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Homework Help: Why can't I get Stoke's Theorem to Work ?

  1. Sep 25, 2007 #1
    Why can't I get Stoke's Theorem to Work!!!?

    1. The problem statement, all variables and given/known data
    This is very similar to the problem I put up last night actually part c of the same problem. I really have put an amazing amount of time into this problem (week and a half/ 5 hrs a day) and it countinues to stump me. At this point I'm spinning mental wheels and could use some help.

    we have a 3-4-5 right triangle lying on the x-y plane with the length three side on the x-axis and the length 4 side on the y-axis. we define a field
    [tex]\vec{u}=x^2\hat{x} + xy\hat{y}[/tex]

    and I am trying to evaluate the integral:


    2. Relevant equations

    I need to solve both:



    [tex]\oint{(\nabla{}x\vec{u}) . \hat{n} dA}[/tex]
    please excuse my clumsy notation, this should be del cross the vector u

    3. The attempt at a solution

    The surface integral after doing del cross u I end up with

    [tex]\frac {1}{2} \oint{ y \hat{k} dxdy} [/tex]
    evaluated from 0 - 3 on the x-axis and 0 - 4 on the y-axis
    Doing this I get 12.
    the 1/2 I added in to this equation is largely a guess on my part, and could be correct or not, just seemed to make some sense for a triagle who's area is 1/2 base*height

    Looking at the line integral

    I placed curve 1 along the y axis from 4 - 0 then paramaterized x = 0 and dx = 0
    so for curve 1 is zero

    Curve 2, I paramaterized y = 0, dy = 0 that left

    [tex]\int{x^2 dx}[/tex] evaluated from 0 - 3 gives me 9

    Curve 3 along the hypotinous (I can't spell ... sorry) paramaterized y = 4 - 4/3 * x dy = -3/4 * dx
    plugging these new values into y and dy and integrating from 3 - 0 gives me -1

    So for the whole curve I have 8

    So somewhere I missed something because last I checked 8 =! 12, or if you ignore my guess of 1/2 in the surface integral 8 =! 24
    so I'm only off by a factor of 3 somewhere, though it could be coincidence that's a pretty nice number to be off by, makes me think I'm just missing a term somewhere.
    If I had to guess I would say somehow [tex] \hat{n} [/tex] is somehow equal to 1/3, or perhaps 1/6 to include my guess about 1/2 in the surface integral

    Thank you for reading this long rambling thing, and thanks for any help

  2. jcsd
  3. Sep 26, 2007 #2


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    I believe your line integrals are all correct [though you have a typo: dy = -(4/3)dx, not -(3/4)dx].

    But your surface integral is wrong. You can't just multiply by a factor of one half. You need to integrate y over the area of the triangle. This integral takes the form

    [tex]\int_a^b dx \int_c^d dy \; y[/tex]

    where you need to figure out what the limits a, b, c, and d should be so that you are covering the triangle. Key point: c and d may depend on x.

    When I do this integral with the correct limits, I get 8, matching your result from the line integrals.
  4. Sep 26, 2007 #3
    Thank you!!!!, armed with your hint I went back and 'carefully' looked at one of my undergrad math books I'm using as a reference, and of course there it was staring me in the face the whole time, but until you came along I was just to dense to see it.

    and thank you for not giving me the answer, and letting me work it out myself.

    I kinda feel bad, because I know these are such easy questions I'm throwing out and I should know this stuff, but I'm stoked now, I can do integrals - Awesome.
  5. Sep 26, 2007 #4


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    Science Advisor

    Glad to help. The easy stuff is sometimes the hardest to see when you're in the middle of a problem.
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