'm being stupid but we've only ever done stokes' law the other way round so I've never had any experience this way!!
  6. Nov 8, 2009 #5
    Can anyone find a way to explain this to me or point me in the right direction? I've been searching online and my books for hours and the only examples that I find are the opposite way round!!!
  7. Nov 8, 2009 #6


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    show your working!

    these are definte integrals, so you should end up with a number for each integral, not a function...

    so for a->b:(0,0,1)->(0,1,1) along the y axis
    notice x = 0, z=1, only y is changing
    [tex] \vec{F} \bullet \vec{dr} = (2z^2,6x,0)\bullet (0,1,0)dy = 6x.dy |_{x=0}[/tex]

    so there is actually no field component parallel to dy in the integral...

    note the dot product method is a slight shortcut in notation, really should parameteris the line in terms of a variable & integrate, though as teh intergals are along axes in this question we can take the shortcut
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