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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.
     
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  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.
     
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