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Computing a line integral

  1. Jan 11, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Let C be the arc [itex]x=t^2, \space y=2t, \space z= \sqrt{4+3t}[/itex] for [itex]t \in [-1,0][/itex]

    Evaluate the line integral :

    [itex]\int_{C} z^2dx + \sqrt{x}dy - 4xyz dz[/itex]

    2. Relevant equations

    [itex]\int_{C} f(P) dx = \int_{a}^{b} f(P(t)) x'(t) dt[/itex] for [itex]t \in [a,b][/itex]

    3. The attempt at a solution

    So I have not computed one of these before. I know I'm supposed to parametrize ( choose a parameter ) and then compute my according integral. I really only need help setting it up, so here goes nothing :

    This question already has a nice parameter t picked out for me on an appropriate interval. Since the curve is an arc, we know it is one to one, i.e there are no self intersections.

    So we compute :

    [itex]dx = 2tdt, \space dy = 2dt, \space dz = \frac{1}{2} (4+3t)^{-1/2} dt[/itex]

    *I'm not sure I should be concerned with smoothness here*.

    Thus our integral becomes :

    [itex]\int_{C} z^2dx + \sqrt{x}dy - 4xyz dz = \int_{-1}^{0} (6t^2+8t)dt + (2t)dt - (4t^3)dt[/itex]

    So after computing everything and setting it all up that's what I think I should get. Then computing the integral is a simple matter. I just want to make sure I'm not jumping too quickly missing anything and if anyone could verify it would be much appreciated.
     
  2. jcsd
  3. Jan 11, 2013 #2

    Dick

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    You've got the right procedure. You don't need to check that it's 1-1. You do want it to be differentiable, but it clearly is. You are missing some details. Check dz again, and I don't think sqrt(x) is t. Remember t is negative.
     
  4. Jan 11, 2013 #3

    Zondrina

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    Silly me, I forgot my chain rule for a second there.

    Also, x = t2, so sqrt(x) = t.
     
  5. Jan 11, 2013 #4

    haruspex

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    You expression for dz is not quite right. Other than that it looks ok.
     
  6. Jan 11, 2013 #5

    Dick

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    t is NEGATIVE. I don't think the square root of t^2 can be negative.
     
  7. Jan 11, 2013 #6

    Zondrina

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    The most common mistake in math I believe... I see why it's -t now.
     
  8. Jan 12, 2013 #7

    HallsofIvy

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    If "t in [-1, 0]" bothers you let s= -t. Then [itex]x= t^2= s^2[/itex], [itex]y= 2t= -2s[/itex], and [itex]z= \sqrt{4+ 3t}= \sqrt{4- 3s}[/itex] with s in [0, 1].
     
  9. Jan 12, 2013 #8

    Zondrina

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    Ohh I understand now. I had a question to prove this in general in my tutorial that I never had a chance to do.
     
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