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Line integral of a vector field.

  1. Nov 17, 2008 #1
    Hi all, i'm new to the forums so if i do something stupid don't hesitate to tell me.

    Anyway i'm struggling with this problem:

    [​IMG]

    I could do part a ok, but part b has me stumped, I am in the second year of a physics degree and this is a from a maths problem sheet, i haven't done line integrals before now and they have me a bit confused, my textbook has a few examples but none of them include vectors and wiki has me even more confused.

    Here is my attempt so far:

    [​IMG]
    (please excuse bad handwriting, i am dyslexic)

    I basically don't know where to begin with it, any help much appreciated.
     
  2. jcsd
  3. Nov 17, 2008 #2

    Dick

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    Use that along your path y=x^3. So r=x*i+x^3*j. (i=x hat and j=y hat). What is the vector dr? Change the y's in V to x^3's as well and integrate dx.
     
  4. Nov 17, 2008 #3
    I think dr is just a small difference in the postion vector r, so dr = dx*i+dy*j ? r isn't mentioned anywhere else in the question.
     
  5. Nov 17, 2008 #4

    Dick

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    Sure it is. But now let's decide to integrate dx. Replace y with x^3. Now what's dr?
     
  6. Nov 17, 2008 #5
    dr=x*i +(x^3)j

    I also have V(r)=(2x-x^6)i + [(6x^6)-(2x^4)]j

    But i am a bit unsure about V.dr, is it {[ (2x^2)-(x^7) ]i + [ (6x^9)-(2x^7) ]j}dx ?
     
  7. Nov 17, 2008 #6

    Dick

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    What happened to the 'd's on the right side of dr? You've got y*j, shouldn't it be d(y)*j? Same for i. Let's get dr right before going to the dot product.
     
  8. Nov 17, 2008 #7
    OK, i jumped the gun a bit with dr, is it d(r) = d(x)*i + d(y)*j = d(x)*i + d(x^3)*j ?
     
  9. Nov 17, 2008 #8

    Dick

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    Right. And d(x^3)=3*x^2*dx, right? So dr=(i+3x^2*j)dx.Now do the dot product and you'll get a single integral dx.
     
  10. Nov 17, 2008 #9
    Ok, V.dr=[(2x-x^6)+(18x^8)-(6x^6)]dx

    When i integrate this with the limits x1=0 and x2=1 i get 2, does this sound about right?
     
  11. Nov 17, 2008 #10

    Dick

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    That's what I got. Not so hard, hmm?
     
  12. Nov 17, 2008 #11
    Yeah thanks that's been a great help.

    Part c seems trivial now but part d seems a bit daunting still.
     
  13. Nov 17, 2008 #12

    Dick

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    Don't be daunted. You know dphi/dx=2x-y^2 (partial derivative). What does that tell you about phi?
     
  14. Nov 17, 2008 #13
    Hmm, after much head scratching i think what you are getting at is that [tex]\nabla[/tex](phi(x,y))=(dphi/dx)i+(dphi/dy)j, which means:

    phi(x,y)=[int(2x-y^2)dx] + [int((6y^2)-2xy)]

    which gives:

    phi=x2 + 2y3 -xy2

    Which does seem to give 2 when i do phi(1,1) - phi(0,0)
     
  15. Nov 17, 2008 #14

    Dick

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    The answer is right. Since dphi/dx=2x-y^2, that means phi=x^2-x*y^2+f(y). Yeah, that is 'integral dx' with f(y) being the constant of integration. That gives dphi/dy=-2xy+f'(y). Since you are supposed to get 6y^2-2xy, you can figure that the f'(y) part must be the 6y^2, so if you put it all together, phi=x^2-xy^2+2y^3.
     
  16. Nov 17, 2008 #15

    Dick

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    I'm not taking your int()+int() formula literally. If I did I would get a -xy^2 term from each one of those. For a total of -2xy^2, which is not right.
     
  17. Nov 17, 2008 #16
    Well i've managed to do the problem, and i could probably do others now using that technique, but i'm not sure i understand why it works, i mean in questions b, c and d the only constant is x1=0, y1=0, x2=1 and y2=1, why doesn't the chosen path have an effect, is it something to do with the curl being 0?
     
  18. Nov 17, 2008 #17

    Dick

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    It has everything to do with the curl being zero. It's one of the conditions you need for V to be conservative. 'Conservative' means that the line integral between two points is independent of the path chosen. If they hadn't given you a conservative V, then the answer would depend on the path.
     
  19. Nov 17, 2008 #18
    Ok i get it now! Thanks for your help, i wouldn't have been able to do this much without it.
     
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