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Line Integrals 2

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate this integral directly

    2. Relevant equations

    [itex]\int cos x sin y dx +sin x cos y dy[/itex] on vertices (0,0), (3,3) and (0,3) for a triangle

    3. The attempt at a solution

    Does this have to evaluated parametrically using [itex]r(t)=(1-t)r_0+tr_1[/itex] for [itex]0 \le t\le 1[/itex]

    or can I just proceed evaluating each line segment in a counter clockwise manner starting with the vertices:

    from (0,0) t (3,3) which gives the line y=x hence dy=dx, substituting in the above

    [itex] \displaystyle I_{ab} = \int_0^3 (cos x sin x + sin x cos x ) dx = \int_0^3 sin 2x dx[/itex]........?
     
  2. jcsd
  3. Jan 21, 2012 #2
    How directly is directly? Because using Green's theorem makes this trivial.
     
  4. Jan 22, 2012 #3
    Yes, i can use green's but i need to verify it directly....
     
  5. Jan 23, 2012 #4
    Any thoughts on how I calculate this directly?
     
  6. Jan 23, 2012 #5
    I guess the best way would be to just split it up over each side and evaluate it parametrically.
     
  7. Jan 23, 2012 #6

    HallsofIvy

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    There is no "or" here- the methods you mention are identical. Of course, because the path itself has "corners", you can't have a single polynomial for the entire path- you need a "piecewise" polynomial which is what you have started with- the line from (0,0) to (3,3). Now, the lline from (3, 3) to (0, 3) is given by x= 3- t, y= 0 so that dx= -dt, dy= 0 with t from 0 to 3 Or, equivalently, x= t, y= 0 so that dx= dt, dy= 0 with t from 3 to 0. Note that integrating from 3 to 0 reverses the sign of the integral, giving the same result as the "-dx" when integrating from 0 to 3.

    Now, how would you handle the path from (0,3) to (0,0)?
     
  8. Jan 23, 2012 #7
    My attempt for (3,3) back to (0,3) is as follows

    THis is a horizontal line at y=3 implies dy=0 therefore the original integral of ∫cos x sin y dx+ sin x cos y dy (and subbing y=3 )reduces to

    [itex]\int_3^0 cos x sin 3 dx[/itex]...? Why isn't this right?
     
  9. Jan 23, 2012 #8
    I have it thank you.
     
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