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Homework Help: Finding Line Integral of Vector Field

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data
    You are given a vector field A= kx2 x.
    a. First, calculate the line integral of A from x=-2 to x=2 along the x axis.
    b. Next, calculate the line integral of A between the same 2 points, but use
    a semicircular path with a center at the origin. Recall that in cylindrical
    coordinates, dl= ds s + sdφ φ + dz z.
    c. Are the 2 results the same? Explain.


    2. Relevant equations



    3. The attempt at a solution

    I believe I did the first part correctly and I got (16/3)k. I am getting confused on the second part though. My value for dl ended up being 2d(phi)phi. This is as far as I could get though.
     
  2. jcsd
  3. Sep 20, 2010 #2

    vela

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    So the integrand is equal to

    [tex]\vec{A}\cdot d\vec{l} = (kx^2\,\hat{x})\cdot(s\,d\phi\,\hat{\phi}) = kx^2s\,d\phi\,(\hat{x}\cdot\hat{\phi})[/tex]

    What is [tex]\hat{x}\cdot\hat{\phi}[/tex] equal to?
     
    Last edited: Sep 20, 2010
  4. Sep 20, 2010 #3
    I'm not sure. Should I change the 'x' into cylindrical form? That would make x=rcos(theta).
     
  5. Sep 20, 2010 #4

    HallsofIvy

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    Is x the "position vector" <x, y>? On the x-axis, of course, that is <x, 0> so the vector field, restricted to the x-axis is <kx^3, 0>. The integral, then, from x=-2 to 2 is
    [tex]\int_{-2}^2 kx^3dx= \frac{k}4}x^4\right|_{-2}^2= 0[/tex]

    Or is x the unit vector in the x direction? (I would use [itex]\vec{i}[/itex] for that. If so, then the vector field is just <kx^2, 0> and the integral is
    [tex]\int_{-2}^2 kx^2 dx= \frac{k}{3}x^3\right|_{-2}^2= \frac{16k}{3}[/tex].

    I would NOT change to cylindrical form. In Cartesian coordinates, the upper semi circle from -2 to 2 can be written in parametric form as x= 2cos(t), y= 2sin(t), t going from [itex]-\pi[/itex] to 0. [itex]d\vec{s}= (-2sin(t)\vec{i}+ 2cos(t)\vec{j}) dt.

    With x= <x, y>, [itex]kx^2[/itex]x is [itex]<x^3, x^2y>[/itex] and on the upper semi-circle that is [itex]8sin^3(t)\vec{i}+ 8sin(t)cos^2(t)\vec{j}[/itex]

    The integral would be
    [tex]16\int_{-\pi}^0 -sin^4(t)+ sin(t)cos^3(t) dt[/tex]
    For the first,[itex]sin^4(t)[/itex], use the trig identity [itex]sin^2(t)= (1/2)(1- cos(2t)[/itex]. For the second, [itex]sin(t)cos^3(t)[/itex], use the substitution u= cos(t), du= -sin(t)dt. When [itex]t= -\pi[/itex], u= -1 and when t= 0, u= 1.
     
  6. Sep 20, 2010 #5

    vela

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    If you're referring to the variable [itex]x[/itex] and not the unit vector [itex]\hat{x}[/itex], then yes, you need to express it in terms of the cylindrical variables in the integrand. You're going to be integrating with respect to [itex]\phi[/itex], so any quantity that varies with [itex]\phi[/itex] has to be expressed in terms of it.

    In your textbook or in your notes, you have probably been told what [tex]\hat{\phi}[/tex] is equal to. Use it to evaluate the dot product.
     
  7. Sep 20, 2010 #6
    HallsofIvy: I see what you are saying, but I think the teacher wants us to convert to cylindrical coordinates. I think that's why he put 'dl' in terms of cylindrical coordinates in the problem.
    Vela: I don't have anything in my notes, but I found this online...Phi-hat= -sin(phi)(x-hat) + cos(phi)(y-hat). Is that the appropriate form?
    If so, if I dot that value with x-hat, I just get -sin(phi)?
     
  8. Sep 20, 2010 #7

    vela

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    Yes, that's exactly what you want. [tex]\hat{\phi}[/tex] points in the direction a point would move if you increase φ infinitesimally. If you try evaluate [tex]\hat{\phi}[/tex] for a few values of φ, you should see that expression is correct.
     
  9. Sep 20, 2010 #8
    Ok so I plugged x=Rcos(phi) and Phi-hat= -sin(phi)(x-hat) + cos(phi)(y-hat) into what I had. Shouldn't I have a dPhi somewhere though?
     
  10. Sep 20, 2010 #9

    vela

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    Yes. What did you do with it?
     
  11. Sep 20, 2010 #10
    Ohhh I found it. Ok so here is my work. I think I got it. Also for part C, I just stated that they were supposed to be equal since line integrals are path independent.
     

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  12. Sep 20, 2010 #11

    vela

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    I couldn't read part of it, but the set-up looked fine. For part (c), remember that line integrals aren't always path-independent. You should show that this vector field satisfies the condition required for path independence.
     
  13. Sep 20, 2010 #12
    Well I ended up getting 16k/3 for both part A and B, so I hope it's correct. And for part C, I know that independence requires a line integral of a vector field to be equal to the gradient of a scalar field.
     
  14. Sep 20, 2010 #13

    vela

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    I didn't mean to cast any doubt on your work. I just didn't want to imply that I had checked all of it carefully. I think you did everything correctly.

    Can you guess a scalar field Φ that satisfies [itex]\vec{A} = \nabla \phi[/itex]?
     
  15. Sep 20, 2010 #14
    Oh ok.

    And I honestly am not sure how you even got that to start with. I understand the general format, but I don't see why it's the gradient of Phi. Phi is the angle that the particle travels, but I am still confused on the relation that allowed you to write gradientPhi
     
  16. Sep 20, 2010 #15

    vela

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    Oops, I'm sorry. I should have used a different symbol. I meant just some scalar function (phi is common notation for such a function). It has nothing to do with the angle.
     
  17. Sep 20, 2010 #16
    Ohhh ok. I understand it now, but I still don't know what the function would be. Would it be
    (kx^3)/3?
     
  18. Sep 20, 2010 #17

    vela

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    Yes, that works.
     
  19. Sep 20, 2010 #18
    So should I say something like this.....


    As long as the relationship of A=gradient((kx^3)/3) holds, then the line integral is path independent.
     
  20. Sep 20, 2010 #19

    vela

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    It would be better to say, "Because the vector field A can be expressed as a gradient of the scalar function kx3/3, blah blah blah."
     
  21. Sep 20, 2010 #20
    Ok I see what you are saying. I wrote this....


    Yes because the vector filed A can be expressed as a gradient of the scalar function kx^3/3. We know that if the line integral of a vector field is equal to the gradient of a scalar field, the line integral is path independent.
     
  22. Sep 20, 2010 #21

    vela

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    The first sentence/fragment is fine, but the second one isn't. It isn't the case that the line integral of a vector field is equal to the gradient of a scalar field; it's that the vector field itself is the gradient of a scalar field.
     
  23. Sep 20, 2010 #22
    Ok so how about this...
    Yes because the vector field A can be expressed as a gradient of the scalar function kx^3/3. We know that the vector field is equal to the gradient of a scalar field, so the line integral is path independent.

    I feel like I am not really explaining why though. Is it just because that's one of the requirements for path independence?
     
  24. Sep 20, 2010 #23

    vela

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    You're explaining why the two results in this problem should be equal. By showing the vector field is of the right form, you've explained why the two results have to be equal. The problem isn't asking you to explain why, in general, if a vector field can be written as the gradient of a scalar field, the line integral between two points is path-independent.

    If you feel like you need to explain more, you could calculate

    [tex]\int_C \vec{A}\cdot d\vec{l} = \int_C (\nabla f)\cdot d\vec{l} = \cdots[/tex]

    where f=kx3/3 and show it only depends on the endpoints and gives the same result 16k/3.
     
  25. Sep 21, 2010 #24
    Oh ok I understand it now. Ok I will just put what I said in my last message, and I think it will be good. Thank you so much again for your help.
     
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