1. Limited time only! Sign up for a free 30min personal 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!

Problem with a simple Stoke's Theorem question

  1. May 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Using Stoke's Theorem, evaluate the contour integral:
    [tex]\oint F.dr[/tex]
    as an integral over an appropriately chosen 2 dimensional surface.

    Use F = [tex](e^{x}y+cos\siny,e^{x}+sinx\cosy,ycosz)[/tex] and take the contour C to be the boundary of the rectangle with the vertices (0,0,0), (A,0,0), (A,B,0), (0,B,0) oriented anticlockwise.

    Then evaluate the same integral directly as a contour integral.

    2. Relevant equations
    Stoke's Theorem,

    [tex]\int\int_{S}(curl F).n\;d^{2}A=\oint F.dr[/tex]
    where n is unit normal vector.

    3. The attempt at a solution

    I got Curl F to be (cos z)i by the standard method using a matrix. I set n as -k as the surface is in the xy plane so the normal vector is along the z direction and I got the negative by the right hand rule.

    This gives:

    [tex]\int^{0}_{A}dx\int^{0}_{B}(-1)k.(cosz)i\dy[/tex]

    (-k).(cos z)i is 0

    which makes the double integral nothing

    However when I solve the same thing directly as a contour integral I get an answer of -B.

    Going anti clockwise from the origin, (so from (0,0,0) to (A,0,0) is I, (A,0,0) to (A,B,0) is II, (A,B,0) to (0,B,0) is III and (0,B,0) to (0,0,0) is IV) I get:
    I=0
    II=[tex]e^{A}B + sinA\;sinB[/tex]
    III=[tex]-(e^{A}B+sinA\;sinB)[/tex]
    IV=[tex]-B[/tex]

    Which, when added together, gives -B.... I think there must be a glaring error somewhere. I can write up my calculations for the contour integral if that would help solve it... Any help with this would be much appreciated.
     
  2. jcsd
  3. May 10, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Technically, the normal to the surface should be +k as given by the right-hand rule when your path is anticlockwise. Luckily, it didn't affect your answer in this case as -1*0 is still zero..

    Your error is in integral III; recheck that calculation and post your work for it if you can't find the error.
     
  4. May 11, 2009 #3
    I got it! Thanks a lot. Ha, I somehow messed up with the right hand rule, which is pretty basic, but that wasn't the problem. I just made a copying error and forgot about one of the [tex]e^{0}B[/tex] which should have become B, not nothing. Thanks a lot!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Problem with a simple Stoke's Theorem question
  1. Stokes theorem problem (Replies: 1)

Loading...