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Line Integrals

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate [tex]\int[/tex]F dr if C = C1 + C2 where C1 is the line segment from P1(-1,pi,-1) to P2(0,0,0) and C2 is the line segment from P2(0,0,0) to P3(2,0,4)

    vectorF= yz i + (xz - e^(z)siny) j + (e^(z)cosy + xy) k



    3. The attempt at a solution
    Im having problems setting up the C1 integral:
    I have it parametrized as x=t, y=-pi t z=t, as t goes 0 to 1

    r(t)= t i - pit j + t k

    r'(t)= i - pi j + k

    F= -pi t^2 i + (t^2-e^(t)sin(pit)) j + (e^(t)cos(pit) + pit^2) k

    F dot r'(t) = -pi t^2 + pi e^(t)sin(pit) + e^tcos(pit)

    I would then take integral dt as t goes from 0 to 1

    I think I have answered C2 correctly, I'm not sure if my parameters are correct
     
  2. jcsd
  3. Dec 9, 2008 #2

    Defennder

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    Big time-saving hint: What is curl F ?
     
  4. Dec 9, 2008 #3
    dF2/dx - dF1/dy =0 z-z=0
    dF3/dy - dF2/dz =0 (-e^zsiny + x) + (-x + e^zsiny) = 0
    dF1/dz - dF3/dx = 0 y-y=0

    Curl test?? how does that help?
     
  5. Dec 9, 2008 #4

    Defennder

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    So what does that tell you about F?
     
  6. Dec 9, 2008 #5
    that it is path independent or conservative?
     
    Last edited: Dec 9, 2008
  7. Dec 9, 2008 #6

    gabbagabbahey

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    Yes, which also tells you that you can write F as the gradient of some scalar [itex]\varphi[/itex]...So if you can find such a scalar, you can easily compute the path integral using the fundamental theorem of gradients.

    However, the way your original question is worded leads e to believe that they want you to evaluate the path integral along that specific path, so I would use the gradient method only as a check.
     
  8. Dec 9, 2008 #7

    gabbagabbahey

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    That doesn't seem right; [itex]r(0)=(0,0,0)[/itex] which is not your starting point, and [itex]r(1)=(1,-\pi,1)[/itex] which is neither your starting point or your finish point!:eek:
     
  9. Dec 9, 2008 #8
    x=t-1
    y= pi t + t
    z=t-1

    t goes from -2 to -1,
    would that work?
     
  10. Dec 9, 2008 #9

    gabbagabbahey

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    Nope; now your starting point is r(-2)=(-3,-2pi-2,-3) and your finish point is just as incorrect.

    Let's think this through logically instead of stabbing blindly in the dark....You know that c1 is the direct line from P1 to P2; so you can write down a general parameterized line for r: [itex]\mathbf{r}(t)=(at+b,ct+d,ft+g)[/itex]....say you let t go from 0 to 1; what must a,b,c,d,f and g be for [itex]\mathbf{r}(0)[/itex] to be P1 and [itex]\mathbf{r}(1)[/itex] to be P2?
     
  11. Dec 10, 2008 #10
    x=t-1
    y=pit^2
    z=t-1
     
  12. Dec 10, 2008 #11

    gabbagabbahey

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    x and z look good, but y isn't even linear in t!...how did you come up with that? Is y(1) really 0?
     
  13. Dec 10, 2008 #12
    its late.....what else can i say
    I don't know what to use for t going from 0 to 1
    y=pi t + pi
    y(0)=pi
    y(1)=2pi

    y=pit - pi
    y(0)=-pi
    y(1)=pi t - pi=0
     
    Last edited: Dec 10, 2008
  14. Dec 10, 2008 #13

    gabbagabbahey

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    well now y is linear, and y(1)=0 :smile: , but y(0)=-pi not +pi:frown:

    Perhaps you should get some rest and leave this until morning? :wink:
     
  15. Dec 10, 2008 #14
    Great idea, thanks for your help
     
  16. Dec 10, 2008 #15
    Here's a big hint for finding x, y and z,
    P1(-1,pi,-1) to P2(0,0,0)
    So, we have (-1, pi, -1) + some t
    (-1, pi,-1)+(1,a, b)t
    So x= -1+t
    I got this by having our original point and taking 0 from our second point and and -1 from the first point.
    So, we have -1+[0-(-1)]t=-1+t
    Follow this in the same manner for finding y and z.
     
  17. Dec 10, 2008 #16

    Defennder

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    Actually my intention wasn't to find the scalar potential which itself could be rather tedious, but to show that one could evaluate the line integral from P1 to P3 directly bypassing P2 to save time.
     
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