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Calculating line integral

  1. Apr 15, 2014 #1
    1. The problem statement, all variables and given/known data

    lineintegrals_zpsc07a6d02.jpg
    2. Relevant equations
    Trued only 1st question..
    Unfortunately I lost my notes about this and cannot find anything relevant to this.
    I think,
    cF.dr = ∫cF.dr/dt dt ..
    also dr/dt isn't it = ∂x/∂ti +∂y/∂tj+∂z/∂tk
    Also it seems that C is with parabolic shape?
    Can someone tell me what is the relationship with y=x^2 and 't'
    Also if there is just dot product of F and ∂x/∂ti +∂y/∂tj+∂z/∂tk and boundaries of integral 0 and 1
    why he have this y=x^2 from (0,0) to (1,1) ?
    I think I miss something general here..

    Thanks
     
  2. jcsd
  3. Apr 15, 2014 #2
    ##t## is the parameter of the curve, so that the curve can be described by ##x = f(t), y = g(t), z = h(t) ##. In your case, you can simply let ##x = t ##.
     
  4. Apr 15, 2014 #3

    pasmith

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    Homework Helper

    The principle is that [itex]C[/itex] can be parametrized by any [itex]\mathbf{r}(t) = (x(t),y(t))[/itex] such that [itex]\mathbf{r}(0) = (0,0)[/itex], [itex]\mathbf{r}(1) = (1,1)[/itex], together with [itex]\|d\mathbf{r}/dt\| > 0[/itex] for all [itex]t \in (0,1)[/itex] and [itex]y(t) = (x(t))^2[/itex] for all [itex]t \in (0,1)[/itex]. Subject to those constraints the integral doesn't depend on the particular choice of [itex]\mathbf{r}(t)[/itex].

    Can you think of a simple choice for [itex]x(t)[/itex] and [itex]y(t)[/itex]?
     
  5. Apr 15, 2014 #4
    I got it it was simpler that I thought .. Thanks for the time :)
    https://www.dropbox.com/s/3az4t1r6sj52rkw/IMG_20140415_224810.jpg
     
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