1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    [tex]\oint{\vec{u}d\vec{l}}[/tex]


    2. Relevant equations

    I need to solve both:

    [tex]\oint{\vec{u}d\vec{l}}[/tex]

    and

    [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

    -Mo
     
  2. jcsd
  3. Sep 26, 2007 #2

    Avodyne

    User Avatar
    Science Advisor

    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

    Avodyne

    User Avatar
    Science Advisor

    Glad to help. The easy stuff is sometimes the hardest to see when you're in the middle of a problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?