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Homework Help: Line integral in vector field

  1. Sep 30, 2009 #1
    First I wanna greet everyone because I am new here.

    I have attended to applied electromagnetic course which seems to be pretty hard to understand and issues came up at very first time after I went at calculations.

    I try to explain this as good as possible.

    1. Vectorfield F(x,y,z) = (y-2x)ux + (y2-x2)ux. Calculate the fields line integral among the parabola y=3x2, from origin to poin r (r=ux-3uy).

    Does it mean that every point of the slope y=3x2 can be expressed in terms of vector field F?

    I have tried the following
    I put F=r to express the amound of x and y is needed in vector field F to reach point r. It gives me the following:
    y-2x = 1
    y2-x2 = 3

    Does this make sense? I know that vector field´s line integral is int(F-dot-d)
    Next I have tried to make this happen and had

    How I implement the y=3x2 in this integral? Shouldn´t the integral be
    int(3x2) from point 0 to r?

    I also have tried to put the function y=3x2 to F(x,y,z) and tried int[(3x2ux-2xux+9x4uy-x2uy)*(dux+duy)
    After this I had 2x2+9x4-2x and tried to put value for X which i got when I did F = r but it doesn´t seem to work.

    Can anyone give me a clue how should I THINK or understant this? How do I make the calculation?

    The answe should be 15/2
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 30, 2009 #2


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    ok so we want to calculate

    [tex]\int_C \vec{F} . d\vec{r} [/tex]

    Where C is curve y=3x2 from (0,0) to i+3j (i is the same as ux and j is the same as uy). So it is basically from (0,0) to (1,3) .

    I think it should be i+3j or (1,3) as (1,-3) does not lie on the curve. So you can easily parameterize the curve by letting x= some function of t.

    so now find [itex]\vec{r}[/itex] for the curve which is in the form [itex] \vec{r} = x(t)\hat{i} + y(t) \hat{j}[/itex].

    So what is your range of values for t on the vector r?

    When you have this. You can easily parameterize F (make x in terms of t and y in terms of t) and you can find dr
    Last edited: Sep 30, 2009
  4. Oct 4, 2009 #3
    Ok, I have tried this but something is still going wrong.

    I replace the y = 3x^2 to in F(x,y,z) and I get F(x,y,z) = (2x2-2x)ux + (9x4-x2)uy

    So I know the y in terms of x in the vector field.

    I tried to somehow force r in terms of x and y and got r = 1ux and 720uy but I think this is not gonna work..

    What it means that I have to take [itex]\vec{r} = x(t)\hat{i} + y(t) \hat{j}[/itex] in terms of t?
    How I express r in values of x and y? Do I need to replace ux and uy or how I erase them
  5. Oct 4, 2009 #4


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    You can always parameterize an equation y = f(x) by letting x = t, y = f(t). In your case you would get:

    r(t) = t i + 3t2j
  6. Oct 4, 2009 #5


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    Well you need to do what LCKurtz suggested. I don't think it matters whether you use ux for i or uy for j.

    When you find [itex]\vec{r}[/itex]. You need to also get [itex]F(\vec{r(t)})}[/itex]
  7. Oct 6, 2009 #6
    Ok, thank you by far. this is makes sense now but I still can´t get the right answer.

    I got got that the following,
    r = ti + 9t2j => dr = d(ti) + d(9t2j)

    And the F would be F = (3t2-2t)i + (9t4-t2)j

    Can I now take dr = (i + 9tj)dt? Ok, I have done this and int(F.dt):

    I had int(3t2-2t+81t5-9t3) with only t:s

    How do I get t:s? i tried to put F(t) = r(t) and got 27t4-2t-10 = 0. i mean I put 3t2-2t = 1 and 9t4-t2 = 3 and got very nonsense with t = about -/+ 0,809... I tried to integrate from 0 to t and got some ********.
    Last edited: Oct 6, 2009
  8. Oct 6, 2009 #7


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    Your r should be ti + 3t2j. When t = 0 the point is at (0,0) and when t = 1 the point is at (1,3) on your parabola. So the t variable in the integral should go from 0 to 1.
  9. Oct 8, 2009 #8
    hmmm. of course. int(3t2-2t+81t5-9t3) that is
    /(t3 - t2 + 13,5t6 - 2,25t4) right?
    Now from 0 to 1 gives me 11,25 and the answer should be 7,5. Somewhere is still error but I can´t find it.
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