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Vector analysis, calculate path

  1. Jan 17, 2016 #1
    1. The problem statement, all variables and given/known data
    I have these vectorfields defined on the specified set. All of them are conservative on their set.
    upload_2016-1-17_12-14-15.png

    Also i have three unit circles C1, C2 and C3 centered respectively on (0,0) , (-2,0) and (-1,0)

    I need to find the line integrals over all of them on H
    upload_2016-1-17_12-17-23.png
    (beregn = calculate, og = and. (I'm Danish))

    2. The attempt at a solution
    Since H is conservative i think both integrals in (b) equals zero because you can just "repair" the hole in the middle. Not sure about (c)
     

    Attached Files:

  2. jcsd
  3. Jan 22, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jan 23, 2016 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    You can draw small circles S2 of small radius r around (-2,0) and S0 of small radius r around (0,0). Suppose S2 intersects C3 at A and B (where A is nearly (-2,r) and B is nearly (-2,-r)),and S0 intersects C3 at C and D (where C is nearly (0,-r) and D is nearly (0,r)). Then your integral in (c) equals
    [tex] I_c = I_1 + I_2 + I_3 + I_4, [/tex]
    where
    [tex] \begin{array}{l}I_1 = \int_D^A \vec{H} \cdot d\vec{r} \\
    I_2 = \int_B^C \vec{H} \cdot d\vec{r} \\
    I_3 = \int_C^D \vec{H} \cdot d\vec{r} \; \text{along C3}\\
    I_4 = \int_A^B \vec{H} \cdot d\vec{r} \; \text{along C3}
    \end{array} [/tex]
    Also, let ##I_3'## be the integral from C to D along the small circle S0 (clockwise, inside the big circle C3) and let ##I_4'## be the integral from A to B along the small circle S2 (clockwise, inside C3). We have ##I_1 + I_2 + I_3' + I_4' = 0## because there are no singularities inside the corresponding closed path. Thus, the answer ##I_c = I_3+I_4-I_3' - I_4'##. We can try to evaluate ##I_3 - I_3'## directly, since inside the small circle S0 the direct path (along C3) from C to D is almost a straight line from (0,-r) to (0,r), so ##I_3 - I_3'## is an integral along a closed semicircle, where the singularity in H at (0,0) does not appear because x = 0 along the vertical line through (0,0); that is, the term ##y dx/(x^2+y^2) = 0## because dx = 0 on the vertical line, and ##x dy/(x^2+y^2) = 0## because ##x=0## on the vertical line.
     
    Last edited: Jan 23, 2016
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